How to construct a matrix of all possible differences of a vector in numpy

Question

I have a one dimensional array, lets say:

import numpy as np
inp_vec = np.array([1, 2, 3])

Now, I would like to construct a matrix of the form

[[1 - 1, 1 - 2, 1 - 3],
 [2 - 1, 2 - 2, 2 - 3],
 [3 - 1, 3 - 2, 3 - 3]])

Of course it can be done with for loops but is there a more elegant way to do this?

Answer 1

这我也找到了一个很好的方法：

np.subtract.outer([1,2,3], [1,2,3])

Answer 2

This seems to work:

In [1]: %paste
import numpy as np
inp_vec = np.array([1, 2, 3])

## -- End pasted text --

In [2]: inp_vec.reshape(-1, 1) - inp_vec
Out[2]: 
array([[ 0, -1, -2],
       [ 1,  0, -1],
       [ 2,  1,  0]])

Explanation:

You first reshape the array to nx1 . When you subtract a 1D array, they are both broadcast to nxn :

array([[ 1,  1,  1],
       [ 2,  2,  2],
       [ 3,  3,  3]])

and

array([[ 1,  2,  3],
       [ 1,  2,  3],
       [ 1,  2,  3]])

Then the subtraction is done element-wise, which yields the desired result.

Answer 3

import numpy as np
inp_vec = np.array([1, 2, 3])

a, b = np.meshgrid(inp_vec, inp_vec)
print(b - a)

Output:

Array([[ 0 -1 -2],
       [ 1  0 -1],
       [ 2  1  0]])

Answer 4

Using np.nexaxis

import numpy as np
inp_vec = np.array([1, 2, 3])

output = inp_vec[:, np.newaxis] - inp_vec

Output

array([[ 0, -1, -2],
       [ 1,  0, -1],
       [ 2,  1,  0]])

Answer 5

This is a quick and simple alternative.

import numpy as np
inp_vec = np.array([1, 2, 3])

N = len(inp_vec)
np.reshape(inp_vec,(N,1)) - np.reshape(inp_vec,(1,N))