Question

也许这已被提出并在其他地方解决，但我还没有找到它。假设我们有一个numpy数组：

a = np.arange(100).reshape(10,10)
b = np.zeros(a.shape)
start = np.array([1,4,7])   # can be arbitrary but valid values
end = np.array([3,6,9])     # can be arbitrary but valid values

start和end都有有效值，因此每个切片对a也有效。我想将a中的子数组值复制到b中的相应位置：

b[:, start:end] = a[:, start:end]   #error

这种语法不起作用，但它相当于：

b[:, start[0]:end[0]] = a[:, start[0]:end[0]]
b[:, start[1]:end[1]] = a[:, start[1]:end[1]]
b[:, start[2]:end[2]] = a[:, start[2]:end[2]]

我想知道是否有更好的方法来代替start和end数组上的显式for循环。

谢谢！

Answer 1

我们可以使用broadcasting创建要编辑的地点的掩码，使用针对start和end数组的两组比较，然后简单地使用boolean-indexing分配矢量化解决方案 -

# Range array for the length of columns
r = np.arange(b.shape[1])

# Broadcasting magic to give us the mask of places
mask = (start[:,None] <= r) & (end[:,None] >= r)

# Boolean-index to select and assign 
b[:len(mask)][mask] = a[:len(mask)][mask]

示例运行 -

In [222]: a = np.arange(50).reshape(5,10)
     ...: b = np.zeros(a.shape,dtype=int)
     ...: start = np.array([1,4,7])
     ...: end = np.array([5,6,9]) # different from sample for variety
     ...: 

# Mask of places to be edited
In [223]: mask = (start[:,None] <= r) & (end[:,None] >= r)

In [225]: print mask
[[False  True  True  True  True  True False False False False]
 [False False False False  True  True  True False False False]
 [False False False False False False False  True  True  True]]

In [226]: b[:len(mask)][mask] = a[:len(mask)][mask]

In [227]: a
Out[227]: 
array([[ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9],
       [10, 11, 12, 13, 14, 15, 16, 17, 18, 19],
       [20, 21, 22, 23, 24, 25, 26, 27, 28, 29],
       [30, 31, 32, 33, 34, 35, 36, 37, 38, 39],
       [40, 41, 42, 43, 44, 45, 46, 47, 48, 49]])

In [228]: b
Out[228]: 
array([[ 0,  1,  2,  3,  4,  5,  0,  0,  0,  0],
       [ 0,  0,  0,  0, 14, 15, 16,  0,  0,  0],
       [ 0,  0,  0,  0,  0,  0,  0, 27, 28, 29],
       [ 0,  0,  0,  0,  0,  0,  0,  0,  0,  0],
       [ 0,  0,  0,  0,  0,  0,  0,  0,  0,  0]])

numpy如何使用数组对数组进行切片索引？

1 个答案: