What range function does to a Python list?

Question

I am not able to figure out what is happening here. Appending reference to range function is kind of creating a recursive list at index 3.

>>> x = range(3)
[0, 1, 2]
>>> x.append(x)
[0, 1, 2, [...]]
>>> x[3][3][3][3][0] = 5
[5, 1, 2, [...]]

Whereas, when I try this:

 >>> x = range(3)
 [0, 1, 2]
 >>> x.append(range(3))
 [0, 1, 2, [0, 1, 2]]

I can easily deduce the reason for the second case but not able to understand what appending reference to range function is doing to the list appended.

Why is this tagged both Python 2 and 3? range behaves differently in 2 and 3. — interjay
– interjay, Commented Apr 3, 2016 at 14:20
Lists are passed by reference in python. This means that in the first scenario, you are appending the pointer to x, not the value of x. Because x now contains a pointer to x, it calls itself recursively at that point. — Jacob H
– Jacob H, Commented Apr 3, 2016 at 14:21
@JacobH: saying "lists are passed by reference" makes it seem like lists might behave differently from other objects. All objects are passed the same way (although people argue about what the right phrase to describe that is.) — DSM
– DSM, Commented Apr 3, 2016 at 14:36

Karoly Horvath · Accepted Answer · 2016-04-03 14:46:25Z

9

In python2, ranges are lists.

lists, and most of the things in python are objects with identities.

li = [0,1]
li[1] = li    # [0, [...]]
              # ^----v
id(li)        # 2146307756
id(li[1])     # 2146307756

Since you're putting the list inside itself, you're creating a recursive data structure.

edited Apr 3, 2016 at 14:46

answered Apr 3, 2016 at 14:22

Karoly Horvath

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3 Comments

MattDMo Over a year ago

Please note that this is Python 2-specific. Ranges are entirely different in Python 3, returning a lazily-evaluated range object, not a list at all.

Rahul Over a year ago

@Karoly thanks for the answer but I am not able to understand why value assignment of 5 is working. Whatever I have learnt is that if I use a[1][0] = 5, then it means assign 5 to 0 index at list of list at index 1.

Karoly Horvath Over a year ago

if you meant my example: li[1] is li, so it's simply li[0]=5.

hilberts_drinking_problem · Accepted Answer · 2016-04-03 14:50:46Z

First this is strange, and you probably should not use this in practice. The issue is not specific to range function and has to do with references. When you call x.append(x), you essentially say that x[-1] is x. So when you modify x[0], you also modify x[-1][0], x[-1][-1][0] etc.

To see that this is not range specific, you could use copy.copy:

from copy import copy

x = range(1)
x.append(x)  # here x[1] is reference to x itself (same object)
print(x[0], x[1][0], x[1][1][0])
x[0] = 1
print(x[0], x[1][0], x[1][1][0])  # all values change

# 

x = range(1)
x.append(copy(x))  # x[1] is a copy of x at previous state (new object)
print(x[0], x[1][0])  # cannot call x[1][1][0] -> x[1][1] is an int
x[0] = 1
print(x[0], x[1][0])  # only the first value changes

Output:

(0, 0, 0)
(1, 1, 1)
(0, 0)
(1, 0)

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