Python 如何使用奇特的索引创建Numpy数组
如何使用numpy的奇特索引创建此,我希望获得最快的性能:Python 如何使用奇特的索引创建Numpy数组,python,arrays,numpy,scipy,Python,Arrays,Numpy,Scipy,如何使用numpy的奇特索引创建此,我希望获得最快的性能: array([[ 1, 2, 3, 4, 16, 31], [ 2, 3, 4, 5, 17, 32], [ 3, 4, 5, 6, 18, 33], [ 4, 5, 6, 7, 19, 34], [ 5, 6, 7, 8, 20, 35], [ 6, 7, 8, 9, 21, 36], [ 7, 8, 9, 10
array([[ 1, 2, 3, 4, 16, 31],
[ 2, 3, 4, 5, 17, 32],
[ 3, 4, 5, 6, 18, 33],
[ 4, 5, 6, 7, 19, 34],
[ 5, 6, 7, 8, 20, 35],
[ 6, 7, 8, 9, 21, 36],
[ 7, 8, 9, 10, 22, 37],
[ 8, 9, 10, 11, 23, 38],
[ 9, 10, 11, 12, 24, 39],
[10, 11, 12, 13, 25, 40]]
从这一点开始:
a = np.arange(0,10)
aa = np.arange(0,50)
y = 1
AA = [(aa[np.array([x+y, 1+x+y, 2+x+y, 3+x+y, 15+x+y, 30+x+y])]) for x, i in enumerate(a)]
我明白了
[array([ 2, 3, 4, 5, 17, 32]),
array([ 3, 4, 5, 6, 18, 33]),
array([ 4, 5, 6, 7, 19, 34]),
array([ 5, 6, 7, 8, 20, 35]),
array([ 6, 7, 8, 9, 21, 36]),
array([ 7, 8, 9, 10, 22, 37]),
array([ 8, 9, 10, 11, 23, 38]),
array([ 9, 10, 11, 12, 24, 39]),
array([10, 11, 12, 13, 25, 40]),
array([11, 12, 13, 14, 26, 41])]
我们可以在这里利用广播:
>>> np.arange(0,10).reshape(-1,1) + np.array([*range(1,5),16,31])
array([[ 1, 2, 3, 4, 16, 31],
[ 2, 3, 4, 5, 17, 32],
[ 3, 4, 5, 6, 18, 33],
[ 4, 5, 6, 7, 19, 34],
[ 5, 6, 7, 8, 20, 35],
[ 6, 7, 8, 9, 21, 36],
[ 7, 8, 9, 10, 22, 37],
[ 8, 9, 10, 11, 23, 38],
[ 9, 10, 11, 12, 24, 39],
[10, 11, 12, 13, 25, 40]])
在这里,我们创建一个范围从0到(不包括)10的10×1矩阵,并使用数据创建一个1×6矩阵[1,2,3,4,16,31]
如果希望y
作为“偏移量”,可以将其写成:
>>> y = 1
>>> np.arange(y,y+10).reshape(-1,1) + np.array([*range(0,4),15,30])
array([[ 1, 2, 3, 4, 16, 31],
[ 2, 3, 4, 5, 17, 32],
[ 3, 4, 5, 6, 18, 33],
[ 4, 5, 6, 7, 19, 34],
[ 5, 6, 7, 8, 20, 35],
[ 6, 7, 8, 9, 21, 36],
[ 7, 8, 9, 10, 22, 37],
[ 8, 9, 10, 11, 23, 38],
[ 9, 10, 11, 12, 24, 39],
[10, 11, 12, 13, 25, 40]])
利用给定的变量
a
、aa
和y
,这里有一个用于外部加法的变量-
offset = np.array([0,1,2,3,15,30])
out = aa[a[:,None] + offset + y]
使用-
越界情况 对于越界情况(aa小于要求),我们可以用
aa
填充零,然后索引到其中-
offset = np.array([0,1,2,3,15,30])
idx = np.add.outer(a , offset + y)
aa_p = np.pad(aa,(0,idx.max()-len(a)+1), 'constant')
out = aa_p[idx]
或者初始化输出数组,然后创建一个用于分配的有效位置的掩码-
offset = np.array([0,1,2,3,15,30])
idx = np.add.outer(a , offset + y)
mask = idx < len(aa)
out = np.zeros(idx.shape, dtype=aa.dtype)
out[mask] = aa[idx[mask]]
使用
broadcasting
:np.array([1,2,3,4,16,31])+np.arange(10)[:,None]
。我需要y,aa变量——这是一个玩具示例。。我实际上是在现实世界中切“aa”。我需要填充“aa”吗?若偏移量的最后一个元素大于“aa”的长度,这会爆炸吗?还是会缠绕。
offset = np.array([0,1,2,3,15,30])
idx = np.add.outer(a , offset + y)
mask = idx < len(aa)
out = np.zeros(idx.shape, dtype=aa.dtype)
out[mask] = aa[idx[mask]]
In [234]: a = np.arange(0,10)
...: aa = np.arange(4,42)
...: y = 6
...:
In [235]: out
Out[235]:
array([[10, 11, 12, 13, 25, 40],
[11, 12, 13, 14, 26, 41],
[12, 13, 14, 15, 27, 0],
[13, 14, 15, 16, 28, 0],
[14, 15, 16, 17, 29, 0],
[15, 16, 17, 18, 30, 0],
[16, 17, 18, 19, 31, 0],
[17, 18, 19, 20, 32, 0],
[18, 19, 20, 21, 33, 0],
[19, 20, 21, 22, 34, 0]])