When I loop over this list via the given code below:
my_list = ['a', 'b', 'c', 'b', 'd', 'm', 'n', 'n']
for i in my_list:
count = my_list.count(i)
if count > 1:
print(i)
it gives me an output ['b','b','n','n'].
How can I modify my code to just give the duplicated letter only, not each letter in the list without using sets?
Efficient Solution:
Here update a dictionary counter, and whenever a duplicate is found it's added to a set.
my_list = ['a', 'b', 'c', 'b', 'd', 'm', 'n', 'n']
char_counter = {}
duplicates = set()
for x in my_list:
if x not in char_counter.keys():
char_counter[x] = 1
else:
char_counter[x] += 1
duplicates |= {x}
print(duplicates)
One liner:
my_list = ['a', 'b', 'c', 'b', 'd', 'm', 'n', 'n']
print(set([x for x in my_list if my_list.count(x) > 1]))
Quick Fix:
my_list = ['a', 'b', 'c', 'b', 'd', 'm', 'n', 'n']
duplicates = []
for i in my_list:
count = my_list.count(i)
if count > 1 and i not in duplicates:
duplicates.append(i)
print(duplicates)
Try Counter
my_list = ['a', 'b', 'c', 'b', 'd', 'm', 'n', 'n']
from collections import Counter
duplicates = [item for item, count in Counter(my_list).items() if count > 1]
OUTPUT:
['b','b','n','n']
