numpy.outer

y = numpy.outer（a,b）:

其中：a,b是两个矢量，函数的作用是: $y = a^{T} * b$ ,如果 $a\epsilon R^{M}, b\epsilon R^{N}$ ,那么 $y\epsilon R^{M*N}$

numpy.outer(a, b, out=None)[source]

Compute the outer product of two vectors.

Given two vectors, a = [a0, a1, ..., aM] and b = [b0, b1, ..., bN], the outer product [R60] is:

[[a0*b0  a0*b1 ... a0*bN ]
 [a1*b0    .
 [ ...          .
 [aM*b0            aM*bN ]]

Parameters:

Parameters:	a : (M,) array_like First input vector. Input is flattened if not already 1-dimensional. b : (N,) array_like Second input vector. Input is flattened if not already 1-dimensional. out : (M, N) ndarray, optional A location where the result is stored New in version 1.9.0.
Returns:	out : (M, N) ndarray `out[i, j] = a[i] * b[j]`

a : (M,) array_like

First input vector. Input is flattened if not already 1-dimensional.

b : (N,) array_like

Second input vector. Input is flattened if not already 1-dimensional.

out : (M, N) ndarray, optional

A location where the result is stored

New in version 1.9.0.

Returns:

out : (M, N) ndarray

out[i, j] = a[i] * b[j]

示例1：

# -- coding: UTF-8 --
import numpy as np
k = np.float32([1,4,6,4,1])
k = np.outer(k,k)
print(k)

输出：

示例2：

# -- coding: UTF-8 --
import numpy as np
x = np.array(['a', 'b', 'c'], dtype=object)
y = np.outer(x, [1, 2, 3])

print(y)

输出：