Question

我有一个大字符串，我想从中删除所有日期字符串子字符串。受约束，日期字符串都遵循以下格式：

年份的月份字符串（例如：2018年9月1日）

假设我的字符串是：

bad_s = "It was a fine day. September 1, 2018 and I had a lot of laughs August 2, 2017"

我想回来 good_s = "It was a fine day. and I had a lot of laughs"

在Python中有一种简便的方法吗？

这是我尝试过的：

reg_ex = """/[\'January\'\,\ \'February\'\,\ \'March\'\,\ \'April\'\,\ \'May\'\,\ \'June\'\,\ \'July\'\,\ \'August\'\,\ \'September\'\,\ \'October\'\,\ \'November\'\,\ \'December\'](?:\^\(\[1\-9\]\|\[12\]\\d\|3\[0\-q\]\)\$)/"""
replaced = re.sub(reg_ex, bad_s, "")

但是，这不能代替我想要的。我仍然以bad_s结尾。

编辑：如果它使任何人都更容易，这里是12个月的清单，因此您不必编写它们： months = ['January', 'February', 'March', 'April', 'May', 'June', 'July', 'August', 'September', 'October', 'November', 'December']

Answer 1

喜欢吗？

(january|february|march|april|may|june|july|august|september|octorber|november|december) ([1-9]|[1-2]\d|3[01]), \d{4}

请不要忘记/i标志或任何与Python等效的标志。

请注意，这并不关心一个月中有多少天，因此February 31, 2017将会匹配，它也不关心leap年。此正则表达式是匹配器而不是验证器。

如果您希望它更通用并且完全忽略日期验证，那么它将起作用：

(january|february|march|april|may|june|july|august|september|octorber|november|december) \d+, \d+

https://regex101.com/r/zVbb0v/5

Answer 2

也许您可以尝试以下方法：

重新导入

bad_s = 'It was a fine day. September 20, 2018 and I had a lot of laughs August 2, 2017'
regex = '([^\s]+ ([1-9]|[12]\d|3[01])\, ([12]\d{3}))'

for x in re.findall(regex, bad_s):
    bad_s = bad_s.replace(x[0], '')

print(bad_s)

结果：

"It was a fine day.  and I had a lot of laughs"

如何删除像2019年7月1日这样的字符串

2 个答案: