Python 在numpy中组合两个不同尺寸的阵列

Python 在numpy中组合两个不同尺寸的阵列,python,arrays,for-loop,Python,Arrays,For Loop,我希望合并两个阵列 a = np.array(["A", "B", "C"]) b = np.array(["1", "2", "3", "4", "5"]) 类似于以下内容的数组(矩阵): [["A", "B", "C", "1"] ["A", "B", "C", "2"] ["A", "B", "C", "3"] ["A", "B", "C", "4"] ["A", "B", "C", "5"]] 我尝试了一个for循环,但似乎不起作用。我是Python新手,任何帮助都将不胜感激

我希望合并两个阵列

a = np.array(["A", "B", "C"])
b = np.array(["1", "2", "3", "4", "5"])
类似于以下内容的数组(矩阵):

[["A", "B", "C", "1"]
 ["A", "B", "C", "2"]
 ["A", "B", "C", "3"]
 ["A", "B", "C", "4"]
 ["A", "B", "C", "5"]]
我尝试了一个for循环,但似乎不起作用。我是Python新手,任何帮助都将不胜感激。 谢谢。

也许可以将Python用于:

或许可以将Python与以下内容结合使用:

np.hstack((np.tile(a,(len(b),1)),b[:,None])) 数组(['A','B','C','1'], [A',B',C',2'], [A',B',C',3'], [A',B',C',4'], [A',B',C',5']],dtype='
>>np.hstack((np.tile(A,(len(B),1)),B[:,None]))
数组(['A','B','C','1'],
[A',B',C',2'],
[A',B',C',3'],
[A',B',C',4'],

[A',B',C',5']],dtype='一种方法是:

import numpy as np

a = np.array(["A", "B", "C"])
b = np.array(["1", "2", "3", "4", "5"])

output=[]

for i in list(b):
    a_list=list(a)
    a_list.append(i)
    output.append(a_list)

output=np.asarray(output)

print(output)
结果如下所示:

[['A' 'B' 'C' '1']
 ['A' 'B' 'C' '2']
 ['A' 'B' 'C' '3']
 ['A' 'B' 'C' '4']
 ['A' 'B' 'C' '5']]
>>> 

一种方法是:

import numpy as np

a = np.array(["A", "B", "C"])
b = np.array(["1", "2", "3", "4", "5"])

output=[]

for i in list(b):
    a_list=list(a)
    a_list.append(i)
    output.append(a_list)

output=np.asarray(output)

print(output)
结果如下所示:

[['A' 'B' 'C' '1']
 ['A' 'B' 'C' '2']
 ['A' 'B' 'C' '3']
 ['A' 'B' 'C' '4']
 ['A' 'B' 'C' '5']]
>>> 

这将实现以下目的:

将numpy导入为np
a=np.数组([“a”、“B”、“C”])
b=np.数组([“1”、“2”、“3”、“4”、“5”])
c=np.hstack([np.broadcast_to(a,shape=(len(b),len(a))),b.重塑(-1,1)])
输出:

['A''B''C''1']
['A''B''C''2']
['A''B''C''3']
['A''B''C''4']
[A''B''C''5']

这将实现以下目的:

将numpy导入为np
a=np.数组([“a”、“B”、“C”])
b=np.数组([“1”、“2”、“3”、“4”、“5”])
c=np.hstack([np.broadcast_to(a,shape=(len(b),len(a))),b.重塑(-1,1)])
输出:

['A''B''C''1']
['A''B''C''2']
['A''B''C''3']
['A''B''C''4']
[A''B''C''5']
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