Question

假设：

a = [[0, 1], [2, 2], [4, 2]]
b = [[0, 1, 2, 3], [2, 2, 3, 4], [4, 2, 3, 3], [2, 3, 3, 3]]

解决方案是：

for (i,j) in zip(a[:, 0], a[:, 1]):
                print b[np.logical_and( a[:, 0] == i,  a[:, 1] == j)]

结果应为

[[0 1 2 3]]
[[2 2 3 4]]
[[4 2 3 3]]

在不使用for - 循环的情况下，是否有解决此问题的方法？

Answer 1

您可以使用NumPy broadcasting作为矢量化解决方案，如此 -

# 2D mask corresponding to all iterations of : 
# "np.logical_and( a[:, 0] == i,  a[:, 1] == j)"
mask = (a[:,None,0] == a[:,0]) & (a[:,None,1] == a[:,1])

# Use column indices of valid ones for indexing into b for final output
_,C_idx = np.where(mask)
out = b[C_idx]

示例运行 -

In [67]: # Modified generic case
    ...: a = np.array([[0, 1], [3, 2], [3, 2]])
    ...: b = np.array([[0, 1, 2, 3], [2, 2, 3, 4], [4, 2, 3, 3], [2, 3, 3, 3]])
    ...: 
    ...: for (i,j) in zip(a[:, 0], a[:, 1]):
    ...:     print b[np.logical_and( a[:, 0] == i,  a[:, 1] == j)]
    ...:     
[[0 1 2 3]]
[[2 2 3 4]
 [4 2 3 3]]
[[2 2 3 4]
 [4 2 3 3]]

In [68]: mask = (a[:,None,0] == a[:,0]) & (a[:,None,1] == a[:,1])
    ...: _,C_idx = np.where(mask)
    ...: out = b[C_idx]
    ...: 

In [69]: out
Out[69]: 
array([[0, 1, 2, 3],
       [2, 2, 3, 4],
       [4, 2, 3, 3],
       [2, 2, 3, 4],
       [4, 2, 3, 3]])

Answer 2

假设：

a = np.array(([0, 1], [2, 2], [4, 2]))
b = np.array(([0, 1, 2, 3], [2, 2, 3, 4], [4, 2, 3, 3], [2, 3, 3, 3]))

计算：

temp = np.in1d(b[:,0], a[:,0]) * np.in1d(b[:,1], a[:,1])
result = b[temp]
print 'result:', result

输出：

result: [[0 1 2 3]
 [2 2 3 4]
 [4 2 3 3]]

过滤numpy数组由另一个不同形状的数组

2 个答案: