我将堆栈生成器表达式：with open(filename) as f_in:

lines = (line.rstrip() for line in f_in) # All lines including the blank ones

lines = (line for line in lines if line) # Non-blank lines

现在，lines都是非空行。这样你就不用在电话线上打两次电话了。如果您想要一个行列表，那么您可以执行以下操作：with open(filename) as f_in:

lines = (line.rstrip() for line in f_in)

lines = list(line for line in lines if line) # Non-blank lines in a list

你也可以用一行代码（不包括with语句）来完成，但它没有效率更高，也更难阅读：with open(filename) as f_in:

lines = list(line for line in (l.strip() for l in f_in) if line)

更新：

我同意这是丑陋的，因为重复的令牌。如果你愿意，你可以写一个生成器：def nonblank_lines(f):

for l in f:

line = l.rstrip()

if line:

yield line

那就叫它：with open(filename) as f_in:

for line in nonblank_lines(f_in):

# Stuff

更新2：with open(filename) as f_in:

lines = filter(None, (line.rstrip() for line in f_in))

在CPython上（使用确定性引用计数）lines = filter(None, (line.rstrip() for line in open(filename)))

在Python 2中，如果需要生成器，请使用itertools.ifilter；在Python 3中，如果需要列表，请将整个过程传递给list。