i need to sort the given list in specific order without using nested for loop-CodePudding

I need this code to pop and insert 0 in the list at the end of the list. I know some alternate method but i need to know what i did wrong in this code.

for i in arr:
    if i == 0:
        arr.insert(-1, arr.pop(arr.index(i)))
print(arr)

result:
[4, 5, 1, 9, 0, 0, 5]

expected result :
[4, 5, 1, 9, 5, 0, 0]

CodePudding user response：

The direct reason is that you misunderstand how insert method works.

See this one: Array: Insert with negative index

The following code works fine instead. But don't change the elements of a list on iteration anyway.

arr = [4, 5, 0, 1, 9, 0, 5] 
print(arr)
for i in arr:
    if i == 0:
        # arr.insert(-1, arr.pop(arr.index(i)))
        arr.append(arr.pop(arr.index(i)))
print(arr)

CodePudding user response：

One possible approach would be just to sort the list using a lambda function to test each value for 0. For example:

print(sorted([4, 5, 0, 1, 9, 0, 5], key=lambda x: x == 0))

This would display:

[4, 5, 1, 9, 5, 0, 0]

Alternatively you could just build two arrays up as follows:

arr = [4, 5, 0, 1, 9, 0, 5]

values = []
zeros = []

for value in arr:
    if value == 0:
        zeros.append(value)
    else:
        values.append(value)
        
output = values   zeros
print(output)

These approaches avoid trying to alter arr whilst iterating over it.

CodePudding user response：

You could rebuild the list in two parts, first filtering out zeros then adding the number of zeros at the end:

arr = [4, 5, 0, 1, 9, 0, 5]

arr = [*filter(None,arr),*[0]*arr.count(0)]

print(arr)
[4, 5, 1, 9, 5, 0, 0]

If you need to do it in-place, you can extend the list with zeros that you pop-out of it going backward (so that indexes remain stable):

arr = [4, 5, 0, 1, 9, 0, 5]

arr.extend(arr.pop(i) for i in reversed(range(len(arr))) if not arr[i])

print(arr)
[4, 5, 1, 9, 5, 0, 0]

Or use the equivalent for-loop:

for i in reversed(range(len(arr))):
    if not arr[i]:
        arr.append(arr.pop(i))