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
start
和 end
两者都有有效值,因此每个切片也对 a
有效.我想复制
a
中子数组的值到 b
中的相应位置:b[:, start:end] = a[:, start:end] #error
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]]
我想知道是否有更好的方法来代替
start
上的显式 for 循环。和 end
数组。谢谢!
最佳答案
我们可以使用 broadcasting
使用与 start
的两组比较来创建要编辑的地点掩码和 end
数组,然后简单地用 boolean-indexing
赋值对于矢量化解决方案 -
# 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]
sample 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]])
关于python - numpy 如何使用数组对数组进行切片索引?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46734116/