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interviews | Everything you need to know to get the job | Learning library

 by   kdn251 Java Version: Current License: MIT

 by   kdn251 Java Version: Current License: MIT

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kandi X-RAY | interviews Summary

interviews is a Java library typically used in Tutorial, Learning, Example Codes, LeetCode applications. interviews has no bugs, it has no vulnerabilities, it has a Permissive License and it has medium support. However interviews build file is not available. You can download it from GitHub.
Your personal guide to Software Engineering technical interviews. Video solutions to the following interview problems with detailed explanations can be found here. Maintainer - Kevin Naughton Jr.
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Quality
Quality
Security
Security
License
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kandi-support Support

  • interviews has a medium active ecosystem.
  • It has 53816 star(s) with 11436 fork(s). There are 2640 watchers for this library.
  • It had no major release in the last 12 months.
  • There are 28 open issues and 18 have been closed. On average issues are closed in 39 days. There are 68 open pull requests and 0 closed requests.
  • It has a neutral sentiment in the developer community.
  • The latest version of interviews is current.
interviews Support
Best in #Learning
Average in #Learning
interviews Support
Best in #Learning
Average in #Learning

quality kandi Quality

  • interviews has 0 bugs and 0 code smells.
interviews Quality
Best in #Learning
Average in #Learning
interviews Quality
Best in #Learning
Average in #Learning

securitySecurity

  • interviews has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
  • interviews code analysis shows 0 unresolved vulnerabilities.
  • There are 0 security hotspots that need review.
interviews Security
Best in #Learning
Average in #Learning
interviews Security
Best in #Learning
Average in #Learning

license License

  • interviews is licensed under the MIT License. This license is Permissive.
  • Permissive licenses have the least restrictions, and you can use them in most projects.
interviews License
Best in #Learning
Average in #Learning
interviews License
Best in #Learning
Average in #Learning

buildReuse

  • interviews releases are not available. You will need to build from source code and install.
  • interviews has no build file. You will be need to create the build yourself to build the component from source.
  • Installation instructions are not available. Examples and code snippets are available.
  • interviews saves you 5461 person hours of effort in developing the same functionality from scratch.
  • It has 11447 lines of code, 777 functions and 516 files.
  • It has medium code complexity. Code complexity directly impacts maintainability of the code.
interviews Reuse
Best in #Learning
Average in #Learning
interviews Reuse
Best in #Learning
Average in #Learning
Top functions reviewed by kandi - BETA

kandi has reviewed interviews and discovered the below as its top functions. This is intended to give you an instant insight into interviews implemented functionality, and help decide if they suit your requirements.

  • Decodes a string .
    • Returns the minimal window .
      • Get the vertical order of the node in the tree
        • Generate matrix .
          • Returns a list of integers .
            • Returns the level order in the tree rooted at the tree
              • Merge k lists .
                • Group an array of Strings
                  • Determine how many substrings are unique
                    • Return the minimum cost of an array of costs .

                      Get all kandi verified functions for this library.

                      Get all kandi verified functions for this library.

                      interviews Key Features

                      Everything you need to know to get the job.

                      Directory Tree

                      copy iconCopydownload iconDownload
                      .
                      ├── Array
                      │   ├── bestTimeToBuyAndSellStock.java
                      │   ├── findTheCelebrity.java
                      │   ├── gameOfLife.java
                      │   ├── increasingTripletSubsequence.java
                      │   ├── insertInterval.java
                      │   ├── longestConsecutiveSequence.java
                      │   ├── maximumProductSubarray.java
                      │   ├── maximumSubarray.java
                      │   ├── mergeIntervals.java
                      │   ├── missingRanges.java
                      │   ├── productOfArrayExceptSelf.java
                      │   ├── rotateImage.java
                      │   ├── searchInRotatedSortedArray.java
                      │   ├── spiralMatrixII.java
                      │   ├── subsetsII.java
                      │   ├── subsets.java
                      │   ├── summaryRanges.java
                      │   ├── wiggleSort.java
                      │   └── wordSearch.java
                      ├── Backtracking
                      │   ├── androidUnlockPatterns.java
                      │   ├── generalizedAbbreviation.java
                      │   └── letterCombinationsOfAPhoneNumber.java
                      ├── BinarySearch
                      │   ├── closestBinarySearchTreeValue.java
                      │   ├── firstBadVersion.java
                      │   ├── guessNumberHigherOrLower.java
                      │   ├── pow(x,n).java
                      │   └── sqrt(x).java
                      ├── BitManipulation
                      │   ├── binaryWatch.java
                      │   ├── countingBits.java
                      │   ├── hammingDistance.java
                      │   ├── maximumProductOfWordLengths.java
                      │   ├── numberOf1Bits.java
                      │   ├── sumOfTwoIntegers.java
                      │   └── utf-8Validation.java
                      ├── BreadthFirstSearch
                      │   ├── binaryTreeLevelOrderTraversal.java
                      │   ├── cloneGraph.java
                      │   ├── pacificAtlanticWaterFlow.java
                      │   ├── removeInvalidParentheses.java
                      │   ├── shortestDistanceFromAllBuildings.java
                      │   ├── symmetricTree.java
                      │   └── wallsAndGates.java
                      ├── DepthFirstSearch
                      │   ├── balancedBinaryTree.java
                      │   ├── battleshipsInABoard.java
                      │   ├── convertSortedArrayToBinarySearchTree.java
                      │   ├── maximumDepthOfABinaryTree.java
                      │   ├── numberOfIslands.java
                      │   ├── populatingNextRightPointersInEachNode.java
                      │   └── sameTree.java
                      ├── Design
                      │   └── zigzagIterator.java
                      ├── DivideAndConquer
                      │   ├── expressionAddOperators.java
                      │   └── kthLargestElementInAnArray.java
                      ├── DynamicProgramming
                      │   ├── bombEnemy.java
                      │   ├── climbingStairs.java
                      │   ├── combinationSumIV.java
                      │   ├── countingBits.java
                      │   ├── editDistance.java
                      │   ├── houseRobber.java
                      │   ├── paintFence.java
                      │   ├── paintHouseII.java
                      │   ├── regularExpressionMatching.java
                      │   ├── sentenceScreenFitting.java
                      │   ├── uniqueBinarySearchTrees.java
                      │   └── wordBreak.java
                      ├── HashTable
                      │   ├── binaryTreeVerticalOrderTraversal.java
                      │   ├── findTheDifference.java
                      │   ├── groupAnagrams.java
                      │   ├── groupShiftedStrings.java
                      │   ├── islandPerimeter.java
                      │   ├── loggerRateLimiter.java
                      │   ├── maximumSizeSubarraySumEqualsK.java
                      │   ├── minimumWindowSubstring.java
                      │   ├── sparseMatrixMultiplication.java
                      │   ├── strobogrammaticNumber.java
                      │   ├── twoSum.java
                      │   └── uniqueWordAbbreviation.java
                      ├── LinkedList
                      │   ├── addTwoNumbers.java
                      │   ├── deleteNodeInALinkedList.java
                      │   ├── mergeKSortedLists.java
                      │   ├── palindromeLinkedList.java
                      │   ├── plusOneLinkedList.java
                      │   ├── README.md
                      │   └── reverseLinkedList.java
                      ├── Queue
                      │   └── movingAverageFromDataStream.java
                      ├── README.md
                      ├── Sort
                      │   ├── meetingRoomsII.java
                      │   └── meetingRooms.java
                      ├── Stack
                      │   ├── binarySearchTreeIterator.java
                      │   ├── decodeString.java
                      │   ├── flattenNestedListIterator.java
                      │   └── trappingRainWater.java
                      ├── String
                      │   ├── addBinary.java
                      │   ├── countAndSay.java
                      │   ├── decodeWays.java
                      │   ├── editDistance.java
                      │   ├── integerToEnglishWords.java
                      │   ├── longestPalindrome.java
                      │   ├── longestSubstringWithAtMostKDistinctCharacters.java
                      │   ├── minimumWindowSubstring.java
                      │   ├── multiplyString.java
                      │   ├── oneEditDistance.java
                      │   ├── palindromePermutation.java
                      │   ├── README.md
                      │   ├── reverseVowelsOfAString.java
                      │   ├── romanToInteger.java
                      │   ├── validPalindrome.java
                      │   └── validParentheses.java
                      ├── Tree
                      │   ├── binaryTreeMaximumPathSum.java
                      │   ├── binaryTreePaths.java
                      │   ├── inorderSuccessorInBST.java
                      │   ├── invertBinaryTree.java
                      │   ├── lowestCommonAncestorOfABinaryTree.java
                      │   ├── sumOfLeftLeaves.java
                      │   └── validateBinarySearchTree.java
                      ├── Trie
                      │   ├── addAndSearchWordDataStructureDesign.java
                      │   ├── implementTrie.java
                      │   └── wordSquares.java
                      └── TwoPointers
                          ├── 3Sum.java
                          ├── 3SumSmaller.java
                          ├── mergeSortedArray.java
                          ├── minimumSizeSubarraySum.java
                          ├── moveZeros.java
                          ├── removeDuplicatesFromSortedArray.java
                          ├── reverseString.java
                          └── sortColors.java
                      
                      18 directories, 124 files
                      

                      How to find most common words from specific rows and column and list how often it occurs at data.csv?

                      copy iconCopydownload iconDownload
                      import pandas as pd
                      import numpy as np    
                      df = pd.read_csv("data.csv")
                      small_df = df[['title','duration_min','description']]
                      result_time = small_df.sort_values('duration_min', ascending=False)
                      print("TOP 10 LONGEST: ")
                      print(result_time.head(n=10))
                      
                      most_common = pd.Series(' '.join(result_time.iloc[0:10]['description']).lower().split()).value_counts()[:20]
                      print("20 Most common words from TOP 10 longest movies: ")
                      print(most_common) 
                      

                      How to calculate cumulative sums in MySQL

                      copy iconCopydownload iconDownload
                      select day,product_count,
                      sum(product_count) over (order by t.day ROWS UNBOUNDED PRECEDING) as cumulative_sum from (
                      SELECT
                        date(purchase_date) as day,
                        count(product_id) as product_count
                        FROM products
                        where day > DATE_SUB(now(), INTERVAL 6 MONTH)
                        AND customer_city = 'Seattle'
                        GROUP BY day 
                        ORDER BY product_count desc
                      )t
                      
                      

                      css grid relayout if element changes height

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                      <head>
                      
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                        <meta name="apple-mobile-web-app-title" content="CodePen">
                      
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                        <div class="grid grid-cols-4 gap-4 services">
                      
                          <div class="flex flex-col rounded-md bg-gray-200 p-8 items-center justify-center js-service-click">
                            <figure class="bg-blue-500 rounded-full flex items-center align-center justify-center w-32 h-32">
                              <img src="http://davidbakerconsulting.test/wp-content/uploads/2022/01/image-12.svg" class="attachment-full size-full" alt="" loading="lazy"> </figure>
                            <h2 class="text-lg font-bold mt-7">Report Writing</h2>
                            <div class="mb-2 text-primary overflow-hidden service-details">DBC can provide reports on any activity for organisations wishing to change any aspect of its operation or strategic direction. David Baker has written consultancy reports for many countries on library and information transformation projects, various
                              institutions seeking to gain taught degree awarding powers (TDAP), leadership and management benchmarking exercises and organisational training provision. Notable recent examples include three reports written for for SCONUL to assess the strategic
                              role of libraries and their leaders from the viewpoint of the universities’ senior management. This resulted in a publication for the commissioning body SCONUL to highlight the expectations of university leadership in the development of relevant
                              library strategies. As a further phase of this work, Alison Allden and David Baker looked at the opportunities and transferable skills relating to international movement amongst library leaders. David Baker has been working with the University
                              of London since March 2020 to develop a new strategy for Senate House Library, along with the School of Advanced Study, Federal Member Institute Libraries and the University of London Worldwide. The initial report was accepted by the University
                              in September 2020. In the light of this, David has created a major Library Transformation Programme (LTP). The work involved in developing and operationalising the report included surveys, benchmarking, focus groups and workshops.</div>
                          </div>
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                              <img src="http://davidbakerconsulting.test/wp-content/uploads/2022/01/image-11.svg" class="attachment-full size-full" alt="" loading="lazy"> </figure>
                            <h2 class="text-lg font-bold mt-7">Mentoring</h2>
                            <div class="mb-2 text-primary overflow-hidden service-details">Professor David Baker has experience of mentoring professionals from different sectors and academics reaching back to 1990. He is currently heavily involved in mentoring for CILIP Chartership, Certification and Fellowship status (the UK’s library
                              and information association). The focus is on developing and nurturing staff while learning new ideas and approaches from other professionals. Mentoring can be undertaken in person, by email, by telephone or through online platforms.</div>
                          </div>
                          <div class="flex flex-col rounded-md bg-gray-200 p-8 items-center justify-center js-service-click">
                            <figure class="bg-blue-500 rounded-full flex items-center align-center justify-center w-32 h-32">
                              <img src="http://davidbakerconsulting.test/wp-content/uploads/2022/01/image-10.svg" class="attachment-full size-full" alt="" loading="lazy"> </figure>
                            <h2 class="text-lg font-bold mt-7">Workshops</h2>
                            <div class="mb-2 text-primary overflow-hidden service-details">DBC has a wealth of experience in delivering face-to-face and online workshops for a range of projects and purposes. These can be for the purpose of stakeholder engagement, data gathering or as part of a communications strategy. Workshops, interviews
                              and focus groups are carried out consistently and coherently to give maximum value of data and information gathered. This is done through agreed pro forma and protocols.</div>
                          </div>
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                            <figure class="bg-blue-500 rounded-full flex items-center align-center justify-center w-32 h-32">
                              <img src="http://davidbakerconsulting.test/wp-content/uploads/2022/01/image-9.svg" class="attachment-full size-full" alt="" loading="lazy"> </figure>
                            <h2 class="text-lg font-bold mt-7">Training</h2>
                            <div class="mb-2 text-primary overflow-hidden service-details">The DBC team can offer project management and change management training using previous live (anonymised) projects. A DBC Associate is PRINCE2 trained. David Baker has delivered training in library management and information systems for senior leaders,
                              library tecnhnicians and assistants for many countries of the world including, Slovenia, Ireland, Kuwait, Hungary, Germany and Portugal and has published several training guides, including on the subject of co-operative training in this area.
                              He has also provided training and development for third world countries such as Ethiopia and Nigeria and has published a book on this.</div>
                          </div>
                          <div class="flex flex-col rounded-md bg-gray-200 p-8 items-center justify-center js-service-click">
                            <figure class="bg-blue-500 rounded-full flex items-center align-center justify-center w-32 h-32">
                              <img src="http://davidbakerconsulting.test/wp-content/uploads/2022/01/image-8.svg" class="attachment-full size-full" alt="" loading="lazy"> </figure>
                            <h2 class="text-lg font-bold mt-7">Benchmarking</h2>
                            <div class="mb-2 text-primary overflow-hidden service-details">In 2022, DBC is undertaking a major international benchmarking exercise based on the Association of Commonwealth Universities (ACU) model. It is being led by David Baker, Caroline Williams (Librarian of the University of Queensland for academic
                              and research libraries in the Australia and South Pacific region), Cliff Wragg and Lucy Ellis. The title is "Benchmarking Library, Information And Education Services: New Strategic Choices In Challenging Times". An Elsevier book publication bearing
                              the same title will be published in early 2023. The benchmarking model can be adapted for any organisational purpose.</div>
                          </div>
                          <div class="flex flex-col rounded-md bg-gray-200 p-8 items-center justify-center js-service-click">
                            <figure class="bg-blue-500 rounded-full flex items-center align-center justify-center w-32 h-32">
                              <img src="http://davidbakerconsulting.test/wp-content/uploads/2022/01/image-7.svg" class="attachment-full size-full" alt="" loading="lazy"> </figure>
                            <h2 class="text-lg font-bold mt-7">Research Services</h2>
                            <div class="mb-2 text-primary overflow-hidden service-details">DBC Associates worked in 2021 on a scoping study commissioned by Research Libraries UK (RLUK) and funded by the Arts and Humanities Research Council (AHRC). It resulted in a wealth of evidence of the role and potential of research libraries as partners
                              and leaders of research, contributing to longer-term strategic development in the process. A range of research techniques were used for this and other consultancies such as surveys, benchmarking, focus groups and workshops. Two senior associates
                              have research backgrounds and PhDs and the Director, Professor Baker, has a substantial and significant track record in research and publishing.</div>
                          </div>
                          <div class="flex flex-col rounded-md bg-gray-200 p-8 items-center justify-center js-service-click">
                            <figure class="bg-blue-500 rounded-full flex items-center align-center justify-center w-32 h-32">
                              <img src="http://davidbakerconsulting.test/wp-content/uploads/2022/01/image-6.svg" class="attachment-full size-full" alt="" loading="lazy"> </figure>
                            <h2 class="text-lg font-bold mt-7">Organisation Development</h2>
                            <div class="mb-2 text-primary overflow-hidden service-details">DBC has been developing strategic plans at organisational and pan-organisational levels for many years, not least through working in chief executive and governance roles as well as high-level consultancy work. We are accustomed to working with governing
                              bodies, steering committees, task forces and other groupings to shape strategic direction and effect major organisational change as a result. Our biographical details demonstrate that we have developed, written and implemented many strategic plans,
                              including for the Higher Education Statistics Agency (HESA) and the Joint Information Systems Committee (Jisc), as well as contributing to strategy development for the Society of College, National and University Libraries (SCONUL) and Research
                              Libraries UK (RLUK).</div>
                          </div>
                          <div class="flex flex-col rounded-md bg-gray-200 p-8 items-center justify-center js-service-click">
                            <figure class="bg-blue-500 rounded-full flex items-center align-center justify-center w-32 h-32">
                              <img src="http://davidbakerconsulting.test/wp-content/uploads/2022/01/image-5.svg" class="attachment-full size-full" alt="" loading="lazy"> </figure>
                            <h2 class="text-lg font-bold mt-7">Consultancy</h2>
                            <div class="mb-2 text-primary overflow-hidden service-details">DBC delivers high-quality consultancy projects in higher education both nationally and internationally, with a long-standing track record, especially in strategy development. Associates have a broad and deep knowledge of the field.</div>
                          </div>
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                      Is it acceptable to pass this amount of JSON information as a parameter to a C# COM DLL method?

                      copy iconCopydownload iconDownload
                      #include <windows.h>
                      #include <memory>
                      #include <string>
                      #include <iostream>
                      #include <comdef.h>
                      
                      using namespace std;
                      
                      int main(int argc, char* argv)
                      {
                          const size_t MEGABYTEPLUSONE = 1024 * 1024 + 1;
                          auto pChar = std::make_unique<OLECHAR[]>(MEGABYTEPLUSONE);
                          for (size_t i = 0; i < MEGABYTEPLUSONE - 2; ++i)
                              pChar[i] = (i % 26) + 65;
                      
                          pChar[MEGABYTEPLUSONE - 1] = 0;
                          
                          _bstr_t bstr(pChar.get());
                          int len = bstr.length();
                          wcout << L"Length of BSTR: " << len << endl;
                          const WCHAR* const pwcStart = bstr.GetBSTR();
                          for (const WCHAR* pwc = pwcStart;  pwc < pwcStart + 52; ++pwc)
                              wcout << *pwc;
                          wcout << endl;
                          
                          return 0;
                      }
                      

                      matrix maximum diagonal pattern match

                      copy iconCopydownload iconDownload
                      import numpy as np
                      from math import ceil
                      from itertools import takewhile
                      
                      def max_sequence(arr):
                          solns = []
                          i = arr.shape[0]
                          for x in range(-i, i+1):
                              values = arr.diagonal(x)
                              N = len(values)
                              possibles = np.where(values == pattern[0])[0]
                              for p in possibles:
                                  check = values[p:p+N]
                                  m = len(list(takewhile(lambda x:len(set(x))==1, zip(pattern,check))))
                                  solns.append(m)
                          return max(solns)
                      
                      def find_longest(arr):
                          if len(arr)>0:
                              return max([max_sequence(x) for x in [arr, np.fliplr(arr), np.flipud(arr), arr[::-1]]])
                          else:
                              return 0
                      
                      arr = np.array([
                          [1,0,2,1,1,1,1,0,0,0,0,0,0],
                          [1,2,2,1,1,1,1,0,0,0,0,0,0],
                          [1,0,0,1,1,1,1,0,0,0,0,0,0],
                          [1,0,0,2,1,1,1,0,0,0,0,0,0],
                          [1,0,0,2,0,1,1,0,0,0,0,0,0],
                          [1,0,0,1,1,2,1,0,0,0,0,0,0],
                          [1,0,0,1,1,1,0,0,0,0,0,0,0],
                          [0,0,0,0,0,0,0,1,0,0,0,0,0],
                          [0,0,0,0,0,0,0,0,2,0,0,0,0],
                          [0,0,0,0,0,0,0,0,0,0,0,0,0],
                          [0,0,0,0,0,0,0,0,0,0,2,0,0],
                          [0,0,0,0,0,0,0,0,0,0,0,0,0],
                      ])
                      
                      arr1 = np.array([
                          [1,0,2,1,1,1,1],
                          [1,2,2,1,1,1,1],
                          [1,0,0,1,1,1,1],
                          [1,0,0,2,1,1,1],
                          [1,0,0,2,0,1,1],
                          [1,0,0,1,1,2,1],
                          [1,0,0,1,1,1,0]
                      ])
                      
                      arr2 = np.array([])
                      
                      pattern = [1, 2, 0, 2, 0, 2, 0]
                      # Make sure pattern repeats longer than the max diagonal
                      pattern = np.tile(pattern,ceil(arr.shape[1] / len(pattern)))
                      
                      for a in [arr, arr1, arr2]:
                          print(find_longest(a))
                      
                      12
                      7
                      0
                      
                      import numpy as np
                      from math import ceil
                      from itertools import takewhile
                      
                      def max_sequence(arr):
                          solns = []
                          i = arr.shape[0]
                          for x in range(-i, i+1):
                              values = arr.diagonal(x)
                              N = len(values)
                              possibles = np.where(values == pattern[0])[0]
                              for p in possibles:
                                  check = values[p:p+N]
                                  m = len(list(takewhile(lambda x:len(set(x))==1, zip(pattern,check))))
                                  solns.append(m)
                          return max(solns)
                      
                      def find_longest(arr):
                          if len(arr)>0:
                              return max([max_sequence(x) for x in [arr, np.fliplr(arr), np.flipud(arr), arr[::-1]]])
                          else:
                              return 0
                      
                      arr = np.array([
                          [1,0,2,1,1,1,1,0,0,0,0,0,0],
                          [1,2,2,1,1,1,1,0,0,0,0,0,0],
                          [1,0,0,1,1,1,1,0,0,0,0,0,0],
                          [1,0,0,2,1,1,1,0,0,0,0,0,0],
                          [1,0,0,2,0,1,1,0,0,0,0,0,0],
                          [1,0,0,1,1,2,1,0,0,0,0,0,0],
                          [1,0,0,1,1,1,0,0,0,0,0,0,0],
                          [0,0,0,0,0,0,0,1,0,0,0,0,0],
                          [0,0,0,0,0,0,0,0,2,0,0,0,0],
                          [0,0,0,0,0,0,0,0,0,0,0,0,0],
                          [0,0,0,0,0,0,0,0,0,0,2,0,0],
                          [0,0,0,0,0,0,0,0,0,0,0,0,0],
                      ])
                      
                      arr1 = np.array([
                          [1,0,2,1,1,1,1],
                          [1,2,2,1,1,1,1],
                          [1,0,0,1,1,1,1],
                          [1,0,0,2,1,1,1],
                          [1,0,0,2,0,1,1],
                          [1,0,0,1,1,2,1],
                          [1,0,0,1,1,1,0]
                      ])
                      
                      arr2 = np.array([])
                      
                      pattern = [1, 2, 0, 2, 0, 2, 0]
                      # Make sure pattern repeats longer than the max diagonal
                      pattern = np.tile(pattern,ceil(arr.shape[1] / len(pattern)))
                      
                      for a in [arr, arr1, arr2]:
                          print(find_longest(a))
                      
                      12
                      7
                      0
                      

                      Cleaner/Simpler way to check if content in array has a value greater than 0

                      copy iconCopydownload iconDownload
                      Link={relatedList.some(v => !!v.length) ? 'Click to go to list' : ''}
                      

                      How to combine columns, group them then get a total count?

                      copy iconCopydownload iconDownload
                      Select [candidate_id]
                            ,B.[Interview_name]
                            ,pass_count   = case when result='Pass' then 1 else 0 end
                            ,reject_count = case when result='Pass' then 0 else 1 end
                       From YourTable A
                       Cross Apply ( values ([interview_1],[result_1])
                                           ,([interview_2],[result_2])
                                   ) B(Interview_name,result)
                      
                      candidate_id    Interview_name  pass_count  reject_count
                      1               Interviewer_A   1           0
                      1               Interviewer_B   1           0
                      2               Interviewer_C   1           0
                      2               Interviewer_D   0           1
                      
                      Select [candidate_id]
                            ,B.[Interview_name]
                            ,pass_count   = case when result='Pass' then 1 else 0 end
                            ,reject_count = case when result='Pass' then 0 else 1 end
                       From YourTable A
                       Cross Apply ( values ([interview_1],[result_1])
                                           ,([interview_2],[result_2])
                                   ) B(Interview_name,result)
                      
                      candidate_id    Interview_name  pass_count  reject_count
                      1               Interviewer_A   1           0
                      1               Interviewer_B   1           0
                      2               Interviewer_C   1           0
                      2               Interviewer_D   0           1
                      
                      WITH combined_set AS(
                      SELECT
                       candidate_id,
                       interviewer_1 as interviewer,
                       result_1 as result
                      from candidate_table
                      
                      UNION
                      
                      SELECT
                       candidate_id,
                       interviewer_2 as interviewer,
                       result_2 as result
                      from candidate_table)
                      
                      SELECT
                       interviewer,
                       count(case when result = 'Pass' then 1 end) as pass_count,
                       count(case when result = 'Reject' then 1 end) as reject_count
                      FROM combined_set
                      

                      formulate capture groups for inconsistently present substrings

                      copy iconCopydownload iconDownload
                      library(tidyr)
                      
                      tst <- c("In: ja COOL;  #00:04:24-6#  ",           
                               "  in den vier, FÜNF wochen, #00:04:57-8# ",
                               "In: jah,  #00:02:07-8# ",
                               "In:     [ja; ] #00:03:25-5# [ja; ] #00:03:26-1#",
                               "    also jA:h; #00:03:16-6# (1.1)",
                               "Bz:        [E::hm;    ]  #00:03:51-4#  (3.0)  ",
                               "Bz:    [mhmh,      ]",
                               "  in den bilLIE da war;")     
                      
                      data.frame(tst) %>%
                        extract(col = tst,
                                into = c("Role", "Utterance", "Timestamp", "Gap"),
                                regex = "^(\\w{2}:\\s|\\s+)([\\s\\S]*?)(?:\\s*#([^#]+)(?:#\\s*(\\([0-9.]+\\))?\\s*)?)?$")
                      
                        Role                      Utterance  Timestamp   Gap
                      1 In:                        ja COOL; 00:04:24-6      
                      2           in den vier, FÜNF wochen, 00:04:57-8      
                      3 In:                            jah, 00:02:07-8      
                      4 In:      [ja; ] #00:03:25-5# [ja; ] 00:03:26-1      
                      5                          also jA:h; 00:03:16-6 (1.1)
                      6 Bz:                    [E::hm;    ] 00:03:51-4 (3.0)
                      7 Bz:                   [mhmh,      ]                 
                      8               in den bilLIE da war; 
                      
                      library(tidyr)
                      
                      tst <- c("In: ja COOL;  #00:04:24-6#  ",           
                               "  in den vier, FÜNF wochen, #00:04:57-8# ",
                               "In: jah,  #00:02:07-8# ",
                               "In:     [ja; ] #00:03:25-5# [ja; ] #00:03:26-1#",
                               "    also jA:h; #00:03:16-6# (1.1)",
                               "Bz:        [E::hm;    ]  #00:03:51-4#  (3.0)  ",
                               "Bz:    [mhmh,      ]",
                               "  in den bilLIE da war;")     
                      
                      data.frame(tst) %>%
                        extract(col = tst,
                                into = c("Role", "Utterance", "Timestamp", "Gap"),
                                regex = "^(\\w{2}:\\s|\\s+)([\\s\\S]*?)(?:\\s*#([^#]+)(?:#\\s*(\\([0-9.]+\\))?\\s*)?)?$")
                      
                        Role                      Utterance  Timestamp   Gap
                      1 In:                        ja COOL; 00:04:24-6      
                      2           in den vier, FÜNF wochen, 00:04:57-8      
                      3 In:                            jah, 00:02:07-8      
                      4 In:      [ja; ] #00:03:25-5# [ja; ] 00:03:26-1      
                      5                          also jA:h; 00:03:16-6 (1.1)
                      6 Bz:                    [E::hm;    ] 00:03:51-4 (3.0)
                      7 Bz:                   [mhmh,      ]                 
                      8               in den bilLIE da war; 
                      
                      library(dplyr)
                      library(tidyr)
                      
                      data.frame(tst) %>%
                        extract(tst, "Gap", "(\\(.*?\\))", remove = FALSE) %>%
                        extract(tst, "Timestamp", "(#.*?#)", remove = FALSE) %>%
                        extract(tst, c("Role", "Utterance"), "^(\\S+:|)([^#]*)") %>%
                        mutate(across(, coalesce, ""), Utterance = trimws(Utterance))
                      
                        Role                 Utterance    Timestamp   Gap
                      1  In:                  ja COOL; #00:04:24-6#      
                      2      in den vier, FÜNF wochen, #00:04:57-8#      
                      3  In:                      jah, #00:02:07-8#      
                      4  In:                    [ja; ] #00:03:25-5#      
                      5                     also jA:h; #00:03:16-6# (1.1)
                      6  Bz:              [E::hm;    ] #00:03:51-4# (3.0)
                      7  Bz:             [mhmh,      ]                   
                      8          in den bilLIE da war;                   
                      
                      library(dplyr)
                      library(tidyr)
                      
                      data.frame(tst) %>%
                        extract(tst, "Gap", "(\\(.*?\\))", remove = FALSE) %>%
                        extract(tst, "Timestamp", "(#.*?#)", remove = FALSE) %>%
                        extract(tst, c("Role", "Utterance"), "^(\\S+:|)([^#]*)") %>%
                        mutate(across(, coalesce, ""), Utterance = trimws(Utterance))
                      
                        Role                 Utterance    Timestamp   Gap
                      1  In:                  ja COOL; #00:04:24-6#      
                      2      in den vier, FÜNF wochen, #00:04:57-8#      
                      3  In:                      jah, #00:02:07-8#      
                      4  In:                    [ja; ] #00:03:25-5#      
                      5                     also jA:h; #00:03:16-6# (1.1)
                      6  Bz:              [E::hm;    ] #00:03:51-4# (3.0)
                      7  Bz:             [mhmh,      ]                   
                      8          in den bilLIE da war;                   
                      

                      SQL Query to find people who have interviews in different rooms inany days

                      copy iconCopydownload iconDownload
                      SELECT name
                      FROM worktable
                      GROUP BY name
                      HAVING COUNT(DISTINCT name) > 1
                      
                      SELECT name
                      FROM worktable
                      GROUP BY name
                      HAVING COUNT(distinct room) > 1;
                      
                      name
                      Sam
                      
                      SELECT name
                      FROM worktable
                      GROUP BY name
                      HAVING COUNT(distinct room) > 1;
                      
                      name
                      Sam
                      

                      Complexity analysis for the permutations algorithm

                      copy iconCopydownload iconDownload
                      T(x) = x * T(x-1) + O(x)     if x > 1
                      T(1) = A.length
                      

                      Community Discussions

                      Trending Discussions on interviews
                      • How to find most common words from specific rows and column and list how often it occurs at data.csv?
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                      • matrix maximum diagonal pattern match
                      • Cleaner/Simpler way to check if content in array has a value greater than 0
                      • How to combine columns, group them then get a total count?
                      • formulate capture groups for inconsistently present substrings
                      • SQL Query to find people who have interviews in different rooms inany days
                      • Complexity analysis for the permutations algorithm
                      Trending Discussions on interviews

                      QUESTION

                      How to find most common words from specific rows and column and list how often it occurs at data.csv?

                      Asked 2022-Mar-03 at 20:14

                      I want to get 20 most common words from the descriptions of top 10 longest movies from data.csv, by using Python. So far, I got top 10 longest movies, however I am unable to get most common words from those specific movies, my code just gives most common words from whole data.csv itself. I tried Counter, Pandas, Numpy, Mathlib, but I have no idea how to make Python look exactly for most common words in the specific rows and column (description of movies) of the data table

                      My code:

                      import pandas as pd
                      import numpy as np
                      df = pd.read_csv("data.csv")
                      small_df = df[['title','duration_min','description']]
                      result_time = small_df.sort_values('duration_min', ascending=False)
                      print("TOP 10 LONGEST: ")
                      print(result_time.head(n=10))
                      
                      most_common = pd.Series(' '.join(result_time['description']).lower().split()).value_counts()[:20]
                      print("20 Most common words from TOP 10 longest movies: ")
                      print(most_common)
                      

                      My output:

                      TOP 10 LONGEST: 
                                                   title  duration_min                                        description
                      6840        The School of Mischief         253.0  A high school teacher volunteers to transform ...
                      4482                No Longer kids         237.0  Hoping to prevent their father from skipping t...
                      3687            Lock Your Girls In         233.0  A widower believes he must marry off his three...
                      5100               Raya and Sakina         230.0  When robberies and murders targeting women swe...
                      5367                        Sangam         228.0  Returning home from war after being assumed de...
                      3514                        Lagaan         224.0  In 1890s India, an arrogant British commander ...
                      3190                  Jodhaa Akbar         214.0  In 16th-century India, what begins as a strate...
                      6497                  The Irishman         209.0  Hit man Frank Sheeran looks back at the secret...
                      3277      Kabhi Khushi Kabhie Gham         209.0  Years after his father disowns his adopted bro...
                      4476  No Direction Home: Bob Dylan         208.0  Featuring rare concert footage and interviews ...
                      20 Most common words from TOP 10 longest movies: 
                      a        10134
                      the       7153
                      to        5653
                      and       5573
                      of        4691
                      in        3840
                      his       3005
                      with      1967
                      her       1803
                      an        1727
                      for       1558
                      on        1528
                      their     1468
                      when      1320
                      this      1240
                      from      1114
                      as        1050
                      is         988
                      by         894
                      after      865
                      dtype: int64
                      

                      ANSWER

                      Answered 2022-Mar-03 at 20:05

                      You can select the first 10 rows of your dataframe with iloc[0:10].

                      In this case, the solution would look like this, with the least modification to your existing code:

                      import pandas as pd
                      import numpy as np    
                      df = pd.read_csv("data.csv")
                      small_df = df[['title','duration_min','description']]
                      result_time = small_df.sort_values('duration_min', ascending=False)
                      print("TOP 10 LONGEST: ")
                      print(result_time.head(n=10))
                      
                      most_common = pd.Series(' '.join(result_time.iloc[0:10]['description']).lower().split()).value_counts()[:20]
                      print("20 Most common words from TOP 10 longest movies: ")
                      print(most_common) 
                      

                      Source https://stackoverflow.com/questions/71343075

                      Community Discussions, Code Snippets contain sources that include Stack Exchange Network

                      Vulnerabilities

                      No vulnerabilities reported

                      Install interviews

                      You can download it from GitHub.
                      You can use interviews like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the interviews component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .

                      Support

                      For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .

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