Question

我有以下代码，我扫描每一行并放入一个列表。如果该行匹配一个字符串“New changes”，我不想把它放在列表中。有关如何实现这一目标的任何建议吗？

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

Answer 1

您可以在列表理解中进行过滤：

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)

如果有人有更多的pythonic方法，请随时告诉我。

Answer 4

理解也可以接受一个条件。尝试：

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

或

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

Answer 5

这里的不同之处是来自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(), [])

阅读文件，跳过不需要的行＆amp;添加到列表中

5 个答案: