我想創(chuàng)建一個(gè)包含另一個(gè) ndarray 比率的新數(shù)組。第一個(gè)簡(jiǎn)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(bào)array([[2. , 2. , 0. , 0.99], [2. , 2. , 0. , 1.99], [0. , 0. , 0. , 0.89]])現(xiàn)在我想計(jì)算一個(gè)包含周 [:,3] 比率的 ndarrayratio = week[:,3].reshape(1,-1).T/ week[:,3]回報(bào)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(gè) 5d 數(shù)組，其中前 4 個(gè)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(bào)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(xiàn)在我想為每個(gè) ndarray 計(jì)算相同的比率轉(zhuǎn)置 5darray 會(huì)返回奇怪的結(jié)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(xiàn)1797條經(jīng)驗(yàn) 獲得超6個(gè)贊
我認(rèn)為循環(huán)是你最好的希望，有一個(gè)緩慢和快速的方法來(lái)做到這一點(diǎn)：
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(huán)數(shù)組很慢，但這numba就是發(fā)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. ]]]]])
添加回答
舉報(bào)
0/150
提交
取消