I know this is probably an easy answer but I can't figure it out. What is the best way in Python to keep the duplicates in a list:
x = [1,2,2,2,3,4,5,6,6,7]
The output should be:
[2,6]
I found this link: Find (and keep) duplicates of sublist in python, but I'm still relatively new to Python and I can't get it to work for a simple list.
I'd use a collections.Counter:
from collections import Counter
x = [1, 2, 2, 2, 3, 4, 5, 6, 6, 7]
counts = Counter(x)
output = [value for value, count in counts.items() if count > 1]
Here's another version which keeps the order of when the item was first duplicated that only assumes that the sequence passed in contains hashable items and it will work back to when set or yeild was introduced to the language (whenever that was).
def keep_dupes(iterable):
seen = set()
dupes = set()
for x in iterable:
if x in seen and x not in dupes:
yield x
dupes.add(x)
else:
seen.add(x)
print list(keep_dupes([1,2,2,2,3,4,5,6,6,7]))
[k for k in OrderedDict.fromkeys(x) if counts[k] > 1].OrderedDict there? why not just [k for k in x if counts[k] > 1]? Actually, that's better than what I have. I'll update...This is a short way to do it if the list is sorted already:
x = [1,2,2,2,3,4,5,6,6,7]
from itertools import groupby
print [key for key,group in groupby(x) if len(list(group)) > 1]
groupby groups consecutive elementsList Comprehension in combination with set() will do exactly what you want.
>>> list(set([i for i in x if x.count(i) > 1]))
[2, 6]
keepin' it simple:
array2 = []
aux = 0
aux2=0
for i in x:
aux2 = i
if(aux2==aux):
array2.append(i)
aux= i
list(set(array2))
That should work
[2,2,6]?
Not efficient but just to get the output, you could try:
import numpy as np
def check_for_repeat(check_list):
repeated_list = []
for idx in range(len(check_list)):
elem = check_list[idx]
check_list[idx] = None
if elem in temp_list:
repeated_list.append(elem)
repeated_list = np.array(repeated_list)
return list(np.unique(repeated_list))
