读取文件，跳过不需要的行并添加到列表中

Question

我有以下代码，我在其中扫描每一行并放入列表中。 如果该行与字符串“ New changes”匹配，则我不想放入列表中。 关于如何实现这一目标的任何建议？

with open('file.txt', 'rb') as f:
    mainlist = [line.strip() for line in f]

Answer 1

您可以在列表理解范围内进行过滤：

mainlist = [line.strip() for line in f if line.strip() !=  "New changes"]

Answer 2

列表推导也可以进行过滤：

mainlist = [line.strip() for line in f if "New changes" not in line]

Answer 3

with open('file.txt', 'rb') as f:
    mainlist = []
    for line in f:
        s = line.strip()
        if s != "New changes":
            mainlist.append(s)

如果有人有更Python化的方法来执行此操作，请随时告诉我。

Answer 4

理解力也可以接受条件。 尝试：

mainlist = [line.strip() for line in f if line != "New changes"]

要么

mainlist = [line.strip() for line in f if "New changes" not in line]

Answer 5

与这里不同的是reduce和regex学校的咒语：

import re
with open('file.txt', 'rb') as f:
  mainlist = reduce(lambda x, y: x+re.findall("^((?!.*New changes).*)\n?$", y), f.readlines(), [])