I have a list like this;
a=[['2019', '08'], ['2018', '10'], ['2019', '08'], ['2019', '08'], ['2018', '10'], ['2019', '02']]
How can I delete duplicates.
[['2019', '08'], ['2018', '10'], ['2019', '02']]
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1Possibly a better dupe target: stackoverflow.com/questions/3819348/…ruohola– ruohola2019-08-02 13:33:08 +00:00Commented Aug 2, 2019 at 13:33
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5 Answers
If order is important (but algorithmic complexity is not):
b = []
for x in a:
if x not in b:
b.append(x)
If the complexity is relevant, here's an O(𝑛) solution:
seen = set()
b = []
for x in a:
t = tuple(x)
if t not in seen:
b.append(x)
seen.add(t)
Comments
You can easily do this with
a=[['2019', '08'], ['2018', '10'], ['2019', '08'], ['2019', '08'], ['2018', '10'], ['2019', '02']]
uniq = []
[uniq.append(x) for x in a if x not in uniq]
uniq
>>[['2019', '08'], ['2018', '10'], ['2019', '02']]
Comments
obviously if you don't know this Method. I highly recommand you start with this Implementation and get to know how this work and then move on to higher level implementation
a=[['2019', '08'], ['2018', '10'], ['2019', '08'], ['2019', '08'], ['2018', '10'], ['2019', '02']]
print(a)
b = []
for l in a:
if l not in b:
b.append(l)
print(b)
try to play here and get to know how this work because it is a simple Implementation which shows the basics of how the work should get done solving a simple problem like this
Comments
try:
a=[['2019', '08'], ['2018', '10'], ['2019', '08'], ['2019', '08'], ['2018', '10'], ['2019', '02']]
a = set(tuple(a) for a in a)
a = [list(a) for a in a]
print(a)
# output : [['2018', '10'], ['2019', '02'], ['2019', '08']]
Comments
a = [['2019', '08'], ['2018', '10'], ['2019', '08'], ['2019', '08'], ['2018', '10'], ['2019', '02']]
a=[list(t) for t in set(tuple(element) for element in a)]
print(a)
Convert elements to tuples then back to a list
Output:
[['2019', '02'], ['2018', '10'], ['2019', '08']]
