is it a 'for' or '.append()' bug in python?

Question

I try to use 'uga' as input(=text). I want to get [['U','g','a'],['u','G','a'],['u','g','A'] as output.

I have tried this code :

text = 'uga'
abc = list(text)
out_list = []

for i in range (len(text)):
    out_list.append(abc)

x = out_list.copy()

for item in range(len(x)):
    x[item][item] = x[item][item].upper()
print(x)

after using that code I got [['U', 'G', 'A'], ['U', 'G', 'A'], ['U', 'G', 'A']] as the output.

If we cut the code...

text = 'uga'
abc = list(text)
out_list = []

for i in range (len(text)):
    out_list.append(abc)

x = out_list.copy()
print(x)

it will give [['u', 'g', 'a'], ['u', 'g', 'a'], ['u', 'g', 'a']] as the output.

after that I try to make the code without append(). I just make a new variable using this list --> [['u', 'g', 'a'], ['u', 'g', 'a'], ['u', 'g', 'a']] which is same as x

a = [['u', 'g', 'a'], ['u', 'g', 'a'], ['u', 'g', 'a']]
for item in range(len(a)):
    a[item][item] = a[item][item].upper()
print(a)

surprisingly I can get result which is [['U', 'g', 'a'], ['u', 'G', 'a'], ['u', 'g', 'A']]

after that I try again using append but manually, like this:

b = []
b.append(['u','g','a'])
b.append(['u','g','a'])
b.append(['u','g','a'])

for i in range (len(b)):
    b[i][i] = b[i][i].upper()

print(b)

the result is right too which is [['U', 'g', 'a'], ['u', 'G', 'a'], ['u', 'g', 'A']]

I don't know why my first code give the wrong result... because when I check the type of x , a, and b it same which is a list. After that I check a==b==x it gives the result True

I also put them in the same code which is:

for i in range (len(something)):
    something[i][i] = something[i][i].upper()

I have thinking a lot and ask many of friends, but no body could know what happened..

I'm sorry because my English is not very good. Hope you could help me...

Thank you

In the first snippet, `out_list[0] is out_list[1]`. In the second, it is merely the case that `out_list[0] == out_list[1]`. See [Is there a difference between “==” and “is”?](https://stackoverflow.com/questions/132988/is-there-a-difference-between-and-is). You need to copy `abc` so you have different sublists, not copy `out_list` (or use deepcopy). Check how your code runs in the visualizer [here](http://www.pythontutor.com/visualize.html#mode=edit) by stepping through it, and it should be abundantly clear. (Not writing an answer, because it will almost certainly be closed as a duplicate.) — Amadan, Feb 10 '20 at 10:49

score 0 · Accepted Answer · answered Feb 10 '20 at 10:50

You can see what happens if you add a print statement in the second loop. The output the looks like this:

[['U', 'g', 'a'], ['U', 'g', 'a'], ['U', 'g', 'a']]
[['U', 'G', 'a'], ['U', 'G', 'a'], ['U', 'G', 'a']]
[['U', 'G', 'A'], ['U', 'G', 'A'], ['U', 'G', 'A']]
[['U', 'G', 'A'], ['U', 'G', 'A'], ['U', 'G', 'A']]

As you can see the values all change at the same time. That happens because you add the abc object to the array 3 times and don't copy it.

is it a 'for' or '.append()' bug in python?

1 Answers1