我想創建一個包含另一個 ndarray 比率的新數組。第一個簡單的例子：import numpy as npweek = np.full((3, 4), 2, dtype=float)week[:,2] = 0week[2,0:2] =0week[0,3] =0.99week[1,3] =1.99week[2,3] =0.89week回報array([[2. , 2. , 0. , 0.99], [2. , 2. , 0. , 1.99], [0. , 0. , 0. , 0.89]])現在我想計算一個包含周 [:,3] 比率的 ndarrayratio = week[:,3].reshape(1,-1).T/ week[:,3]回報array([[1. , 0.497, 1.112], [2.01 , 1. , 2.236], [0.899, 0.447, 1. ]])正是我想要的。更一般的情況 一個 5d 數組，其中前 4 個維度可以改變weeks_5d= np.full((1,1,2, 3, 4), 2, dtype=float)weeks_5d[:,:,:,:,2] = 0weeks_5d[:,:,0,2,0:2] =0weeks_5d[:,:,1,1,0:2] =0weeks_5d[:,:,:,0,3] = 0.99weeks_5d[:,:,:,1,3] = 1.99weeks_5d[:,:,:,2,3] = 0.89weeks_5d回報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]]]]])現在我想為每個 ndarray 計算相同的比率轉置 5darray 會返回奇怪的結果。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. ]]]]])
1 回答
FFIVE
TA貢獻1797條經驗 獲得超6個贊
我認為循環是你最好的希望，有一個緩慢和快速的方法來做到這一點：
緩慢的方式：
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
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. ]]]]])
顯然，在 python 中循環數組很慢，但這numba就是發明的目的：
快速（呃）方式
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
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. ]]]]])
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