Python If == true语句仅在readline的最后一行有效 [英] Python If == true statement only working on last line of readline

问题描述

我的函数只说单词文件中的最后一个单词是一个字谜(第一个辅助函数).但是文件中的每个单词都是我测试过的单词的一个字谜，通过主函数之外的helper函数独立返回true.我不确定它是否与/n 是字符串的一部分有关，然后才考虑了这一点，但是我尝试放入一个if语句，说要删除它是否在字符串中那里，那也不起作用.我还进行了测试，以确保它遍历 .txt 文件中的每个单词，并且确实如此.

  def is_anagram(string1，string2):如果两个字符串彼此相同，则"返回True.str，str->布尔"如果sorted(string1)== sorted(string2):返回True别的:返回Falsedef find_anagrams(word):最终= []内容=打开("small_list.txt")content.close而True:行= content.readline()打印(行)如果is_anagram(word，line)== True:打印("bruh")final.append(行)省略号行=='':休息返回最终 

解决方案

根据您用于读取行的方法( file.readline )，这是预期的.从文档:

f.readline()从文件中读取一行；换行符( \ n )留在字符串的末尾，仅在末尾省略如果文件未以换行符结尾，则为该行.

您的 line 带有尾随换行符，但是 word 当然没有.因此，最后，您需要更改的只是:

  line = content.readline().rstrip() 

好吧，这就是您要做的所有更改，以使其正常工作.另外，我还建议使用 with ... as 上下文管理器来处理文件I/O.这是一个好习惯，您会为此感到感谢的.

 ，其中open("small_list.txt")为f:对于f中的行:如果is_anagram(word，line.rstrip()):...#在这里做点事 

最好使用 for 循环遍历文件的行(而不是 while ，这更干净).此外，在使用上下文管理器时，也无需显式调用 f.close()(您当前未在执行此操作，仅引用该方法而未实际调用它).

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并入@Christian Dean的

解决方案

This is expected, based on the method you use to read a line (file.readline). From the documentation:

f.readline() reads a single line from the file; a newline character (\n) is left at the end of the string, and is only omitted on the last line of the file if the file doesn’t end in a newline.

Your line has a trailing newline, but word certainly does not. So, in the end, all you'd need to change is:

line = content.readline().rstrip()

Well, that's all you'd need to change to get it working. Additionally, I'd also recommend using the with...as context manager to handle file I/O. It's good practice, and you'll thank yourself for it.

with open("small_list.txt") as f:
    for line in f:
        if is_anagram(word, line.rstrip()):
            ... # do something here

It's better to use a for loop to iterate over the lines of a file (rather than a while, it's cleaner). Also, there's no need to explicitly call f.close() when you use a context manager (you're not currently doing it, you're only referencing the method without actually calling it).

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Incorporating @Christian Dean's suggestion in this answer, you can simplify your anagram function as well - call sorted and return the result in a single line:

def is_anagram(a, b):
    return sorted(a) == sorted(b)

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