我该如何将ISO 8601持续时间字符串转换为datetime.timedelta？
我尝试使用持续时间字符串和格式字符串实例化timedelta，但是出现了异常：
>>> from datetime import timedelta
>>> timedelta("PT1H5M26S", "T%H%M%S")
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: unsupported type for timedelta seconds component: str
如果要进行反向操作，请参见如何在Python中将datetime.timedelta转换为ISO 8601持续时间？。
如果你正在使用Pandas，你可以使用pandas.Timedelta。该构造函数接受一个ISO 8601字符串，同时pandas.Timedelta.isoformat可以将实例格式化回字符串：
>>> import pandas as pd
>>> dt = pd.Timedelta("PT1H5M26S")
>>> dt
Timedelta('0 days 01:05:26')
>>> dt.isoformat()
'P0DT1H5M26S'
import datetime
def get_isosplit(s, split):
if split in s:
n, s = s.split(split)
else:
n = 0
return n, s
def parse_isoduration(s):
# Remove prefix
s = s.split('P')[-1]
# Step through letter dividers
days, s = get_isosplit(s, 'D')
_, s = get_isosplit(s, 'T')
hours, s = get_isosplit(s, 'H')
minutes, s = get_isosplit(s, 'M')
seconds, s = get_isosplit(s, 'S')
# Convert all to seconds
dt = datetime.timedelta(days=int(days), hours=int(hours), minutes=int(minutes), seconds=int(seconds))
return int(dt.total_seconds())
> parse_isoduration("PT1H5M26S")
3926
from datetime import timedelta
def parse_isoduration(isostring, as_dict=False):
"""
Parse the ISO8601 duration string as hours, minutes, seconds
"""
separators = {
"PT": None,
"W": "weeks",
"D": "days",
"H": "hours",
"M": "minutes",
"S": "seconds",
}
duration_vals = {}
for sep, unit in separators.items():
partitioned = isostring.partition(sep)
if partitioned[1] == sep:
# Matched this unit
isostring = partitioned[2]
if sep == "PT":
continue # Successful prefix match
dur_str = partitioned[0]
dur_val = float(dur_str) if "." in dur_str else int(dur_str)
duration_vals.update({unit: dur_val})
else:
if sep == "PT":
raise ValueError("Missing PT prefix")
else:
# No match for this unit: it's absent
duration_vals.update({unit: 0})
if as_dict:
return duration_vals
else:
return tuple(duration_vals.values())
dur_isostr = "PT3H2M59.989333S"
dur_tuple = parse_isoduration(dur_isostr)
dur_dict = parse_isoduration(dur_isostr, as_dict=True)
td = timedelta(**dur_dict)
s = td.total_seconds()
⇣
>>> dur_tuple
(0, 0, 3, 2, 59.989333)
>>> dur_dict
{'weeks': 0, 'days': 0, 'hours': 3, 'minutes': 2, 'seconds': 59.989333}
>>> td
datetime.timedelta(seconds=10979, microseconds=989333)
>>> s
10979.989333
def parse_isoduration(s):
""" Parse a str ISO-8601 Duration: https://en.wikipedia.org/wiki/ISO_8601#Durations
Originally copied from:
https://dev59.com/eVoU5IYBdhLWcg3w5p2a
:param s:
:return:
"""
# ToDo [40]: Can't handle legal ISO3106 ""PT1M""
def get_isosplit(s, split):
if split in s:
n, s = s.split(split, 1)
else:
n = '0'
return n.replace(',', '.'), s # to handle like "P0,5Y"
s = s.split('P', 1)[-1] # Remove prefix
s_yr, s = get_isosplit(s, 'Y') # Step through letter dividers
s_mo, s = get_isosplit(s, 'M')
s_dy, s = get_isosplit(s, 'D')
_, s = get_isosplit(s, 'T')
s_hr, s = get_isosplit(s, 'H')
s_mi, s = get_isosplit(s, 'M')
s_sc, s = get_isosplit(s, 'S')
n_yr = float(s_yr) * 365 # These are approximations that I can live with
n_mo = float(s_mo) * 30.4 # But they are not correct!
dt = datetime.timedelta(days=n_yr+n_mo+float(s_dy), hours=float(s_hr), minutes=float(s_mi), seconds=float(s_sc))
return dt # int(dt.total_seconds()) # original code wanted to return as seconds, we don't.
def parse_isoduration(str):
## https://dev59.com/eVoU5IYBdhLWcg3w5p2a
## Parse the ISO8601 duration as years,months,weeks,days, hours,minutes,seconds
## Returns: milliseconds
## Examples: "PT1H30M15.460S", "P5DT4M", "P2WT3H"
def get_isosplit(str, split):
if split in str:
n, str = str.split(split, 1)
else:
n = '0'
return n.replace(',', '.'), str # to handle like "P0,5Y"
str = str.split('P', 1)[-1] # Remove prefix
s_yr, str = get_isosplit(str, 'Y') # Step through letter dividers
s_mo, str = get_isosplit(str, 'M')
s_wk, str = get_isosplit(str, 'W')
s_dy, str = get_isosplit(str, 'D')
_, str = get_isosplit(str, 'T')
s_hr, str = get_isosplit(str, 'H')
s_mi, str = get_isosplit(str, 'M')
s_sc, str = get_isosplit(str, 'S')
n_yr = float(s_yr) * 365 # approx days for year, month, week
n_mo = float(s_mo) * 30.4
n_wk = float(s_wk) * 7
dt = datetime.timedelta(days=n_yr+n_mo+n_wk+float(s_dy), hours=float(s_hr), minutes=float(s_mi), seconds=float(s_sc))
return int(dt.total_seconds()*1000) ## int(dt.total_seconds()) | dt
import datetime
import re
def parse_iso8601_duration(duration: str) -> datetime.timedelta:
pattern = r"^P(?:(?P<days>\d+\.\d+|\d*?)D)?T?(?:(?P<hours>\d+\.\d+|\d*?)H)?(?:(?P<minutes>\d+\.\d+|\d*?)M)?(?:(?P<seconds>\d+\.\d+|\d*?)S)?$"
match = re.match(pattern, duration)
if not match:
raise ValueError(f"Invalid ISO 8601 duration: {duration}")
parts = {k: float(v) for k, v in match.groupdict("0").items()}
return datetime.timedelta(**parts)
datetime.timedelta对象。标准库没有解析时间差的功能，但是在PyPI上有一些包可以通过pip install安装。如果你想知道如何自己做，我认为这个问题太广泛了，不适合在SO上提问；你应该自己尝试一下，看看在哪里（如果有的话）卡住了无法继续。 - jonrsharpedatetime.timedelta是什么？它是一个包的一部分吗？你能举个具体的例子说明你想要的和你得到的有什么不同吗？ - SherylHohman