Filter a Python list using Dictionary keys and values

Question

GOAL: Filter a list of lists using dictionary as reference in Python 3.8+

CASE USE: When reviewing a nested list -- a series of survey responses -- filtering out responses based on control questions. In the dictionary, the responses to questions 3 (index 2 in list) and 7 (index 6) should both be of corresponding value 5. If both answers for a response are not 5, they should not be populated in the filtered_responses list.

Open to interpretation on how to solve for this. I have reviewed several resources touching on filtering dictionaries using lists. This method is preferred as some survey responses many contain the same array of values, therefore the list element is retained.

no_of_survey_questions = 10
no_of_participants = 5
min_score = 1
max_score = 10

control_questions = {3: 5,
                     7: 5, }

unfiltered_responses = [[4, 5, 4, 5, 4, 5, 4, 5, 4, 5],  # omit
                        [9, 8, 7, 6, 5, 4, 3, 2, 1, 1],  # omit
                        [5, 5, 5, 5, 5, 5, 5, 5, 5, 5],  # include
                        [5, 2, 5, 2, 5, 2, 5, 9, 1, 1],  # include
                        [1, 2, 5, 1, 2, 1, 2, 1, 2, 1]]  # omit

for response in unfiltered_responses:
    print(response)

print()

filtered_responses = []  # should contain only unfiltered_responses values marked 'include'
for response in filtered_responses:
    # INSERT CODE HERE
    print(response)

Thanks in advance!

@JonSG too many attempts to list, but each web search query returned a response of "filtering dictionary." Primarily, I tried to equate each dict key to the corresponding index, but was unable to find an effective solution that iterated through the dictionary for a given element. — paaskanama
– paaskanama, Commented Jan 13, 2023 at 19:32

Andrej Kesely · Accepted Answer · 2023-01-13 20:30:46Z

2

You can use list comprehension + all():

control_questions = {3: 5,
                     7: 5}

unfiltered_responses = [[4, 5, 4, 5, 4, 5, 4, 5, 4, 5],  # omit
                        [9, 8, 7, 6, 5, 4, 3, 2, 1, 1],  # omit
                        [5, 5, 5, 5, 5, 5, 5, 5, 5, 5],  # include
                        [5, 2, 5, 2, 5, 2, 5, 9, 1, 1],  # include
                        [1, 2, 5, 1, 2, 1, 2, 1, 2, 1]]  # omit

filted_questions = [subl  for subl in unfiltered_responses if all(subl[k-1] == v for k, v in control_questions.items())]
print(filted_questions)

Prints:

[
   [5, 5, 5, 5, 5, 5, 5, 5, 5, 5], 
   [5, 2, 5, 2, 5, 2, 5, 9, 1, 1]
]

edited Jan 13, 2023 at 20:30

answered Jan 13, 2023 at 17:42

Andrej Kesely

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2 Comments

paaskanama Over a year ago

with a slight modification to subl[k] to subl[k-1] to align the literal dictionary key to the list index, it looks like this works. Thank you!

Andrej Kesely Over a year ago

@paaskanama Yep, subl[k-1], I've pasted the wrong code :) Fixed.

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