在1900年之前的Python中格式化日期字符串? [英] Formatting date string in Python for dates prior to 1900?
问题描述
任何人都可以解释在Python中格式化日期时间字符串的最佳方法,其中日期值在1900年之前? strftime
需要晚于1900年的日期。
Can anyone explain the best way to format a date time string in Python where the date value is prior to the year 1900? strftime
requires dates later than 1900.
推荐答案
这有点麻烦,但它起作用(至少在python的稳定版本中):
It's a bit cumbersome, but it works (at least in stable versions of python):
>>> ts = datetime.datetime(1895, 10, 6, 16, 4, 5)
>>> '{0.year}-{0.month:{1}}-{0.day:{1}} {0.hour:{1}}:{0.minute:{1}}'.format(ts, '02')
'1895-10-06 16:04'
注意 str
仍然会产生一个可读的字符串:
note that str
would still produce a readable string:
>>> str(ts)
'1895-10-06 16:04:05'
编辑
模拟默认行为的最接近的方式是对字典进行硬编码,例如:
edit
The closest possible way to emulate the default behaviour is to hard-code the dictionary such as:
>>> d = {'%Y': '{0.year}', '%m': '{0.month:02}'} # need to include all the formats
>>> '{%Y}-{%m}'.format(**d).format(ts)
'1895-10'
您需要使用简单的正则表达式将所有格式说明符括入大括号中:
You'll need to enclose all format specifiers into the curly braces with the simple regex:
>>> re.sub('(%\w)', r'{\1}', '%Y-%m-%d %H sdf')
'{%Y}-{%m}-{%d} {%H} sdf'
最后我们来看简单的代码:
and at the end we come to simple code:
def ancient_fmt(ts, fmt):
fmt = fmt.replace('%%', '%')
fmt = re.sub('(%\w)', r'{\1}', fmt)
return fmt.format(**d).format(ts)
def main(ts, format):
if ts.year < 1900:
return ancient_format(ts, fmt)
else:
return ts.strftime(fmt)
其中 d
是一个全局字典,其中与 strftime
表。
where d
is a global dictionary with keys corresponding to some specifiers in strftime
table.
编辑2
要澄清:此方法仅适用于以下说明符:%Y,%m,%d,%H,% M,%S,%f
,即那些是数字的,如果您需要文本信息,最好使用babel或任何其他解决方案。
edit 2
To clarify: this approach will work only for the following specifiers: %Y, %m, %d, %H, %M, %S, %f
, i.e., those that are numeric, if you need textual information, you'd better off with babel or any other solution.
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