list.append() in loops can only return final list

Question

For example:

a = {}; 
b = [];
for i in range(20):
    b.append(i)
    a[i] = b

After loops, all keys of a is linked to the same final list b [0,1,2,...,19]. However, what I want is that a[0] = [0], a[1] = [0,1], a[2] = [0,1,2]. I have no idea what is wrong here.

`a[i] = b` you keep setting the value of your dict at i to *the same list* `b`, use `b.copy()` instead to get a copy... — juanpa.arrivillaga, Feb 21 '19 at 23:53
you fall into copy a list issue in python , actually all list you assign into `a[i]` are all have same ids (try with `id(Object)`) so changing `b` will change the previous `a[i]`s , try to copy list `b` to `a[i]` with `copy.copy` method. — ᴀʀᴍᴀɴ, Feb 21 '19 at 23:54
@AubreydaCunha no, this is a duplicate. This same issue pops up many times every day in the python tag — juanpa.arrivillaga, Feb 21 '19 at 23:54

score 0 · Answer 1 · answered Feb 21 '19 at 23:53

0

a = {}; 
b = [];
for i in range(20):
    b=[k for k in range(i)]
    a[i] = b

Hope it helps

answered Feb 21 '19 at 23:53

TavoGLC

889
11
14

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`[k for k in range(i)]` should probably just be `list(range(i))`. Also, this doesn't really explain to the OP what is wrong, or what you answer is doing to fix it. – juanpa.arrivillaga Feb 21 '19 at 23:54

list.append() in loops can only return final list

1 Answers1