Question

for循环中是否存在用于以下操作的简短numpy命令？

import numpy as np


a= np.array([1.0,2.0,3.0,4.0,5.0,6.0])
b= np.array([10.0,20.0,30.0])
c= np.array([100.0,200.0,300.0,900.0])
y=np.linspace(0,2,50)
m=np.array([0.2,0.1,0.3])

A,C,B,Y = np.meshgrid(a,c,b,y,indexing="ij")

print Y

for i in range(0,len(a)):
  for j in range(0,len(c)):
    for k in range(0,len(b)):
      Y[i][j][k]=Y[i][j][k]*m[k]


print "--------"
print Y

摘要来说，我有$ Y_ {ijkl} $，并且我想将$ Y_ {ij0l} $与$ m_0 $相乘，并将$ Y_ {ij1l} $与$ m_1 $相乘，依此类推...

非常感谢！

Answer 1

要删除循环，只需在这里einsum。

np.einsum('ijkl,k->ijkl', Y, m)

或者只是广播乘法：

Y * m[:, None]

但是，如果您不想首先创建网状网格，则可以首先广播Y，以提高存储效率。

np.einsum(
    "ijkl,k->ijkl",
    np.broadcast_to(y, a.shape + c.shape + b.shape + y.shape),
    m,
)

或：

np.broadcast_to(y, a.shape + c.shape + b.shape + y.shape) * m[:, None]

如果您还需要A，C，B，则可以继续使用当前的方法。

性能

In [44]: %%timeit
    ...: np.einsum(
    ...:     "ijkl,k->ijkl",
    ...:     np.broadcast_to(y, (a.shape[0], c.shape[0], b.shape[0], y.shape[0])),
    ...:     m,
    ...: )
    ...:
21.1 µs ± 121 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [45]: %%timeit
    ...: A,C,B,Y = np.meshgrid(a,c,b,y,indexing="ij")
    ...: for i in range(0,len(a)):
    ...:   for j in range(0,len(c)):
    ...:     for k in range(0,len(b)):
    ...:       Y[i][j][k]=Y[i][j][k]*m[k]
    ...:
420 µs ± 1.58 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

将数组索引与数字相乘

1 个答案: