使用count方法对文本文件中的某个单词进行计数

Question

我正在尝试计算单词“ the”在另存为文本文件的两本书中出现的次数。 我正在运行的代码为每本书返回零。

这是我的代码：

def word_count(filename):
    """Count specified words in a text"""
    try:
        with open(filename) as f_obj:
            contents = f_obj.readlines()
            for line in contents:
                word_count = line.lower().count('the')
            print (word_count)

    except FileNotFoundError:
        msg = "Sorry, the file you entered, " + filename + ", could not be     found."
    print (msg)

dracula = 'C:\\Users\\HP\\Desktop\\Programming\\Python\\Python Crash   Course\\TEXT files\\dracula.txt'
siddhartha = 'C:\\Users\\HP\\Desktop\\Programming\\Python\\Python Crash Course\\TEXT files\\siddhartha.txt'

word_count(dracula)
word_count(siddhartha)

我在这里做错了什么？

Answer 1

您将为每次迭代重新分配word_count 。 这意味着，在最后这将是相同的出现次数the在文件的最后一行。 你应该得到的总和。 另一件事：应该there匹配吗？ 可能不是。 您可能要使用line.split() 。 同样，您可以直接遍历文件对象。 不需要.readlines() 。 最后，使用生成器表达式进行简化。 我的第一个示例没有生成器表达式； 第二个是：

def word_count(filename):
    with open(filename) as f_obj:
        total = 0
        for line in f_obj:
            total += line.lower().split().count('the')
        print(total)

def word_count(filename):
    with open(filename) as f_obj:
        total = sum(line.lower().split().count('the') for line in f_obj)
        print(total)

Answer 2

除非单词“ the”出现在每个文件的最后一行，否则您将看到零。

您可能希望将word_count变量初始化为零，然后使用增强加法（ += ）：

例如：

def word_count(filename):
    """Count specified words in a text"""
    try:
        word_count = 0                                       # <- change #1 here
        with open(filename) as f_obj:
            contents = f_obj.readlines()
            for line in contents:
                word_count += line.lower().count('the')      # <- change #2 here
            print(word_count)

    except FileNotFoundError:
        msg = "Sorry, the file you entered, " + filename + ", could not be     found."
    print(msg)

dracula = 'C:\\Users\\HP\\Desktop\\Programming\\Python\\Python Crash   Course\\TEXT files\\dracula.txt'
siddhartha = 'C:\\Users\\HP\\Desktop\\Programming\\Python\\Python Crash Course\\TEXT files\\siddhartha.txt'

word_count(dracula)
word_count(siddhartha)

增强添加不是必需的，只是有帮助。 这行：

word_count += line.lower().count('the')

可以写成

word_count = word_count + line.lower().count('the')

但是，您也不需要一次将所有行读入内存。 您可以直接从文件对象遍历各行。 例如：

def word_count(filename):
    """Count specified words in a text"""
    try:
        word_count = 0
        with open(filename) as f_obj:
            for line in f_obj:                     # <- change here
                word_count += line.lower().count('the')
        print(word_count)

    except FileNotFoundError:
        msg = "Sorry, the file you entered, " + filename + ", could not be     found."
        print(msg)

dracula = 'C:\\Users\\HP\\Desktop\\Programming\\Python\\Python Crash Course\\TEXT files\\dracula.txt'
siddhartha = 'C:\\Users\\HP\\Desktop\\Programming\\Python\\Python Crash Course\\TEXT files\\siddhartha.txt'

word_count(dracula)
word_count(siddhartha)

Answer 3

其他方式：

with open(filename) as f_obj:
    contents = f_obj.read()
    print("The word 'the' appears " + str(contents.lower().count('the')) + " times")

Answer 4

import os
def word_count(filename):
    """Count specified words in a text"""
    if os.path.exists(filename):
        if not os.path.isdir(filename):
            with open(filename) as f_obj:
                print(f_obj.read().lower().count('t'))
        else:
            print("is path to folder, not to file '%s'" % filename)
    else:
        print("path not found '%s'" % filename)