如何在 Python 中制作列表的浅拷贝 [英] How to make a shallow copy of a list in Python

问题描述

我正在尝试在 Python 中实现一种算法来生成列表的所有排列.但是我在我的 for 循环中，我希望保持原始前缀和其余列表完好无损，因此我试图使用 newprefix 和 newrest 制作这些列表的副本，但是在每次迭代时打印变量 rest 时，我看到即使是可变休息正在被修改！如何在 Python 中制作列表的浅拷贝?或者我尝试的逻辑还有其他问题吗?

I am trying to implement an algorithm in Python to generate all Permutations of a list. But I In my for loop I wish to keep the original prefix and rest lists intact, and therefore I am trying to make a copy of those lists using newprefix and newrest, however on printing the variable rest at each iteration, I see that even the variable rest is getting modified! How can I make a shallow copy of the list in Python? Or is there another issue with my attempted logic?

def perm(prefix, rest):
    if len(rest) == 0:
        print prefix 
    for i in range(len(rest)):
        #prints in the for loop are just for debugging
        print "rest:", rest
        print "i=", i
        newprefix = prefix
        newprefix.append(rest[i])
        newrest = rest
        newrest.pop(i)
        print "old pre : ", prefix
        print "newpre=", newprefix
        print "newrest=", newrest
        perm(newprefix, newrest)


perm([], ['a','b','c'])

推荐答案

要制作浅拷贝，可以将列表切片:

To make a shallow copy, you can slice the list:

newprefix = prefix[:]

或者将其传递到list构造函数中:

Or pass it into the list constructor:

newprefix = list(prefix)

另外，我认为您可以稍微简化一下代码:

Also, I think you can simplify your code a little:

def perm(prefix, rest):
    print prefix, rest

    for i in range(len(rest)):
        perm(prefix + [rest[i]], rest[:i] + rest[i + 1:])

perm([], ['a','b','c'])

这篇关于如何在 Python 中制作列表的浅拷贝的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！