numpy如何使用数组对数组进行切片索引？

Perhaps this has been raised and addressed somewhere else but I haven't found it. Suppose we have a numpy array:

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 and end both have valid values so that each slicing is also valid for a. I wanted to copy value of subarrays in a to corresponding spots in in b:

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

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

this syntax doesn't work, but it's equivalent to:

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]]

I wonder if there is a better way of doing this instead of an explicit for-loop over the start and end arrays.

我想知道是否有更好的方法来做这个而不是在开始和结束数组上的显式for循环。

Thanks!

1 个解决方案

#1

We can use broadcasting to create a mask of places to be edited with two sets of comparisons against start and end arrays and then simply assign with boolean-indexing for a vectorized solution -

我们可以使用广播创建一个要编辑的场所的掩码,使用两组比较开始和结束数组,然后简单地为一个矢量化解决方案分配布尔索引 -

# 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]

Sample run -

样品运行 -

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]])