Why is an element not printing?

Question

I am learning the list and list assignments and using while and for loops with lists. I wrote the following code just to test my understanding.

I=[1,4, 7, 6, 7,8,9]
IU=I

while IU !=[]:
    print ('I: ',I)
    print('IU :' ,IU)
    for i in I:
        print(i)
        if i*i<=36:
            IU.remove(i)
            print(IU)
    if i>=9:
        break

This gives the following output:

I:  [1, 4, 7, 6, 7, 8, 9]
IU : [1, 4, 7, 6, 7, 8, 9]
1
[4, 7, 6, 7, 8, 9]
7
6
[4, 7, 7, 8, 9]
8
9

My understanding is that it would print 4 after it removed 1 from IU. However it is not printing 4. Any hints on why this is happening?

Does this answer your question? [How to remove items from a list while iterating?](https://stackoverflow.com/questions/1207406/how-to-remove-items-from-a-list-while-iterating) — 0x5453, Oct 18 '22 at 18:05

Xin Cheng · Accepted Answer · 2022-10-19T02:05:44.033

3

Issue

The issued line is IU=I

This line does not create a new list IU. Python creates a reference 'IU' to the original list I, which means IU and I point to the same list. Thus, when you remove 4 from your I, IU get affected too.

Solution

There are different ways of copying of I. For example, you can

IU = [_ for _ in I]

Or

IU = I.copy()

Or

IU = I[:]

demo

Here is a demo show the difference

L = [1,2]
Ref = L
Cpy = L.copy()

L.append(3)
print(Ref)
print(Cpy)

Output
```
[1, 2, 3]
[1, 2] 
```

edited Oct 19 '22 at 02:05

answered Oct 18 '22 at 18:02

Xin Cheng

1,432
10
17

1

Another efficient list copy method is to use slicing, as: `copy = list_[:]`. – S3DEV Oct 18 '22 at 19:12
1

@S3DEV, Thanks for the reminder. That's a good point. – Xin Cheng Oct 18 '22 at 20:37

Saptarshi Das · Answer 2 · 2022-10-18T18:09:45.127

1

You don't see 4 printed because when the program removed 1 initially, the loop pointer i was pointing to 1, and so, once 1 got removed, the i pointer moved to the next element, 4. And at the end of the loop, since the for loop goes over to the next element (i++ that we do in c++), it points at 7 at the starting of next iteration. Hope I could solve your doubt.

edited Oct 18 '22 at 18:09

answered Oct 18 '22 at 18:09

Saptarshi Das

11
2

score 1 · Answer 3 · answered Oct 18 '22 at 18:21

List in python is implemented using Dynamic Array

Here, after the removal of 1 from the list, the list on the left moves backward(not exactly but somewhat similar), so i now points to 4.

Now i gets incremented, thus now i points to the next memory location.

Please check the link to see how the list gets implemented inside python.

Why is an element not printing?

3 Answers3

Issue

Solution

demo