Some notes on learning Python.

I want to delete certain Python elements “in-place”; no copies. To start, I will try to delete every element in the array:

# An array of integers 0 through 9.
array = [*range(10)]

The approach below will not work:

for i in array:
    del i

del is like pop, minus the return of the removed element. This will only delete the iterator copied locally to the scope of the loop.

Trying a more old-school loop with indices:

for i in range(len(array)):
    del array[i]

This throws an IndexError: list assignment out of range. Some investigation through printing:

for i in range(len(array)):
    print(i,'before del:',array)
    del array[i]
    print(i,'after del:',array,'\n')

Gives the following output:

0 before del: [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
0 after del: [1, 2, 3, 4, 5, 6, 7, 8, 9] 

1 before del: [1, 2, 3, 4, 5, 6, 7, 8, 9]
1 after del: [1, 3, 4, 5, 6, 7, 8, 9] 

2 before del: [1, 3, 4, 5, 6, 7, 8, 9]
2 after del: [1, 3, 5, 6, 7, 8, 9] 

3 before del: [1, 3, 5, 6, 7, 8, 9]
3 after del: [1, 3, 5, 7, 8, 9] 

4 before del: [1, 3, 5, 7, 8, 9]
4 after del: [1, 3, 5, 7, 9] 

5 before del: [1, 3, 5, 7, 9]

And then finally comes to a halt and throws the same IndexError.

It now becomes clear… the last line printed shows what went wrong, and just how exactly we’re out of range with index 5 in an array 5 elements long.

Removing Based On Certain Conditions

Now I want to remove only odd numbers from my array:

for i in range(len(array)):
    if array[i] % 2 != 0:
        del array[i]

Again, IndexError. Maybe some investigative printing would help…

for i in range(len(array)):
    print(i,'before del:',array)
    
    if array[i] % 2 != 0:
        del array[i]
        
    print(i,'after del:',array,'\n')

Well, this doesn’t even get to run. Unlike earlier, we don’t even get to see on which iteration of the loop this fails; it throws an error right out of the gate.

This error points to our if statement, but if we replace the del with some other expression, this code will run fine:

for i in range(len(array)):    
    if array[i] % 2 != 0:
        print(array[i])

The above will print all odd numbers in the array. The issue must then lie in the del operation? It almost feels like its being “hoisted”, in javascript parlance….

Managing The Array Yourself

If you want something done right, you do it yourself! From a C and C++ perspective, this would be the only way to do most things. With a dynamic language like Python, this feels odd. But it works and there is no mystery:

length = len(array)
i = 0
while i < length:
    if array[i] % 2 != 0:
        del array[i]
        length -=1 
    i+=1

This is a bit more to type, and uses two variables that are useless beyond the scope of this operation (a waste). But when can cut that down to one variable; we don’t have to store length:

i = 0
while i < len(array):
    if array[i] % 2 != 0:
        del array[i]
    i+=1

The length of the array updates as elements are deleted, and len returns the new length every time, so we only need to manage where we are in the array.

The Easiest Way

for i in array:
    if i % 2 != 0:
        array.remove(i)

The remove method! This will do it. Not pop, but remove.