Question

我得到了列表的2D矩阵，其字符串如下：

somelist = [["12","10","0"]
            ["0","33","60"]]

我需要将所有str转换为int。到目前为止我的代码是：

for i in somelist:
        for j in i:
            j = int(j)

Answer 1

>>> [[int(i) for i in j] for j in somelist]
[[12, 10, 0], [0, 33, 60]]

Answer 2

或使用地图：

1526299672891   geckodriver     INFO    geckodriver 0.20.1
1526299672895   geckodriver     INFO    Listening on 127.0.0.1:48641
1526299673899   mozrunner::runner       INFO    Running command: "/usr/bin/firefox" "-marionette" "-profile" "/tmp/rust_mozprofile.YNIc2l05QiIr"
Running Firefox as root in a regular user's session is not supported.  ($HOME is /home/test which is owned by test.)

这里是您的固定代码：

>>> [list(map(int, x)) for x in somelist]
[[12, 10, 0], [0, 33, 60]]

输出：

somelist = [["12","10","0"],
            ["0","33","60"]]

for n, i in enumerate(somelist):
    for k, j in enumerate(i):
        somelist[n][k] = int(j)

print(somelist)

Answer 3

您的代码无效，因为Python中的int是不可变类型。如果我更改一个int变量，它不会更改其地址中的值，但它会更改整个变量。因此，以前持有它的任何列表或对象都不会被更改。以下是一些例子：

In [1]: a_list = [1,2,3]

In [2]: var = a_list[0]

In [3]: var
Out[3]: 1

In [4]: id(var)
Out[4]: 4465792624

In [5]: var = 30

In [6]: id(var) # A different address
Out[6]: 4465793552

In [7]: a_list
Out[7]: [1, 2, 3]

In [8]: id(a_list[0])
Out[8]: 4465792624

In [9]: a_list[0] = 30

In [10]: a_list
Out[10]: [30, 2, 3]

这就是为什么如果你做这个小改动，你的代码就会起作用：

for i in somelist:
    for ind, j in enumerate(i):
        i[ind] = int(j)

Python - 如何将字符串矩阵转换为int？

3 个答案: