How to make a shallow copy of a list in Python

Question

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'])

Unless this is for your own enrichment, you probably should use itertools.permutations(). — dstromberg
– dstromberg, Commented Apr 29, 2013 at 3:16

Blender · Accepted Answer · 2013-04-29 02:26:59Z

55

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

newprefix = prefix[:]

Or pass it into the list constructor:

newprefix = list(prefix)

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'])

answered Apr 29, 2013 at 2:26

Blender

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4 Comments

BlackVegetable Over a year ago

Your first one (slicing) is my preferred way of accomplishing this. I just thought I'd let them know about the copy module as well. +1

Gershom Maes Over a year ago

Just wondering - isn't the slice operation non-declarative? Isn't it best to clearly declare what you're doing? copied_list = original_list[:] doesn't declare that a copy is occurring, whereas copied_list = copy.copy(original_list) is highly declarative.

Jonathan Benn Over a year ago

@user11580406 this is incorrect. A simple test at the Python command line will prove that list(original) will create a shallow copy of original

E. Kaufman Over a year ago

If s is a string, why does s[:] produce an object with the same memory address as s according to here. In other words, s is s[:] if s is a string, but if s is an array, s is not s[:]

BlackVegetable · Accepted Answer · 2019-01-03 23:23:15Z

20

import copy

a = [somestuff]
b = copy.copy(a) # Shallow copy here.
c = copy.deepcopy(a) # Deep copy here.

The copy module is worth knowing about. https://docs.python.org/3/library/copy.html

(Python 2) http://docs.python.org/2/library/copy.html

edited Jan 3, 2019 at 23:23

answered Apr 29, 2013 at 2:27

BlackVegetable

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