How to remove elements from a list of lists where an element is repeated?

Question

I have a list of lists

my_list = [["a", "b", "c"],
           ["d", "f", "g"],
           ["h", "b", "i"]]

I want to remove all elements in my_list where element[1] is not unique. So the result after performing this operation is either

my_list = [["d", "f", "g"]]

or a new list also works

new_list = [["d", "f", "g"]]

My current method is:

from collections import Counter

y_list = []

for element in my_list:
    x, y, z = element
    y_list.append(y)

count_dict = dict(Counter(y_list))

unique_y_list = []

for y, count in count_dict.items():
    if count == 1:
        unique_y_list.append(y)
 

valid_elements = []

for element in my_list:
    x, y, z = element
    if y in unique_y_list:
        valid_elements.append(element)

This works but doesn't seem efficient. Help with a more pythonic way to achieve my desired result? Preserving order is not important.

I mean any element where index[1] of that element occurs more than once in my_list. my_list [0] [1] == "b" ; my_list [2] [1] == "b" ; remove these elements — McQuestion
– McQuestion, Commented Sep 26, 2020 at 19:22
The only thing inefficint is how you re-construct the valid elements. Don't create a list, unique_list, and check if y in unique_y_list:, instead, use if count_dict[y] == 1: valid_elements.append(element) — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Sep 26, 2020 at 19:28
also, dict(Counter(y_list)) is unecessary. a Counter is a dict already — juanpa.arrivillaga
– juanpa.arrivillaga, Commented Sep 26, 2020 at 19:28

Mureinik · Accepted Answer · 2020-09-26 19:15:46Z

2

I'd create a Counter of elements in the index 1, and then use it to check if each list's element at index 1 is unique:

from collections import Counter
c = Counter(x[1] for x in my_list)
result = [l for l in my_list if c[l[1]] == 1]

answered Sep 26, 2020 at 19:15

Mureinik

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2 Comments

juanpa.arrivillaga Over a year ago

This is pretty much what the OP is doing.

McQuestion Over a year ago

@Mureinik thank you, exactly what I was looking for, fewer lines of code

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