对于大数组（n>1e8），有没有比np.isin检查是否有相同元素更快的方法？
我尝试了几种方法，例如 pandas isin、 cython ，但所有这些方法都比np.isin
示例：（测试一维数组的每个元素是否也存在于第二个数组中）
num = int(1e8)
a = np.random.rand(int(num))
b = np.random.rand(int(num))
ref=time.time()
ainb = np.isin(a,b)
print(a[ainb])
print(time.time()-ref,'sec')
>>> [0.23591019 0.46102523]
>>> 65.45570135116577 sec
如果您需要一个插件（用于您的用例）但可能更快地替换np.isin()，您可以使用 Pythonset()进行检查并加速 Numba 中的显式循环：
import numpy as np
import numba as nb
@nb.jit
def is_in_set_nb(a, b):
shape = a.shape
a = a.ravel()
n = len(a)
result = np.full(n, False)
set_b = set(b)
for i in range(n):
if a[i] in set_b:
result[i] = True
return result.reshape(shape)
请注意，有一些（便宜的）额外代码可以使其适用于 N-dim 数组，如果您只需要 1D，您可能会忽略这些代码。
这甚至可以通过添加进一步的并行化来更快：
import numpy as np
import numba as nb
@nb.jit(parallel=True)
def is_in_set_pnb(a, b):
shape = a.shape
a = a.ravel()
n = len(a)
result = np.full(n, False)
set_b = set(b)
for i in nb.prange(n):
if a[i] in set_b:
result[i] = True
return result.reshape(shape)
这比没有 Numba 加速的np.isin(),set()交叉点和解决方案快得多：is_in_set()
def is_in_set(a, b):
set_b = set(b)
return np.array([x in set_b for x in a])
输入大小为一千万个元素：
n = 10 ** 7
k = n // 3
np.random.seed(0)
# note: I used `int`s because I wanted to be able to control the collisions
a = np.random.randint(0, k * n, n)
b = np.random.randint(0, k * n, n)
%timeit ainb = np.isin(a, b); a[ainb]
# 1 loop, best of 3: 3.94 s per loop
%timeit ainb = is_in_set_nb(a, b); a[ainb]
# 1 loop, best of 3: 814 ms per loop
%timeit ainb = is_in_set_pnb(a, b); a[ainb]
# 1 loop, best of 3: 740 ms per loop
%timeit ainb = is_in_set(a, b); a[ainb]
# 1 loop, best of 3: 7.69 s per loop
%timeit set(a).intersection(b) # not a drop-in replacement
# 1 loop, best of 3: 6.79 s per loop
%timeit set(a) & set(b) # not a drop-in replacement
# 1 loop, best of 3: 8.98 s per loop
并且有数亿个元素（最后两种方法最终填满了所有内存，因此被省略）：
n = 10 ** 8
k = n // 3
np.random.seed(0)
a = np.random.randint(0, k * n, n)
b = np.random.randint(0, k * n, n)
%timeit ainb = np.isin(a, b); a[ainb]
# 1 loop, best of 3: 1min 4s per loop
%timeit ainb = is_in_set_nb(a, b); a[ainb]
# 1 loop, best of 3: 13.1 s per loop
%timeit ainb = is_in_set_pnb(a, b); a[ainb]
# 1 loop, best of 3: 11.4 s per loop
%timeit ainb = is_in_set(a, b); a[ainb]
# 1 loop, best of 3: 2min 5s per loop
a为较小的输入添加更多时间，但和的所有长度组合b：
funcs = np.isin, is_in_set_nb, is_in_set_pnb
sep = ' '
print(f'({"n=len(a)":>9s},{"m=len(b)":>9s})', end=sep)
for func in funcs:
print(f'{func.__name__:15s}', end=sep)
print()
I, J = 7, 7
for i in range(I):
for j in range(J):
n = 10 ** i
m = 10 ** j
a = np.random.randint(0, m * n, n)
b = np.random.randint(0, m * n, m)
print(f'({n:9d},{m:9d})', end=sep)
for func in funcs:
result = %timeit -q -o func(a, b)
print(f'{result.best * 1e3:12.3f} ms', end=sep)
print()
( n=len(a), m=len(b)) isin is_in_set_nb is_in_set_pnb
( 1, 1) 0.011 ms 0.001 ms 0.047 ms
( 1, 10) 0.048 ms 0.001 ms 0.023 ms
( 1, 100) 0.050 ms 0.002 ms 0.027 ms
( 1, 1000) 0.102 ms 0.007 ms 0.041 ms
( 1, 10000) 0.766 ms 1.028 ms 1.122 ms
( 1, 100000) 9.717 ms 3.426 ms 3.356 ms
( 1, 1000000) 105.154 ms 43.642 ms 40.734 ms
( 10, 1) 0.010 ms 0.001 ms 0.023 ms
( 10, 10) 0.030 ms 0.001 ms 0.023 ms
( 10, 100) 0.053 ms 0.002 ms 0.027 ms
( 10, 1000) 0.100 ms 0.007 ms 0.055 ms
( 10, 10000) 0.961 ms 1.031 ms 1.154 ms
( 10, 100000) 9.772 ms 3.595 ms 3.761 ms
( 10, 1000000) 105.802 ms 54.260 ms 50.265 ms
( 100, 1) 0.010 ms 0.001 ms 0.024 ms
( 100, 10) 0.030 ms 0.002 ms 0.025 ms
( 100, 100) 0.054 ms 0.002 ms 0.026 ms
( 100, 1000) 0.105 ms 0.008 ms 0.045 ms
( 100, 10000) 0.751 ms 1.076 ms 1.158 ms
( 100, 100000) 9.824 ms 3.253 ms 3.329 ms
( 100, 1000000) 105.697 ms 57.993 ms 55.285 ms
( 1000, 1) 0.012 ms 0.005 ms 0.028 ms
( 1000, 10) 0.038 ms 0.006 ms 0.029 ms
( 1000, 100) 0.119 ms 0.007 ms 0.033 ms
( 1000, 1000) 0.180 ms 0.014 ms 0.063 ms
( 1000, 10000) 0.821 ms 1.074 ms 1.169 ms
( 1000, 100000) 9.920 ms 3.392 ms 3.532 ms
( 1000, 1000000) 104.666 ms 57.845 ms 54.603 ms
( 10000, 1) 0.020 ms 0.041 ms 0.092 ms
( 10000, 10) 0.089 ms 0.088 ms 0.158 ms
( 10000, 100) 0.967 ms 0.112 ms 0.182 ms
( 10000, 1000) 1.017 ms 0.161 ms 0.249 ms
( 10000, 10000) 1.633 ms 1.137 ms 1.283 ms
( 10000, 100000) 10.754 ms 3.027 ms 3.302 ms
( 10000, 1000000) 101.926 ms 48.062 ms 49.117 ms
( 100000, 1) 0.071 ms 0.409 ms 0.455 ms
( 100000, 10) 0.575 ms 0.916 ms 0.803 ms
( 100000, 100) 16.304 ms 1.201 ms 0.940 ms
( 100000, 1000) 15.185 ms 1.566 ms 1.181 ms
( 100000, 10000) 15.914 ms 1.454 ms 1.252 ms
( 100000, 100000) 23.719 ms 4.820 ms 4.313 ms
( 100000, 1000000) 119.668 ms 56.863 ms 54.570 ms
( 1000000, 1) 0.774 ms 4.347 ms 3.407 ms
( 1000000, 10) 6.207 ms 8.793 ms 5.957 ms
( 1000000, 100) 178.498 ms 13.104 ms 8.544 ms
( 1000000, 1000) 169.022 ms 16.198 ms 10.283 ms
( 1000000, 10000) 177.986 ms 13.243 ms 8.973 ms
( 1000000, 100000) 177.989 ms 19.856 ms 13.898 ms
( 1000000, 1000000) 283.207 ms 97.118 ms 84.332 ms
这表明 Numba 和并行化对于较大的输入非常有利，而对于较小的输入，效率会略微降低。np.isin()但是，它们在上述大多数测试中仍然表现出色。
我不确定这是否适合您的需求，但集合 ( &) 的交集看起来更快。缺点是你只会得到元素，而不是它们的索引。
import numpy as np
import time
num = int(1e7)
a = np.random.rand(int(num))
b = np.random.rand(int(num))
a_set = set(a)
b_set = set(b)
时间设置交点：
ref=time.time()
print(a_set & b_set)
print(time.time()-ref,'sec')
>> set()
0.6486170291900635 sec
时间np.isin()：
ref=time.time()
ainb = np.isin(a, b)
print( a[ainb] )
print(time.time()-ref,'sec')
>> []
4.06556510925293 sec