python：检查numpy数组是否包含另一个数组的任何元素

Question

检查numpy数组是否包含另一个数组的任何元素的最佳方法是什么？

例如：

array1 = [10,5,4,13,10,1,1,22,7,3,15,9] 
array2 = [3,4,9,10,13,15,16,18,19,20,21,22,23]` 

我希望得到一个True如果array1包含array2任何值，否则False。

Answer 1

使用熊猫，你可以使用isin：

a1 = np.array([10,5,4,13,10,1,1,22,7,3,15,9]) 
a2 = np.array([3,4,9,10,13,15,16,18,19,20,21,22,23]) 

>>> pd.Series(a1).isin(a2).any() 
True 

并采用in1d numpy的功能（每从@Norman评论）：

>>> np.any(np.in1d(a1, a2)) 
True 

对于小数组，如在这个例子中，使用集合的解决方案是明显的赢家。对于较大的不相似的阵列（即不重叠），熊猫和Numpy解决方案更快。但是，np.intersect1d似乎擅长更大的阵列。

小阵列（12-13元素）

%timeit set(array1) & set(array2) 
The slowest run took 4.22 times longer than the fastest. This could mean that an intermediate result is being cached 
1000000 loops, best of 3: 1.69 µs per loop 

%timeit any(i in a1 for i in a2) 
The slowest run took 12.29 times longer than the fastest. This could mean that an intermediate result is being cached 
100000 loops, best of 3: 1.88 µs per loop 

%timeit np.intersect1d(a1, a2) 
The slowest run took 10.29 times longer than the fastest. This could mean that an intermediate result is being cached 
100000 loops, best of 3: 15.6 µs per loop 

%timeit np.any(np.in1d(a1, a2)) 
10000 loops, best of 3: 27.1 µs per loop 

%timeit pd.Series(a1).isin(a2).any() 
10000 loops, best of 3: 135 µs per loop 

使用阵列由100K元件（无重叠）：

a3 = np.random.randint(0, 100000, 100000) 
a4 = a3 + 100000 

%timeit np.intersect1d(a3, a4) 
100 loops, best of 3: 13.8 ms per loop  

%timeit pd.Series(a3).isin(a4).any() 
100 loops, best of 3: 18.3 ms per loop 

%timeit np.any(np.in1d(a3, a4)) 
100 loops, best of 3: 18.4 ms per loop 

%timeit set(a3) & set(a4) 
10 loops, best of 3: 23.6 ms per loop 

%timeit any(i in a3 for i in a4) 
1 loops, best of 3: 34.5 s per loop 

Answer 2

你可以试试这个

>>> array1 = [10,5,4,13,10,1,1,22,7,3,15,9] 
>>> array2 = [3,4,9,10,13,15,16,18,19,20,21,22,23] 
>>> set(array1) & set(array2) 
set([3, 4, 9, 10, 13, 15, 22]) 

如果你得到的结果意味着有两个数组中的共同要素。

如果结果为空意味着没有公共元素。

Answer 3

可以使用any内置函数和列表理解：

>>> array1 = [10,5,4,13,10,1,1,22,7,3,15,9] 
>>> array2 = [3,4,9,10,13,15,16,18,19,20,21,22,23] 
>>> any(i in array2 for i in array1) 
True