Python将矩阵转换为列表

我有一个python矩阵，如下所示

[
                [1603179035, "1"],
                [1603179095, "2"],
                [1603179155, "3"],
                [1603179215, "4"],
                [1603179275, "5"]
]

如何将其转换为两个列表

list1 = [1603179035,1603179095,1603179155,1603179215,1603179275]
list2 = ["1","2","3","4',"5"]

这将有助于您：

lst = [
                [1603179035, "1"],
                [1603179095, "2"],
                [1603179155, "3"],
                [1603179215, "4"],
                [1603179275, "5"]
]

list1 = [element[0] for element in lst]
list2 = [element[1] for element in lst]

print(list1)
print(list2)

输出：

[1603179035, 1603179095, 1603179155, 1603179215, 1603179275]
['1', '2', '3', '4', '5']

---

如果是numpy数组，则更容易

import numpy as np

lst = [
                [1603179035, "1"],
                [1603179095, "2"],
                [1603179155, "3"],
                [1603179215, "4"],
                [1603179275, "5"]
]
lst = np.matrix(lst)
A, B = lst[:,0], lst[:,1]

# A = [1603179035, 1603179095, 1603179155, 1603179215, 1603179275]
# B = ['1', '2', '3', '4', '5']

如果您计划使用numpy矩阵，希望这对您有所帮助。

---

矩阵可以在numpy mat[行索引，列索引]中索引。

您可以使用这一行代码，无需安装外部库。。。。基本上，您只需解压缩矩阵：

lst = [
                [1603179035, "1"],
                [1603179095, "2"],
                [1603179155, "3"],
                [1603179215, "4"],
                [1603179275, "5"]
]

lst1, lst2 = map(list, zip(*lst))

以下是输出：

>>> lst1
[1603179035, 1603179095, 1603179155, 1603179215, 1603179275]
>>> lst2
['1', '2', '3', '4', '5']

下面是另一种方法：

x = [
            [1603179035, "1"],
            [1603179095, "2"],
            [1603179155, "3"],
            [1603179215, "4"],
            [1603179275, "5"]
    ]
list1 = []
list2 = []
for elem in x:
    list1.append(elem[0])
    list2.append(elem[1])

print(list1)
print(list2)