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是所有非空白行。这将使您不必在线路上两次打电话。如果需要行列表，则可以执行以下操作：
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
您也可以采用单行代码（排除
with语句）来执行此操作，但是它不再高效且更难阅读：
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，由于令牌的重复，这很丑陋。您可以根据需要编写一个生成器：
def nonblank_lines(f): for l in f: line = l.rstrip() if line: yield line
然后像这样调用它：
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))
以及在CPython上（具有确定性引用计数）
lines = filter(None, (line.rstrip() for line in open(filename)))
itertools.ifilter如果需要生成器，请在Python 2中使用；如果需要列表，请在Python 3
list中使用。
