如何计算Python中的一个特定单词？

Question

我想计算文件中的特定单词.

例如,"apple"出现在文件中的次数.我试过这个:

#!/usr/bin/env python
import re 

logfile = open("log_file", "r") 

wordcount={}
for word in logfile.read().split():
    if word not in wordcount:
        wordcount[word] = 1
    else:
        wordcount[word] += 1
for k,v in wordcount.items():
    print k, v

将'word'替换为'apple',但它仍会计算我文件中所有可能的单词.

任何建议将不胜感激.:)

Answer 1

您可以使用,str.count()因为您只关心单个单词的出现:

with open("log_file") as f:
    contents = f.read()
    count = contents.count("apple")

但是,为了避免一些极端情况,例如错误地计算单词"applejack",我建议你使用正则表达式:

import re

with open("log_file") as f:
    contents = f.read()
    count = sum(1 for match in re.finditer(r"\bapple\b", contents))

\b在正则表达式中确保模式在单词边界上开始和结束(而不是在较长字符串中的子字符串).

Answer 2

如果您只关心一个单词,那么您不需要创建一个字典来跟踪每个字数.您可以逐行遍历文件并查找您感兴趣的单词的出现位置.

#!/usr/bin/env python

logfile = open("log_file", "r") 

wordcount=0
my_word="apple"
for line in logfile:
    if my_word in line.split():
        wordcount += 1

print my_word, wordcount

但是,如果您还要计算所有单词,并且只打印您感兴趣的单词的单词计数,那么对代码的这些微小更改应该有效:

#!/usr/bin/env python
import re 

logfile = open("log_file", "r") 

wordcount={}
for word in logfile.read().split():
    if word not in wordcount:
        wordcount[word] = 1
    else:
        wordcount[word] += 1
# print only the count for my_word instead of iterating over entire dictionary
my_word="apple"
print my_word, wordcount[my_word]