wordfreq | Text corpus calculation in Javascript

by timdream HTML Version: 0.1.0-dev License: MIT

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kandi X-RAY | wordfreq Summary

wordfreq is a HTML library. wordfreq has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. You can download it from GitHub.

Text corpus calculation in Javascript. Supports Chinese, English. See demo. This library is a spin-off project from HTML5 Word Cloud.

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wordfreq has a low active ecosystem.

It has 74 star(s) with 19 fork(s). There are 7 watchers for this library.

It had no major release in the last 12 months.

There are 4 open issues and 6 have been closed. On average issues are closed in 50 days. There are no pull requests.

It has a neutral sentiment in the developer community.

The latest version of wordfreq is 0.1.0-dev

Quality

wordfreq has no bugs reported.

Security

wordfreq has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

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.

Reuse

wordfreq releases are not available. You will need to build from source code and install.

Installation instructions are not available. Examples and code snippets are available.

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wordfreq Key Features

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wordfreq Examples and Code Snippets

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Community Discussions

Trending Discussions on wordfreq

word frequency in multiple documents

How do you count the number of words in a column while ignoring blank lines

Stream API collect() use personal class Word intsead of Map

TypeError: unhashable type: 'list' for text summarization

How to use functor as custom comparator in priority_queue

Create Spark UDF of a function that depends on other resources

IndexError: list index out of range when using tuples

Print the maximum occurence of the anagrams and the anagram words itself among the input anagrams

Python: word count from WordCloud

Group the word count of a data frame according to the diferent labels stored in a column

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

Use a defaultdict

Source https://stackoverflow.com/questions/64416381

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

TL;DR version:

Source https://stackoverflow.com/questions/63621067

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

QUESTION

Group the word count of a data frame according to the diferent labels stored in a column

Asked 2020-Feb-11 at 10:02

I would like to know which are the most representative words for two classes on all the cells of a dataframe

...

ANSWER

Answered 2020-Feb-10 at 17:56

You can do:

Source https://stackoverflow.com/questions/60155605

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

Vulnerabilities

No vulnerabilities reported

Install wordfreq

You can download it from GitHub.

Support

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