Standard loop works when list comprehension does not

Question

Why the below code in where I am using list comprehension doesn't work, but the common way does it?

exams = [1,2,3,1,2,3,4,1,5,1]
repeatedExams = []

# EXCPECTED OUPUT
# repeatedExams = [1,2,3]

# IF I USE repeatedExams.append(i)
repeatedExams = [repeatedExams.append(i) for i in exams if (exams.count(i)>1) and (i not in repeatedExams)]
    # OUTPUT 
    # repeatedExams = [None,None,None]

# IF I ONLY USE i
repeatedExams = [i for i in exams if (exams.count(i)>1) and (i not in repeatedExams)]
    # OUTPUT
    # repeatedExams = [1,2,3,1,2,3,4,1,5,1]

# WORK 
for i in exams:
    if exams.count(i) > 1 and i not in repeatedExams:
        repeatedExams.append(i)

    # OUTPUT
    # repeatedExams = [1,2,3]

List comprehensions `append` to a new list for you; that's their entire purpose. `append` doesn't return anything, so your first way doesn't work. — Carcigenicate, Jun 11 '21 at 13:59
Does this answer your question? [Why does append() always return None in Python?](https://stackoverflow.com/questions/16641119/why-does-append-always-return-none-in-python) — Carcigenicate, Jun 11 '21 at 14:00

score 0 · Answer 1 · answered Jun 11 '21 at 14:02

list.append() returns None

Comprehension returns a collection for you; you do not need to pre-define one

As you are building one, you shouldn't be using it, so not in repeatedExams within the comprehension isn't going to do what you want

If you want to count repeats, use a Counter

from collections import Counter

exams = [1,2,3,1,2,3,4,1,5,1]
c = Counter(exams)
repeatedExams = [k for k,v in c.items() if v > 1]

ibarrond · Answer 2 · 2021-06-11T17:52:24.443

0

If you really want to go with list comprehension, then you need to split the loop into two steps:

# Keep only repetated elements
repeatedExams = [i for i in exams if exams.count(i)>1] 
# Remove duplicates of repeated elements       
repeatedExams = set(repeatedExams)

edited Jun 11 '21 at 17:52

answered Jun 11 '21 at 14:03

ibarrond

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Use a `set` to avoid the second list comprehension. – user12339314 Jun 11 '21 at 14:05
@user12339314 That'll remove duplicates, not count them – OneCricketeer Jun 11 '21 at 14:05
This answer has quadratic runtime, btw – OneCricketeer Jun 11 '21 at 14:05
I know :) I'm just plugging everything into list comprehensions. – ibarrond Jun 11 '21 at 14:05
`... for i in repeatedExams if i not in repeatedExams` Isn't that just an expensive no-op? – Ture Pålsson Jun 11 '21 at 17:27

score 0 · Answer 3 · answered Jun 11 '21 at 14:15

The list comprehension syntax tells Python to take an iterable, and for each element of that iterable, evaluate an expression, and then build a list out of all the results returned by that expression. append is what's known as an "in place" operator. What that means is that it operates not by returning a result, but by modifying the associated object. The actual value of inplace operators is generally None. For instance, repeatedExams.append(i) modifies the object repeatedExams. The modified version of repeatedExams is now stored where ther original repeatedExams was, and repeatedExams.append(i) returns None. Your syntax then stores all those Nones in a list.

It may be informative to do

for i in exams:
    if (exams.count(i)>1) and (i not in repeatedExams):
        print(repeatedExams.append(i))

and see what you get.

score 0 · Answer 4 · answered Jun 11 '21 at 14:27

0

Try this -

repeatedExams = [i for i in exams if exams.count(i)>1] 
repeatedExams = set(repeatedExams)

Or-

repeatedExams = set([i for i in exams if exams.count(i)>1])

answered Jun 11 '21 at 14:27

Does this help you? – Jun 11 '21 at 14:27

mightbesimon · Answer 5 · 2021-06-26T06:29:22.187

0

repeatedExams = {exam for exam in exams if exams.count(exam)>1}

similar to user16198685's answer, but the [] brackets and 'set()' are redundant

exam for exam in exams if exams.count(exam)>1 is a generator set comprehension takes generators as arguments, as well as listcomps. So set([]) brackets can be replaced with '{}'

edited Jun 26 '21 at 06:29

answered Jun 25 '21 at 15:04

mightbesimon

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Standard loop works when list comprehension does not

5 Answers5