How to Reverse a List and Find an Element's Index Without Modifying the Original List?

Chaining .reverse with .index on a List

This code fails when attempting to combine .reverse and .index methods on a list:

k = ['F', ' ', 'B', 'F']

def solution(formation):
    return ((formation.index(bCamel) > (len(formation) - 1 - (formation.reverse()).index(fCamel))))

solution(k)

The issue arises because .reverse returns None, effectively making it impossible to chain .index on the reversed list.

Avoiding Separate Reverse Statements

To avoid using a separate statement for reversing the list before indexing, you can use slicing to obtain the reversed list:

formation[::-1]

This returns a new copy of the list in reverse order, preserving the original order within the existing list.

With this adjustment, the corrected solution becomes:

k = ['F', ' ', 'B', 'F']

def solution(formation):
    return ((formation.index(bCamel) > (len(formation) - 1 - (formation[::-1]).index(fCamel))))

solution(k)

This code successfully uses the reversed list to compare the positions of fCamel and bCamel within the formation without modifying the original list.

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