也许这已经在其他地方提出并解决了,但我还没有找到。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
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start并且end两者都具有有效值,因此每个切片也对 有效a。我想将子数组的值复制a到 in 中的相应点b:
b[:, start:end] = a[:, start:end] #error
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此语法不起作用,但等效于:
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]]
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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!
我们可以使用broadcasting两组比较来创建要编辑的位置掩码start和end数组,然后简单地boolean-indexing为矢量化解决方案分配 with -
# 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]
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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]])
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