Question

我想迭代每个文件，修剪每行的空格，如果返回的字符串为空则删除该行。有没有办法避免在下面的列表推导中重复.strip（）调用？它不是性能关键，但感觉不对。

sub main():
    fname = "foo.txt"
    lns = []

    with open(fname, 'r') as file:
        lns = file.readlines()

    newlns = [i.strip() + "\n" for i in lns if i.strip()]
    #I want this to look like the following, which doesn't work:
    #newlns = [y + "\n" for i in lns if i.strip() as y]

    with open("out.txt", 'w') as file:
        file.writelines(newlns)

Answer 1

您可以使用嵌套列表推导（在这种情况下，生成器表达式以避免实际构建列表）：

newlns = [i + "\n" for i in (line.strip() for ln in lns) if i]

实际上你真的不应该先费心去读取文件，只是把它放在那里：迭代文件会产生它的行。

with open(fname, 'r') as file:
    newlns = [i + "\n" for i in (line.strip() for ln in file) if i]

在Python中的列表理解中避免重复函数调用

1 个答案: