Python 搜索文本文件并计算指定字符串的出现次数

Question

我正在尝试使用 python 搜索文本文件并计算用户定义的单词出现的次数。但是，当我运行下面的代码而不是获取该唯一单词在文件中出现的次数总和时，我得到的是该文件中包含该单词的行数的计数。

示例：单词“bob”在文本文件中出现了 56 次，出现在总共 63 行文本中的 19 行中。当我运行代码时，控制台打印“19”。

我猜我需要对我的分割方法做一些不同的事情？我正在运行 Python 2.7.10。

user_search_value = raw_input("Enter the value or string to search for: ")

count = 0    

with open(file.txt, 'r') as f:
    for word in f.readlines():
        words = word.lower().split()
        if user_search_value in words:
            count += 1
    print count

Answer 1

执行此操作的一种方法是在分割行并为每个匹配单词递增

count

后循环遍历单词：

user_search_value = raw_input("Enter the value or string to search for: ")

count = 0    

with open(file.txt, 'r') as f:
    for line in f.readlines():
        words = line.lower().split()
        for word in words:
            if word == user_search_value:
                count += 1
print count

Answer 2

正如我在上面的评论中提到的，在玩了很长一段时间后，我发现了这一点。我的代码如下。

#read file
f = open(filename, "r")
lines = f.readlines()
f.close()
#looking for patterns
for line in lines:
    line = line.strip().lower().split()
    for words in line:
        if words.find(user_search_value.lower()) != -1:
            count += 1
print("\nYour search value of '%s' appears %s times in this file" % (user_search_value, count))

Answer 3

如果“指定字符串”是带有空格的短语，则以下是有效的：

#!/usr/bin/python

import sys
import os

def count_words_in_file(filepath, words, action=None):
    with open(filepath) as f:
        data = f.read()
        for key,val in words.items():
            print "key is " + key + "\n"
            ct = data.count(key)
            words[key] = ct
        if action:
             action(filepath, words)


def print_summary(filepath, words):
    print(filepath)
    for key,val in sorted(words.items()):
        print('{0}:\t{1}'.format(
            key,
            val))


filepath = sys.argv[1]
keys = ["Hello how are you",
"Another phrase with spaces",
"A phrase with spaces and some punctuation."]
words = dict.fromkeys(keys,0)

count_words_in_file(filepath, words, action=print_summary)