A faster strptime?
Is factor 7 lot enough?
datetime.datetime.strptime(a, '%Y-%m-%d').date() # 8.87us
datetime.date(*map(int, a.split('-'))) # 1.28us
EDIT: great idea with explicit slicing:
datetime.date(int(a[:4]), int(a[5:7]), int(a[8:10])) # 1.06us
that makes factor 8.
Python 3.7+: fromisoformat()
Since Python 3.7, the datetime
class has a method fromisoformat
. It should be noted that this can also be applied to this question:
Performance vs. strptime()
Explicit string slicing may give you about a 9x increase in performance compared to normal strptime
, but you can get about a 90x increase with the built-in fromisoformat
method!
%timeit isofmt(datelist)
569 µs ± 8.45 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
%timeit slice2int(datelist)
5.51 ms ± 48.9 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit normalstrptime(datelist)
52.1 ms ± 1.27 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
from datetime import datetime, timedelta
base, n = datetime(2000, 1, 1, 1, 2, 3, 420001), 10000
datelist = [(base + timedelta(days=i)).strftime('%Y-%m-%d') for i in range(n)]
def isofmt(l):
return list(map(datetime.fromisoformat, l))
def slice2int(l):
def slicer(t):
return datetime(int(t[:4]), int(t[5:7]), int(t[8:10]))
return list(map(slicer, l))
def normalstrptime(l):
return [datetime.strptime(t, '%Y-%m-%d') for t in l]
print(isofmt(datelist[0:1]))
print(slice2int(datelist[0:1]))
print(normalstrptime(datelist[0:1]))
# [datetime.datetime(2000, 1, 1, 0, 0)]
# [datetime.datetime(2000, 1, 1, 0, 0)]
# [datetime.datetime(2000, 1, 1, 0, 0)]
Python 3.8.3rc1 x64 / Win10