按值复制列表列表，而不是引用 [英] Copy a list of list by value and not reference

问题描述

要了解为什么我在程序中遇到错误（试图查找行列式的次要变量），我编写了一个简单的程序，因为变量被弄乱了。下面的该函数将2 * 2矩阵作为输入，并返回包含其行的列表（我知道这毫无意义，效率很低，但是我试图理解其背后的原理）。

To understand why I was getting an error in a program , in which I tried to find the "minor" of a determinant, I wrote a simpler program because my variables were messed up. This function below takes in a 2 * 2 matrix as an input, and returns a list containing its rows (pointless and inefficient, I know, but I'm trying to understand the theory behind this).

def alpha(A):   #where A will be a 2 * 2 matrix
    B = A       #the only purpose of B is to store the initial value of A, to retrieve it later
    mylist = []
    for i in range(2):
        for j in range(2):
            del A[i][j]
        array.append(A)
        A = B
    return mylist

但是，这里看来B是动态分配给A 的值，因为我无法将A的初始值存储在B中以便以后使用。为什么？

However, here it seems that B is assigned the value of A dynamically, in the sense that I'm not able to store the initial value of A in B to use it later. Why is that?

推荐答案

因为python通过引用传递列表

Because python passes lists by reference

这意味着，当您写 b = a时，您说的是a和b是同一对象，并且当您更改b时，您也更改了a，反之亦然

This means that when you write "b=a" you're saying that a and b are the same object, and that when you change b you change also a, and viceversa

一种通过值复制列表的方法：

A way to copy a list by value:

new_list = old_list[:]

如果列表中包含对象并且您也要复制它们，请使用通用copy.deepcopy（）：

If the list contains objects and you want to copy them as well, use generic copy.deepcopy():

import copy
new_list = copy.deepcopy(old_list)

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