A faster strptime?

Solution 1:

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.

Solution 2:

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