wordcloud python 提取英文_强制wordcloud python模块包含所有单词

看着wordcloud.py，如果stopwords参数为None，则它使用内置的stopwords集-因此您不会禁止使用stopwords。尝试用stopwords=set()调用它。在

中的内置标记化wordcloud.py将一个单词识别为一系列字母数字字符(因此fa-so被拆分为fa-so)，忽略大小写，同时合并简单的复数(例如dogs到dog)并忽略单个数字。如果您想绕过这个问题，您需要构建一个元组列表，每个元组都包含一个单词及其频率，然后调用WordCloud.generate_from_频率(频率)。在

我无法安装wordcloud，但在wordfreq函数中使用\S+(即，它将连续的非空白字符识别为一个单词)进行的这种简化标记化绝对有效：import re

from operator import itemgetter

words = "do do do do do do do do do do re re re re re mi mi fa-so fa fa fa fa fa fa fa fa fa-so fa-so fa-so fa-so fa-so so la ti do"

item1 = itemgetter(1)

def wordfreq(text):

d = {}

for word in re.findall(r"\S+", text):

# for word in re.findall(r"\w[\w']*", text):

if word.isdigit():

continue

word_lower = word.lower()

# Look in lowercase dict.

if word_lower in d:

d2 = d[word_lower]

else:

d2 = {}

d[word_lower] = d2

# Look in any case dict.

d2[word] = d2.get(word, 0) + 1

d3 = {}

for d2 in d.values():

# Get the most popular case.

first = max(d2.items(), key=item1)[0]

d3[first] = sum(d2.values())

return d3.items()

freqs = wordfreq(words)

print freqs

# prints: [('do', 11), ('la', 1), ('fa-so', 6), ('mi', 2), ('fa', 8), ('so', 1), ('ti', 1), ('re', 5)]

# pass freqs to WordCloud.generate_from_frequencies()

# maybe something like:

# wordcloud = WordCloud(ranks_only = True,stopwords=set()).generate_from_frequencies(freqs)

您可以查看的源代码wordcloud.py-您可以直接或更安全地修改它，并且可以像这个例子一样扩展/修改行为。在