将字符串中的 ISO 8601 替换为本地日期或日期时间

Question

我尝试用本地日期或日期时间 (+02:00) 替换 python 3.x

中字符串中的 ISO 8601 "YYYY-MM-DDTHH:MI:SSZ" (UTC)

字符串示例：

x = "This is first example with dates 2019-07-01 21:30:20 and 2019-07-02 21:30:20"

我的代码，工作正常但不完美：

def date_to_iso(m):
    date_string = (dateutil.parser.parse(m.group(0))).astimezone(pytz.UTC).strftime("%Y-%m-%d" + "T" + "%H:%M:%S" + "Z")
    return iso_8601

y = re.sub(r"\d{4}(?:-\d{2}){2}" + r" \d{2}(?::\d{2}){2}", date_to_iso, x)

第一个例子的结果很好：

Out[269]: 'This is first example with dates 2019-07-01T19:30:20Z and 2019-07-02T19:30:20Z'

我的问题是如何修改它以便有不同格式的日期。例如：

x = "This is second example with dates 2019-07-01 and 2019-07-02 21:30:20, but there are dates 2019-07-10 07:00 and 2019-07-10 09 too"

它应该 return 像这样：

Out[269]: 'This is second example with dates 2019-06-30T22:00:00Z and 2019-07-02T19:30:20Z, but there are dates 2019-07-10T05:00:00Z or 2019-07-10T07:00:00Z'

Answer 1

我发现了。这是代码，对我有好处：

import dateutil.parser
import pytz
import re

x = "This is second example with dates 2019-07-01 and 2019-07-02 21:30:20, but there are dates 2019-07-10 07:00 and 2019-07-10 09 too"

def datetime_to_iso(m):
    datetime_iso = (dateutil.parser.parse(m.group(0))).astimezone(pytz.UTC).strftime("%Y-%m-%d" + "T" + "%H:%M:%S" + "Z")
    return datetime_iso

x = re.sub(r"\d{4}(?:-\d{1,2}){2}" + r"( \d{1,2}(?::\d{1,2}(?::\d{1,2})?)?)?", datetime_to_iso, x)


print(x)
This is second example with dates 2019-06-30T22:00:00Z and 2019-07-02T19:30:20Z, but there are dates 2019-07-10T05:00:00Z and 2019-07-10T07:00:00Z too

将字符串中的 ISO 8601 替换为本地日期或日期时间

Substitution local date or datetime for ISO 8601 in string

python

regex

datetime

substitution

python-dateutil