Python NumPy Sum（带轴）如何工作？
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How does NumPy Sum (with axis) work?
提问by James
I've taken it upon myself to learn how NumPyworks for my own curiosity.
我已经开始学习如何NumPy满足我自己的好奇心。
It seems that the simplest function is the hardest to translate to code (I understand by code). It's easy to hard code each axis for each case but I want to find a dynamic algorithm that can sum in any axis with n-dimensions. The documentation on the official website is not helpful (It only shows the result not the process) and it's hard to navigate through Python/C code.
似乎最简单的函数最难转化为代码（我是通过代码理解的）。为每种情况对每个轴进行硬编码很容易，但我想找到一种动态算法，该算法可以在具有 n 维的任何轴上求和。官方网站上的文档没有帮助（它只显示结果而不显示过程）并且很难浏览 Python/C 代码。
Note:I did figure out that when an array is summed, the axis specified is "removed", i.e. Sum of an array with a shape of (4, 3, 2) with axis 1 yields an answer of an array with a shape of (4, 2)
注意：我确实发现当对数组求和时，指定的轴被“删除”，即形状为 (4, 3, 2) 的数组与轴 1 的总和产生形状为(4, 2)
回答by piRSquared
Setup
设置
consider the numpy array a
考虑 numpy 数组 a
a = np.arange(30).reshape(2, 3, 5)
print(a)
[[[ 0 1 2 3 4]
[ 5 6 7 8 9]
[10 11 12 13 14]]
[[15 16 17 18 19]
[20 21 22 23 24]
[25 26 27 28 29]]]
Where are the dimensions?
维度在哪里？
The dimensions and positions are highlighted by the following
尺寸和位置由以下突出显示
p p p p p
o o o o o
s s s s s
dim 2 0 1 2 3 4
| | | | |
dim 0 ↓ ↓ ↓ ↓ ↓
----> [[[ 0 1 2 3 4] <---- dim 1, pos 0
pos 0 [ 5 6 7 8 9] <---- dim 1, pos 1
[10 11 12 13 14]] <---- dim 1, pos 2
dim 0
----> [[15 16 17 18 19] <---- dim 1, pos 0
pos 1 [20 21 22 23 24] <---- dim 1, pos 1
[25 26 27 28 29]]] <---- dim 1, pos 2
↑ ↑ ↑ ↑ ↑
| | | | |
dim 2 p p p p p
o o o o o
s s s s s
0 1 2 3 4
Dimension examples:
维度示例：
This becomes more clear with a few examples
举几个例子就更清楚了
a[0, :, :] # dim 0, pos 0
[[ 0 1 2 3 4]
[ 5 6 7 8 9]
[10 11 12 13 14]]
a[:, 1, :] # dim 1, pos 1
[[ 5 6 7 8 9]
[20 21 22 23 24]]
a[:, :, 3] # dim 2, pos 3
[[ 3 8 13]
[18 23 28]]
sum
sum
explanation of sumand axisa.sum(0)is the sum of all slices along dim 0
的解释sum和axisa.sum(0)是所有切片的总和dim 0
a.sum(0)
[[15 17 19 21 23]
[25 27 29 31 33]
[35 37 39 41 43]]
same as
与...一样
a[0, :, :] + \
a[1, :, :]
[[15 17 19 21 23]
[25 27 29 31 33]
[35 37 39 41 43]]
a.sum(1)is the sum of all slices along dim 1
a.sum(1)是所有切片的总和 dim 1
a.sum(1)
[[15 18 21 24 27]
[60 63 66 69 72]]
same as
与...一样
a[:, 0, :] + \
a[:, 1, :] + \
a[:, 2, :]
[[15 18 21 24 27]
[60 63 66 69 72]]
a.sum(2)is the sum of all slices along dim 2
a.sum(2)是所有切片的总和 dim 2
a.sum(2)
[[ 10 35 60]
[ 85 110 135]]
same as
与...一样
a[:, :, 0] + \
a[:, :, 1] + \
a[:, :, 2] + \
a[:, :, 3] + \
a[:, :, 4]
[[ 10 35 60]
[ 85 110 135]]
default axis is -1
this means all axes. or sum all numbers.
默认轴是-1
这意味着所有轴。或对所有数字求和。
a.sum()
435
回答by tinyhare
I use a nested loop operation to explain it.
我使用嵌套循环操作来解释它。
import numpy as np
n = np.array(
[[[1, 2, 3],
[4, 5, 6],
[7, 8, 9]],
[[2, 4, 6],
[8, 10, 12],
[14, 16, 18]],
[[1, 3, 5],
[7, 9, 11],
[13, 15, 17]]])
print(n)
print("============ sum axis=None=============")
sum = 0
for i in range(3):
for j in range(3):
for k in range(3):
sum += n[k][i][j]
print(sum) # 216
print('------------------')
print(np.sum(n)) # 216
print("============ sum axis=0 =============")
for i in range(3):
for j in range(3):
sum = 0
for axis in range(3):
sum += n[axis][i][j]
print(sum,end=' ')
print()
print('------------------')
print("sum[0][0] = %d" % (n[0][0][0] + n[1][0][0] + n[2][0][0]))
print("sum[1][1] = %d" % (n[0][1][1] + n[1][1][1] + n[2][1][1]))
print("sum[2][2] = %d" % (n[0][2][2] + n[1][2][2] + n[2][2][2]))
print('------------------')
print(np.sum(n, axis=0))
print("============ sum axis=1 =============")
for i in range(3):
for j in range(3):
sum = 0
for axis in range(3):
sum += n[i][axis][j]
print(sum,end=' ')
print()
print('------------------')
print("sum[0][0] = %d" % (n[0][0][0] + n[0][1][0] + n[0][2][0]))
print("sum[1][1] = %d" % (n[1][0][1] + n[1][1][1] + n[1][2][1]))
print("sum[2][2] = %d" % (n[2][0][2] + n[2][1][2] + n[2][2][2]))
print('------------------')
print(np.sum(n, axis=1))
print("============ sum axis=2 =============")
for i in range(3):
for j in range(3):
sum = 0
for axis in range(3):
sum += n[i][j][axis]
print(sum,end=' ')
print()
print('------------------')
print("sum[0][0] = %d" % (n[0][0][0] + n[0][0][1] + n[0][0][2]))
print("sum[1][1] = %d" % (n[1][1][0] + n[1][1][1] + n[1][1][2]))
print("sum[2][2] = %d" % (n[2][2][0] + n[2][2][1] + n[2][2][2]))
print('------------------')
print(np.sum(n, axis=2))
print("============ sum axis=(0,1)) =============")
for i in range(3):
sum = 0
for axis1 in range(3):
for axis2 in range(3):
sum += n[axis1][axis2][i]
print(sum,end=' ')
print()
print('------------------')
print("sum[1] = %d" % (n[0][0][1] + n[0][1][1] + n[0][2][1] +
n[1][0][1] + n[1][1][1] + n[1][2][1] +
n[2][0][1] + n[2][1][1] + n[2][2][1] ))
print('------------------')
print(np.sum(n, axis=(0,1)))
result：
结果：
[[[ 1 2 3]
[ 4 5 6]
[ 7 8 9]]
[[ 2 4 6]
[ 8 10 12]
[14 16 18]]
[[ 1 3 5]
[ 7 9 11]
[13 15 17]]]
============ sum axis=None=============
216
------------------
216
============ sum axis=0 =============
4 9 14
19 24 29
34 39 44
------------------
sum[0][0] = 4
sum[1][1] = 24
sum[2][2] = 44
------------------
[[ 4 9 14]
[19 24 29]
[34 39 44]]
============ sum axis=1 =============
12 15 18
24 30 36
21 27 33
------------------
sum[0][0] = 12
sum[1][1] = 30
sum[2][2] = 33
------------------
[[12 15 18]
[24 30 36]
[21 27 33]]
============ sum axis=2 =============
6 15 24
12 30 48
9 27 45
------------------
sum[0][0] = 6
sum[1][1] = 30
sum[2][2] = 45
------------------
[[ 6 15 24]
[12 30 48]
[ 9 27 45]]
============ sum axis=(0,1)) =============
57 72 87
------------------
sum[1] = 72
------------------
[57 72 87]
