我想创建一个包含另一个ndarray比率的新数组。
第一个简单示例：
import numpy as np
week = np.full((3, 4), 2, dtype=float)
week[:,2] = 0
week[2,0:2] =0
week[0,3] =0.99
week[1,3] =1.99
week[2,3] =0.89
week
返回
array([[2. , 2. , 0. , 0.99],
[2. , 2. , 0. , 1.99],
[0. , 0. , 0. , 0.89]])
现在我要计算一个包含周比率的ndarray [：，3]
ratio = week[:,3].reshape(1,-1).T/ week[:,3]
返回
array([[1. , 0.497, 1.112],
[2.01 , 1. , 2.236],
[0.899, 0.447, 1. ]])
正是我想要的。
更一般的情况 前4个维度可以更改的5d数组
weeks_5d= np.full((1,1,2, 3, 4), 2, dtype=float)
weeks_5d[:,:,:,:,2] = 0
weeks_5d[:,:,0,2,0:2] =0
weeks_5d[:,:,1,1,0:2] =0
weeks_5d[:,:,:,0,3] = 0.99
weeks_5d[:,:,:,1,3] = 1.99
weeks_5d[:,:,:,2,3] = 0.89
weeks_5d
返回
array([[[[[2. , 2. , 0. , 0.99],
[2. , 2. , 0. , 1.99],
[0. , 0. , 0. , 0.89]],
[[2. , 2. , 0. , 0.99],
[0. , 0. , 0. , 1.99],
[2. , 2. , 0. , 0.89]]]]])
现在我想为每个ndarray计算相同的比率
转置5darray会返回奇怪的结果。
我需要的是
array([[[[[1. , 0.497, 1.112],
[2.01 , 1. , 2.236],
[0.899, 0.447, 1. ]]],
[[1. , 0.497, 1.112],
[2.01 , 1. , 2.236],
[0.899, 0.447, 1. ]]]]])
答案 0 :(得分：1)
我认为循环是您的最大希望，并且有一种缓慢而快速的方法：
#define MY_MACRO(isTrue) ((isTrue) ? do() : (void)0)
打印
static conditional_do(bool isTrue) { if (isTrue) { do(); } }显然在python中遍历数组很慢，但这就是发明def get_ratios(arr):
ni, nj, nk = arr.shape[:3]
last_dim = arr.shape[3]
new_arr = np.zeros(shape=(ni, nj, nk, last_dim, last_dim),
dtype=np.float64)
for i in range(ni):
for j in range(nj):
for k in range(nk):
week = arr[i, j, k]
ratio = week[:, 3].reshape(-1, 1) / week[:, 3]
new_arr[i, j, k] = ratio
return new_arr
get_ratios(weeks_5d)
的目的：
array([[[[[1. , 0.49748744, 1.11235955],
[2.01010101, 1. , 2.23595506],
[0.8989899 , 0.44723618, 1. ]],
[[1. , 0.49748744, 1.11235955],
[2.01010101, 1. , 2.23595506],
[0.8989899 , 0.44723618, 1. ]]]]])
打印
numba