Question

I have a workaround to the following question. That workaround would be a for loop with a test for inclusion in the output like the following:

#!/usr/bin/env python

def rem_dup(dup_list):
    reduced_list = []
    for val in dup_list:
        if val in reduced_list:
            continue
        else:
            reduced_list.append(val)

    return reduced_list

I am asking the following question, because I am curious to see if there is a list comprehension solution.

Given the following data:

reduced_vals = []
vals = [1, 2, 3, 3, 2, 2, 4, 5, 5, 0, 0]

Why does

reduced_vals = = [x for x in vals if x not in reduced_vals]

produce the same list?

>>> reduced_vals
[1, 2, 3, 3, 2, 2, 4, 5, 5, 0, 0]

I think it has something to do with checking the output (reduced_vals) as part of an assignment to a list. I am curious, though as to the exact reason.

Thank you.

Answer 1

The list comprehension creates a new list, while reduced_vals points to the empty list all the time during the evaluation of the list comprehension.

The semantics of assignments in Python are: Evaluate the right-hand side and bind the resulting object to the name on the left-hand side. An assignment to a bare name never mutates any object.

By the way, you should use set() or collections.OrderedDict.fromkeys() to remove duplicates in an efficient way (depending on whether you need to preserve order or not).