Question

在indexing的科学演讲笔记中

创建此数组存在一个示例问题。

[[0., 0., 0., 0., 0.],
 [2., 0., 0., 0., 0.],
 [0., 3., 0., 0., 0.],
 [0., 0., 4., 0., 0.],
 [0., 0., 0., 5., 0.],
 [0., 0., 0., 0., 6.]]

对我来说，问题在于顶部多余的空白行。如何实现示例？

这是我当前的代码。

d =np.zeros([5,],dtype=int) + np.diag(arange(2,7,1))

d
Out[66]: 
array([[2, 0, 0, 0, 0],
       [0, 3, 0, 0, 0],
       [0, 0, 4, 0, 0],
       [0, 0, 0, 5, 0],
       [0, 0, 0, 0, 6]])

Answer 1

您可以使用diag进行此操作，并使用k的{{1}}参数来单独建立索引：

diag

给予：

np.diag(np.arange(2,7), k = -1)

那几乎是正确的。您只需要丢失可以用切片完成的最后一列：

array([[0, 0, 0, 0, 0, 0],
       [2, 0, 0, 0, 0, 0],
       [0, 3, 0, 0, 0, 0],
       [0, 0, 4, 0, 0, 0],
       [0, 0, 0, 5, 0, 0],
       [0, 0, 0, 0, 6, 0]])

给出所需的结果：

np.diag(np.arange(2,7), k = -1)[:, :-1]

Answer 2

使用np.append：

>>> zero_row = np.zeros((1,5))
>>> matrix   = np.diag(np.arange(2,7,1))
>>> np.append(zero_row, matrix, axis=0)
<<< array([[0., 0., 0., 0., 0.],
           [2., 0., 0., 0., 0.],
           [0., 3., 0., 0., 0.],
           [0., 0., 4., 0., 0.],
           [0., 0., 0., 5., 0.],
           [0., 0., 0., 0., 6.]])

Answer 3

这是不使用diag的情况下生成数组的一种方法。相反，我在zeros数组中索引了对角线的元素集：

In [167]: x = np.zeros((6,5))                                                     
In [168]: x[np.arange(1,6), np.arange(5)] = np.arange(2,7)                        
In [169]: x                                                                     
Out[169]: 
array([[0., 0., 0., 0., 0.],
       [2., 0., 0., 0., 0.],
       [0., 3., 0., 0., 0.],
       [0., 0., 4., 0., 0.],
       [0., 0., 0., 5., 0.],
       [0., 0., 0., 0., 6.]])

Answer 4

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我们可以使用import numpy as np m = np.zeros([5,]) n = np.diag(np.arange(2,7,1)) m = np.vstack((m,n)) print(m)

Answer 5

您也可以像这样使用reshape：

out = np.zeros((6, 5))                                                              
out.reshape(5, 6)[:, 5] = np.arange(2, 7)                                            
out                                                                                                         
# array([[0., 0., 0., 0., 0.],
#        [2., 0., 0., 0., 0.],
#        [0., 3., 0., 0., 0.],
#        [0., 0., 4., 0., 0.],
#        [0., 0., 0., 5., 0.],
#        [0., 0., 0., 0., 6.]])

或非常相似：

out = np.zeros((6, 5))
out.reshape(-1)[5::6] = np.arange(2, 7)
out
# array([[0., 0., 0., 0., 0.],
#        [2., 0., 0., 0., 0.],
#        [0., 3., 0., 0., 0.],
#        [0., 0., 4., 0., 0.],
#        [0., 0., 0., 5., 0.],
#        [0., 0., 0., 0., 6.]])

这两种方法都比迄今为止发布的所有方法都快：

import numpy as np
from timeit import timeit

def od_hpj():
    out = np.zeros((6, 5))
    out[np.arange(1,6), np.arange(5)] = np.arange(2,7)                        
    return out

def od_mm():
    return np.diag(np.arange(2,7), k = -1)[:, :-1]

def od_ks():
    m = np.zeros([5,])
    n = np.diag(np.arange(2,7,1))
    return np.vstack((m,n))

def od_as():
    zero_row = np.zeros((1,5))
    matrix   = np.diag(np.arange(2,7,1))
    return np.append(zero_row, matrix, axis=0)

def od_pp1():
    out = np.zeros((6, 5))
    out.reshape(5, 6)[:, 5] = np.arange(2, 7)
    return out

def od_pp2():
    out = np.zeros((6, 5))
    out.reshape(-1)[5::6] = np.arange(2, 7)
    return out

for n, o in list(globals().items()):
    if n.startswith("od_"):
        print(f"{n.replace('od_', ''):3s}: {timeit(o):.3f} us")

样品运行：

hpj: 3.379 us
mm : 2.952 us
ks : 7.804 us
as : 5.222 us
pp1: 1.735 us
pp2: 2.418 us

scipy讲义，创建数组示例

5 个答案: