使用 for in 时，'int' 类型的 python 参数不可迭代答案

【问题标题】：python argument of type 'int' is not iterable when using for in使用 for in 时，'int' 类型的 python 参数不可迭代
【发布时间】：2018-03-29 19:49:10
【问题描述】：

我正在尝试编写从命令行接受文件名的代码，并且打印出以下属性：

行数
字符数
字数
“该”的数量
“a/an”的数量

我不断收到错误消息

“'int'类型的参数不可迭代”

为if 'the' in words: 行。

我该如何解决这个问题？

import sys
import string

file_name=sys.argv[0]

char= words = lines = theCount = aCount= 0

with open(file_name,'r') as in_file:
    for line in in_file:
        lines +=1
        words +=len(line.split())
        char +=len(line)
        if 'the' in words:
            theCount +=1
        if 'a' in words:
            a +=1
        if 'an' in words:
            a +=1

print("Filename:", file_name)
print("Number of lines:", lines)
print("Number of characters:", char)
print("Number of 'the'", theCount)
print("Number of a/an:", aCount)

【问题讨论】：

请参阅官方文档中的 this Counter 配方，其中展示了如何做大部分您想做的事情。

标签： python python-3.x

【解决方案1】：

如果您尝试收集实际单词，而不仅仅是它们的计数，那么您可能需要将单词初始化为一个空列表：

words = []

改变

words += len(line.split())

到

words += line.split()

【讨论】：

【解决方案2】：

您的代码中有一些错误，请阅读此片段中的 cmets：

import sys
#import string   #not sure if this is needed

file_name=sys.argv[0]

char= words = lines = theCount = aCount= 0


with open(file_name,'r') as in_file:
    for line in in_file:
        lines +=1
        x = line.split() #use a variable to hold the split words
                         #so that you can search in it
        words +=len(x)
        char +=len(line)
        if 'the' in x:   #your original code was using "words" variable
                         #that holds the "number of words" in the line,
                         #therefore ints are not iterable
            theCount +=1
        if 'a' in x:
            aCount +=1   #your original code using "a" variable 
                         #which did not initialized, 
                         #you have initialized "aCount" variable
        if 'an' in x:
            aCount +=1   #same as above

print("Filename:", file_name)
print("Number of lines:", lines)
print("Number of characters:", char)
print("Number of 'the'", theCount)
print("Number of a/an:", aCount)

https://repl.it/Mnwz/0

【讨论】：