CODEWITHSUNDEEP
pythonstringlistcpu-wordword-cloud
Pyd
Pyd

Reputation: 6159

generating word cloud for items in a list in python

 my_list=["one", "one two", "three"]

and I am generating a word cloud for this list by using

 wordcloud = WordCloud(width = 1000, height = 500).generate(" ".join(my_list))

As I am converting all the items into string it is generating word cloud for 

   "one","two","three"

 But I want to generate word cloud for the values, "one","one two","three"

help me for generating word cloud for items in a list

Upvotes: 17

Views: 36557

Answers (2)

Pyd
Pyd

Reputation: 6159

one way of doing,

import matplotlib.pyplot as plt
from wordcloud import WordCloud

#convert list to string and generate
unique_string=(" ").join(my_list)
wordcloud = WordCloud(width = 1000, height = 500).generate(unique_string)
plt.figure(figsize=(15,8))
plt.imshow(wordcloud)
plt.axis("off")
plt.savefig("your_file_name"+".png", bbox_inches='tight')
plt.show()
plt.close()

Another way by creating Counter Dictionary,

#convert it to dictionary with values and its occurences
from collections import Counter
word_could_dict=Counter(my_list)
wordcloud = WordCloud(width = 1000, height = 500).generate_from_frequencies(word_could_dict)

plt.figure(figsize=(15,8))
plt.imshow(wordcloud)
plt.axis("off")
#plt.show()
plt.savefig('yourfile.png', bbox_inches='tight')
plt.close()

Upvotes: 20

ikkuh
ikkuh

Reputation: 4603

The WordCloud takes regular expression as argument. Using this we can make the splitting character a + instead of a space.

regexp=r"\w[\w' ]+"

The list of words then needs to be joined on a + as well as each this is now used to split words. Resulting in the following code:

wordcloud = WordCloud(width=1000, height=500, regexp=r"\w[\w' ]+").generate("+".join(my_list))

Upvotes: 1

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