Python的字数降序[重复]

Question

这个问题已经在这里有了答案：

• 如何按降序打印频率？ 4个答案
• Python中的单词频率分析返回字母频率 3个答案

我正在使用以下代码来计算文本文件中单词出现的频率：

#!/usr/bin/python
file=open("out1.txt","r+")
wordcount={}
for word in file.read().split():
    if word not in wordcount:
        wordcount[word] = 1
    else:
        wordcount[word] += 1
for k,v in wordcount.items():
    print k, v

如何按频率编号降序打印输出？

Answer 1

使用Counter.most_common而不指定值来获取单词频率的降序列表。

from collections import Counter

word_count = Counter()

with open("out1.txt","r+") as file:
    word_count.update((word for word in file.read().split()))

for word, count in word_count.most_common():
    print word, count

>>> the 6
Lorem 4
of 4
and 3
Ipsum 3
text 2
type 2

Answer 2

您可以创建一个元组列表并对其进行排序。 这是一个例子。

wordcount = {'cat':1,'dog':2,'kangaroo':20}

ls = [(k,v) for (k,v) in wordcount.items()]

ls.sort(key=lambda x:x[1],reverse=True)

for k,v in ls:
    print k, v

...输出...

kangaroo 20
dog 2
cat 1

Answer 3

这是代码：

file=open("out1.txt","r+")
wordcount={}
for word in file.read().split():
    word = word.lower()
    if word.isalpha == True:
        if word not in wordcount:
            wordcount[word] = 1
        else:
            wordcount[word] += 1
copy = []
for k,v in wordcount.items():
    copy.append((v, k))


copy = sorted(copy, reverse=True)

for k in copy:
        print '%s: %d' %(k[1], k[0])

Out1.txt ：

hello there I am saying hello world because Bob is here and I am saying hello because John is here

运行方式为 ：

hello: 3
saying: 2
is: 2
here: 2
because: 2
am: 2
I: 2
world: 1
there: 1
and: 1
John: 1
Bob: 1

Answer 4

使用Counter模块。

from collections import Counter

s = "This is a sentence this is a this is this"

c = Counter(s.split())
#s.split() is an array of words, it splits it at each space if no parameter is given to split on

print c

>>> Counter({'is': 3, 'this': 3, 'a': 2, 'This': 1, 'sentence': 1})

但是，这对于句点和大写字母无法“正确”运行。 您可以简单地删除单词末尾的句点以正确计数，并使所有内容都小写/大写以使其不区分大小写。

您可以使用以下方法消除这两个问题：

s1 = "This is a sentence. This is a. This is. This."
s2 = ""

for word in s1.split():
    #punctuation checking, you can make this more robust through regex if you want
    if word.endswith('.') or word.endswith('!') or word.endswith('?'):
        s2 += word[:-1] + " "
    else:
        s2 += word + " "

c = Counter(s2.lower().split())

print c

>>> Counter({'this': 4, 'is': 3, 'a': 2, 'sentence': 1})