Simple python code about double loop

Question

I tested the following python code on Spyder IDE. Thinking it would output 2d array q as increasing number as 0..31 from q[0][0] to q[3][7]. But it actually returns q as:

[[24, 25, 26, 27, 28, 29, 30, 31], [24, 25, 26, 27, 28, 29, 30, 31], [24, 25, 26, 27, 28, 29, 30, 31], [24, 25, 26, 27, 28, 29, 30, 31]].

The code:

q=[[0]*8]*4 
for i in range(4): 
    for j in range(8): 
        q[i][j] = 8*i+j 
print q

Any idea of what's happening here? I debugged step by step. It shows the updates of every row will sync with all other rows, quite different from my experience of other programing languages.

unutbu · Accepted Answer · 2012-09-02 13:23:01Z

4

q=[somelist]*4

creates a list with four identical items, the list somelist. So, for example, q[0] and q[1] reference the same object.

Thus, in the nested for loop q[i] is referencing the same list regardless of the value of i.

To fix:

q = [[0]*8 for _ in range(4)]

The list comprehension evaluates [0]*8 4 distinct times, resulting in 4 distinct lists.

Here is a quick demonstration of this pitfall:

In [14]: q=[[0]*8]*4

You might think you are updating only the first element in the second row:

In [15]: q[1][0] = 100

But you really end up altering the first element in every row:

In [16]: q
Out[16]: 
[[100, 0, 0, 0, 0, 0, 0, 0],
 [100, 0, 0, 0, 0, 0, 0, 0],
 [100, 0, 0, 0, 0, 0, 0, 0],
 [100, 0, 0, 0, 0, 0, 0, 0]]

edited Sep 2, 2012 at 13:23

answered Sep 2, 2012 at 13:03

unutbu

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4 Comments

user1640608 Over a year ago

How about j in this case? Is it supposed to be all q[i][0] and q[i][1] reference the same object as well?

Serdalis Over a year ago

@user1640608 when you are using primitives (integer, etc), the reference isn't repeated.

unutbu Over a year ago

Lists are mutable, integers are not. So if q[0] and q[1] point to the same list, then q[0][j] = 100 will modify q[1][j] as well. Although [0]*8 does create a list with 8 references to the same integer, it does not cause a problem because we aren't mutating the integers (an impossibility); we only alter the list's reference to point at other values.

Martijn Pieters Over a year ago

@user1640608: see this old answer of mine: Python list doesn't reflect variable change, new to python; perhaps it'll help you understand what is going on.

zenpoy · Accepted Answer · 2012-09-02 13:22:08Z

1

As explained the problem is caused due to * operation on lists, which create more references to the same object. What you should do is to use append:

q=[]
for i in range(4): 
    q.append([])
    for j in range(8): 
        q[i].append(8*i+j)
print q

[[0, 1, 2, 3, 4, 5, 6, 7], [8, 9, 10, 11, 12, 13, 14, 15], [16, 17, 18, 19, 20, 21, 22, 23], [24, 25, 26, 27, 28, 29, 30, 31]]

answered Sep 2, 2012 at 13:22

zenpoy

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Comments

MAK · Accepted Answer · 2012-09-03 06:23:29Z

0

When you do something like l = [x]*8 you are actually creating 8 references to the same list, not 8 copies.

To actually get 8 copies, you have to use l = [[x] for i in xrange(8)]

>>> x=[1,2,3]
>>> l=[x]*8
>>> l
[[1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3], [1, 2, 3]]
>>> l[0][0]=10
>>> l
[[10, 2, 3], [10, 2, 3], [10, 2, 3], [10, 2, 3], [10, 2, 3], [10, 2, 3], [10, 2, 3], [10, 2, 3]]
>>> l = [ [x] for i in xrange(8)]
>>> l
[[[10, 2, 3]], [[10, 2, 3]], [[10, 2, 3]], [[10, 2, 3]], [[10, 2, 3]], [[10, 2, 3]], [[10, 2, 3]], [[10, 2, 3]]]
>>> l[0][0] = 1
>>> l
[[1], [[10, 2, 3]], [[10, 2, 3]], [[10, 2, 3]], [[10, 2, 3]], [[10, 2, 3]], [[10, 2, 3]], [[10, 2, 3]]]

edited Sep 3, 2012 at 6:23

answered Sep 2, 2012 at 13:05

MAK

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1 Comment

Serdalis Over a year ago

Not correct because what you are saying is [0]*8 will create repeated references, which it will not, only something like [[0]]*8 will do such a thing because the outer [] will contain a reference and not a primitive.

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