How can I convert an ISO 8601 duration string to datetime.timedelta?
I tried just instantiating timedelta with the duration string and a format string, but I get an exception:
>>> 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
For the reverse, see Convert a datetime.timedelta into ISO 8601 duration in Python?.
I found isodate library to do exactly what I want
isodate.parse_duration('PT1H5M26S')
If you're using Pandas, you could use pandas.Timedelta. The constructor accepts an ISO 8601 string, and pandas.Timedelta.isoformat you can format the instance back to a string:
>>> import pandas as pd
>>> dt = pd.Timedelta("PT1H5M26S")
>>> dt
Timedelta('0 days 01:05:26')
>>> dt.isoformat()
'P0DT1H5M26S'
Here's a solution without a new package, but only works if you're dealing with a max duration expressed in days. That limitation makes sense though, because as others have pointed out (1):
Given that the timedelta has more than "a month's" worth of days, how would you describe it using the ISO8601 duration notation without referencing a specific point in time? Conversely, given your example, "P3Y6M4DT12H30M5S", how would you convert that into a timedelta without knowing which exact years and months this duration refers to? Timedelta objects are very precise beasts, which is almost certainly why they don't support "years" and "months" args in their constructors.
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
Great question, obviously the "right" solution depends on your expectations for the input (a more reliable data source doesn't need as much input validation).
My approach to parse an ISO8601 duration timestamp only checks that the "PT" prefix is present and will not assume integer values for any of the units:
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
Based on @r3robertson a more complete, yet not perfect, version
def parse_isoduration(s):
""" Parse a str ISO-8601 Duration: https://en.wikipedia.org/wiki/ISO_8601#Durations
Originally copied from:
https://stackoverflow.com/questions/36976138/is-there-an-easy-way-to-convert-iso-8601-duration-to-timedelta
: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.
This is my modification(Martin, rer answers) to support weeks attribute and return milliseconds. Some durations may use PT15.460S fractions.
def parse_isoduration(str):
## https://stackoverflow.com/questions/36976138/is-there-an-easy-way-to-convert-iso-8601-duration-to-timedelta
## 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
