Confirm that the result is correct in 7 ways
python
import numpy as np
from more_itertools import ilen
from itertools import takewhile
def find_index(a, ths):
for i,v in enumerate(a):
if v >= th:
break
return i
n = 10000000
x = np.arange(n)
th = n / 10
print((x < th).argmin())
print(np.searchsorted(x, th))
print(len(x[x < th]))
print((x < th).sum())
print(np.where(x>=th)[0][0])
print(find_index(x, th))
print(ilen(takewhile(lambda i: i < th, x)))
>>>
1000000
1000000
1000000
1000000
1000000
1000000
1000000
python
%timeit np.searchsorted(x, th)
%timeit (x < th).argmin()
%timeit len(x[x < th])
%timeit (x < th).sum()
%timeit np.where(x >= th)[0][0]
%timeit find_index(x, th)
%timeit ilen(takewhile(lambda i: i < th, x))
>>>
3.84 µs ± 347 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
10.2 ms ± 215 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
12.5 ms ± 62.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
21.2 ms ± 411 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
43.1 ms ± 635 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
3.36 µs ± 19.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
5.74 µs ± 75.5 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
python
th = 10
%timeit np.searchsorted(x, th)
%timeit (x < th).argmin()
%timeit len(x[x < th])
%timeit (x < th).sum()
%timeit np.where(x >= th)[0][0]
%timeit find_index(x, th)
%timeit ilen(takewhile(lambda i: i < th, x))
>>>
3.31 µs ± 24.8 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
9.86 ms ± 28.5 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
13.1 ms ± 530 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
21.7 ms ± 761 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
46.9 ms ± 1.8 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
3.76 µs ± 310 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
6.01 µs ± 231 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
--search sorted is good --If it's an argmin and a for loop, it doesn't have to be sorted
that's all
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