Creating a fallback list in python

Question

I want to back up another list in python. Here is a sample code.

x = [1,2,3]  
y = x  
x.pop(0)  

print y

This, however, gives the result y = [2,3]when I want it to give [1,2,3]. How can I make list y independent of list x?

Answer 1

A common idiom for this is y = x[:]. This makes a shallow copy xand saves it in y.

Please note that if it xcontains links to objects, yit will also contain links to the same objects. This may or may not be what you want. If not, take a look copy.deepcopy().

Answer 2

Here is one way to do this:

import copy

x = [1,2,3]
y = copy.deepcopy(x)
x.pop(0)
print x
print y

docs here

Answer 3

AIX , :

y = list(x)

, . , , . - ( ).

, NOTHING y, , . :

y = tuple(x)

:

y = [a for a in x]

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