ndarray上的Numpy divide

Question

我想创建一个包含另一个数组的比率的新数组。你知道吗

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

退货

array([[2.  , 2.  , 0.  , 0.99],
       [2.  , 2.  , 0.  , 1.99],
       [0.  , 0.  , 0.  , 0.89]])

现在我要计算一个n数组，它包含周[：，3]的比率

ratio =  week[:,3].reshape(1,-1).T/ week[:,3]

退货

array([[1.   , 0.497, 1.112],
       [2.01 , 1.   , 2.236],
       [0.899, 0.447, 1.   ]])

正是我想要的。你知道吗

更一般的情况 前四个维度可以改变的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

退货

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]]]]])

现在我要为每个数组计算相同的比率

变换5darray返回奇怪的结果。你知道吗

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.   ]]]]])

Answer 1

我认为循环是你最大的希望，有一个缓慢和快速的方法：

慢方法：

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)

印刷品

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.        ]]]]])

显然，在python中循环数组的速度很慢，但这正是numba发明的目的：

快的方法

from numba import njit

@njit
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, :, 3]
                for d1 in range(last_dim):
                    for d2 in range(last_dim):
                        new_arr[i, j, k, d1, d2] = week[d1] / week[d2]

    return new_arr

get_ratios(weeks_5d)

印刷品

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.        ]]]]])