wordfreq | word frequencies , in various natural languages | Natural Language Processing library

 by   LuminosoInsight Python Version: v2.2 License: MIT

kandi X-RAY | wordfreq Summary

kandi X-RAY | wordfreq Summary

wordfreq is a Python library typically used in Artificial Intelligence, Natural Language Processing applications. wordfreq has no bugs, it has no vulnerabilities, it has build file available, it has a Permissive License and it has low support. You can install using 'pip install wordfreq' or download it from GitHub, PyPI.

wordfreq is a Python library for looking up the frequencies of words in many languages, based on many sources of data.
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            kandi-support Support

              wordfreq has a low active ecosystem.
              It has 424 star(s) with 34 fork(s). There are 45 watchers for this library.
              OutlinedDot
              It had no major release in the last 12 months.
              There are 6 open issues and 13 have been closed. On average issues are closed in 73 days. There are 1 open pull requests and 0 closed requests.
              It has a neutral sentiment in the developer community.
              The latest version of wordfreq is v2.2

            kandi-Quality Quality

              wordfreq has 0 bugs and 0 code smells.

            kandi-Security Security

              wordfreq has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
              wordfreq code analysis shows 0 unresolved vulnerabilities.
              There are 0 security hotspots that need review.

            kandi-License License

              wordfreq is licensed under the MIT License. This license is Permissive.
              Permissive licenses have the least restrictions, and you can use them in most projects.

            kandi-Reuse Reuse

              wordfreq releases are available to install and integrate.
              Deployable package is available in PyPI.
              Build file is available. You can build the component from source.
              Installation instructions, examples and code snippets are available.
              wordfreq saves you 413 person hours of effort in developing the same functionality from scratch.
              It has 980 lines of code, 71 functions and 17 files.
              It has low code complexity. Code complexity directly impacts maintainability of the code.

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            wordfreq Key Features

            No Key Features are available at this moment for wordfreq.

            wordfreq Examples and Code Snippets

            How do I avoid printing " " in my tokenize function?
            Pythondot img1Lines of Code : 64dot img1License : Strong Copyleft (CC BY-SA 4.0)
            copy iconCopy
            def tokenize(lines):
                words = []
                for line in lines:
                    line = line.strip()
                    start = 0
                    while start < len(line):
            
                        while start < len(line) and line[start].isspace():
                            start = start +
            word frequency in multiple documents
            Pythondot img2Lines of Code : 6dot img2License : Strong Copyleft (CC BY-SA 4.0)
            copy iconCopy
            from collections import Counter
            bigLst = [['hello', 'my', 'friend'], ['jim', 'is', 'cool'], ['peter', 'is', 'nice']]
            print(Counter([word for lst in bigLst for word in lst]))
            
            Counter({'is': 2, 'hello': 1, 'my': 1, '
            How do you count the number of words in a column while ignoring blank lines
            Pythondot img3Lines of Code : 6dot img3License : Strong Copyleft (CC BY-SA 4.0)
            copy iconCopy
            for word in text.strip().split('\n'):
                # skip this word if it is blank
                if not word:
                    continue
                wordfreq[word] = wordfreq.get(word, 0) +1
            
            TypeError: unhashable type: 'list' for text summarization
            Pythondot img4Lines of Code : 10dot img4License : Strong Copyleft (CC BY-SA 4.0)
            copy iconCopy
            from nltk.corpus import indian
            from collections import defaultdict
            
            sentence_score=defaultdict(int)
            #word=nltk.word_tokenize(text)
            for sent in sentences:
                word_count_in_sentence = (len(nltk.word_tokenize(sentence)))
                if word in wordf
            copy iconCopy
            from collections import defaultdict
            words = ['ab','absa','sbaa','basa','ba']
            wordToAnagram= defaultdict(list) 
            # word vs list anagram 
            # loop below will create {aabs:  ['absa', 'sbaa', 'basa']}
            for word in words:
                s = "".join(sorted(wor
            python word count(defaultdict) column not showing
            Pythondot img6Lines of Code : 11dot img6License : Strong Copyleft (CC BY-SA 4.0)
            copy iconCopy
            >>> df = pd.DataFrame.from_dict(word_freq, orient='index')
            >>> df = df.rename(columns={0: 'WordFreq'})
            >>> df.index.name = 'Word'
            >>> df
                     WordFreq
            Word
            france          2
            spain           3
            beaches
            python word count(defaultdict) column not showing
            Pythondot img7Lines of Code : 21dot img7License : Strong Copyleft (CC BY-SA 4.0)
            copy iconCopy
            from collections import Counter
            
            
            text_list = ['france', 'spain', 'spain beaches', 'france beaches', 'spain best beaches']
            
            counter_dict = Counter([split_word for word in text_list for split_word in word.split()]
            #Counter({'france': 2, 'sp
            Find the best representative substrings from many strings
            Pythondot img8Lines of Code : 21dot img8License : Strong Copyleft (CC BY-SA 4.0)
            copy iconCopy
            from collections import Counter
            
            def represent(group):
                groupWords = [ expr.split(" ") for expr in group ]
                wordFreq   = Counter(word for words in groupWords for word in words)
                weights    = [ sum(wordFreq[word] for word in words)
            Python: word count from WordCloud
            Pythondot img9Lines of Code : 3dot img9License : Strong Copyleft (CC BY-SA 4.0)
            copy iconCopy
            >>> WordCloud().process_text('penn penn penn penn penn state state state state uni uni uni college college university states vice president vice president vice president vice president vice president vice president vice president'
            Simple Dictionary
            Pythondot img10Lines of Code : 15dot img10License : Strong Copyleft (CC BY-SA 4.0)
            copy iconCopy
            import string
            from collections import Counter
            txt = "Why's it always the nice guys?"
            
            counted = Counter(
                word if not word[-1] in string.punctuation else word[:-1] for word in txt.split()
            )
            print(counted)
            
            >>> Counter({"Why's":

            Community Discussions

            QUESTION

            How do I avoid printing " " in my tokenize function?
            Asked 2021-Sep-24 at 10:03

            I'm supposed to create a word counting program in Python, which checks the kinds of words in a given text and the frequency of those words.

            As part of the program, certain stop words should not be in the count, and neither should spaces and special characters (+-??:"; etc).

            The first part of the program is to create a tokenize function (I will later test my function, which should go through the following test):

            ...

            ANSWER

            Answered 2021-Sep-17 at 05:59

            Everything looks right to me except the last else where i think you missed an if condition. I also added a line.strip() to start with before any of the logic.

            The condition [" "], [] is failing, because if you don't strip the empty sentences, the final result will be [''] and the test case fails because [] not equal to ['']

            Source https://stackoverflow.com/questions/69218225

            QUESTION

            word frequency in multiple documents
            Asked 2021-Jun-13 at 15:46

            i have a dataframe with the columns title and tokenized words. Now I read in all tokenized words into a list called vcabulary looking like this:

            [['hello', 'my', 'friend'], ['jim', 'is', 'cool'], ['peter', 'is', 'nice']]

            now I want to go through this list of lists and count every word for every list.

            ...

            ANSWER

            Answered 2021-Jun-13 at 15:32

            Convert your 2D list, into a normal list, then use collections.Counter() to return a dictionary of each words occurrence count.

            Source https://stackoverflow.com/questions/67959902

            QUESTION

            How do you count the number of words in a column while ignoring blank lines
            Asked 2021-May-03 at 03:08

            I have the following list in a text file:

            banana

            egg

            balloon

            green giant

            How do I create a dictionary that counts the words, ignoring the blank lines

            my code so far:

            ...

            ANSWER

            Answered 2021-May-03 at 02:59

            After getting the current word, but before assigning it in the dict, check if it is empty, and if so, skip it.

            Source https://stackoverflow.com/questions/67362992

            QUESTION

            Stream API collect() use personal class Word intsead of Map
            Asked 2021-Apr-20 at 09:54

            How can i transform my map to class Word which contain word and his frequency

            ...

            ANSWER

            Answered 2021-Apr-20 at 09:54

            You can use Stream.map(..) method. In your case that would be:

            Source https://stackoverflow.com/questions/67176000

            QUESTION

            TypeError: unhashable type: 'list' for text summarization
            Asked 2020-Oct-18 at 18:33
            from nltk.corpus import indian
            
            sentence_score={}
            #word=nltk.word_tokenize(text)
            for sent in sentences:
                word_count_in_sentence = (len(nltk.word_tokenize(sentence)))
                if word in wordfreq.keys():
                    if sent not in sentence_score.keys():
                        sentence_score[sent]=wordfreq[word]
                    else:
                        senetence_score[sent]+=wordfreq[word]
            
            ...

            ANSWER

            Answered 2020-Oct-18 at 18:33

            QUESTION

            How to use functor as custom comparator in priority_queue
            Asked 2020-Aug-27 at 18:12

            I am trying to create a functor as custom comparator for my priority_queue which takes in an unordered_map as a parameter for constructor. I am not sure how to call the functor when declaring the priority_queue as I am getting the error:

            "Line 22: Char 48: error: template argument for template type parameter must be a type priority_queue pq;"

            ...

            ANSWER

            Answered 2020-Aug-27 at 18:12

            QUESTION

            Create Spark UDF of a function that depends on other resources
            Asked 2020-Aug-17 at 22:51

            I have a code for tokenizing a string.

            But that tokenization method uses some data which is loaded when my application starts.

            ...

            ANSWER

            Answered 2020-Aug-17 at 22:51

            At this point, if you deploy this code Spark will try to serialize your DataProviderUtil, you would need to mark as serializable that class. Another possibility is to declare you logic inside an Object. Functions inside objects are considered static functions and they are not serialized.

            Source https://stackoverflow.com/questions/63457993

            QUESTION

            IndexError: list index out of range when using tuples
            Asked 2020-Aug-07 at 21:45

            I'm very confused. I get an error on line 43 saying that the list index is out of range. Any help is appreciated.

            ...

            ANSWER

            Answered 2020-Aug-07 at 19:28
            def printTopMost(frequencies, n):
                listOfTuples = sorted(frequencies.items(), key=lambda x:x[1], reverse=True)
                n=min(n,len(listOfTuples))
                for x in range(n):
                    pair = listOfTuples[x]
                    word = pair[0]
                    frequency = str(pair[1])
                    print(word.ljust(20), frequency.rjust(5))
            

            Source https://stackoverflow.com/questions/63307111

            QUESTION

            Print the maximum occurence of the anagrams and the anagram words itself among the input anagrams
            Asked 2020-Jun-07 at 15:54
            a = ['ab', 'absa', 'sbaa', 'basa', 'ba']
            res = []
            s = 0
            for i in range(len(a)):
                b=a[i]
                c = ''.join(sorted(b))
                res.append(c)
            res.sort(reverse=False)
            wordfreq = [res.count(p) for p in res]
            d = dict(zip(res, wordfreq))
            all_values = d.values()  #all_values is a list
            max_value = max(all_values)
            print(max_value)
            max_key = max(d, key=d.get)
            print(max_key)
            
            ...

            ANSWER

            Answered 2020-Jun-07 at 15:40

            You can create a dictionary of word v/s list of anagrams

            and then print out the word which contains the maximum number of elements in the anagram list

            Source https://stackoverflow.com/questions/62234078

            QUESTION

            Python: word count from WordCloud
            Asked 2020-Mar-06 at 02:27

            I am using WordCloud on a body of text, and I would like to see the actual counts for each word in the cloud. I can see the weighted frequencies using .words_ but I was wondering if there is an easy way to see the actual counts?

            ...

            ANSWER

            Answered 2020-Mar-06 at 02:27

            Just use WordCloud().process_text(text):

            Source https://stackoverflow.com/questions/60234036

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

            Vulnerabilities

            No vulnerabilities reported

            Install wordfreq

            wordfreq requires Python 3 and depends on a few other Python modules (msgpack, langcodes, and regex). You can install it and its dependencies in the usual way, either by getting it from pip:.
            Chinese, Japanese, and Korean have additional external dependencies so that they can be tokenized correctly. They can all be installed at once by requesting the 'cjk' feature:. Tokenizing Chinese depends on the jieba package, tokenizing Japanese depends on mecab-python3 and ipadic, and tokenizing Korean depends on mecab-python3 and mecab-ko-dic. As of version 2.4.2, you no longer have to install dictionaries separately.

            Support

            This data comes from a Luminoso project called Exquisite Corpus, whose goal is to download good, varied, multilingual corpus data, process it appropriately, and combine it into unified resources such as wordfreq.
            Find more information at:

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