Mixing
Remove first duplicate element of a list
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I need to remove the duplicate occurrence of the 1st element in the list which is duplicate (present more than once) while preserving the order of the input list. For eg: for the input of
in = [2, 3, 4, 5, 3, 6, 4, 1]
output should be
out = [2, 3, 4, 5, 6, 4, 1]
I have tried below and is giving correct result , I just wanted to check with the community if there is a better or more pythonic solution to this
input = [2, 3, 4, 5, 3, 6, 4, 1]
first_dupe = None
for elem in input:
if input.count(elem) > 1:
first_dupe = elem
break
flg = True
new_list =
for x in input:
if x != first_dupe or flg is True:
new_list.append(x)
if x == first_dupe:
flg = False
print(new_list)
python list
asked Jan 3 at 2:45
ShanilShanil
5717
add a comment |
I need to remove the duplicate occurrence of the 1st element in the list which is duplicate (present more than once) while preserving the order of the input list. For eg: for the input of
in = [2, 3, 4, 5, 3, 6, 4, 1]
output should be
out = [2, 3, 4, 5, 6, 4, 1]
I have tried below and is giving correct result , I just wanted to check with the community if there is a better or more pythonic solution to this
input = [2, 3, 4, 5, 3, 6, 4, 1]
first_dupe = None
for elem in input:
if input.count(elem) > 1:
first_dupe = elem
break
flg = True
new_list =
for x in input:
if x != first_dupe or flg is True:
new_list.append(x)
if x == first_dupe:
flg = False
print(new_list)
python list
asked Jan 3 at 2:45
ShanilShanil
5717
add a comment |
I need to remove the duplicate occurrence of the 1st element in the list which is duplicate (present more than once) while preserving the order of the input list. For eg: for the input of
in = [2, 3, 4, 5, 3, 6, 4, 1]
output should be
out = [2, 3, 4, 5, 6, 4, 1]
I have tried below and is giving correct result , I just wanted to check with the community if there is a better or more pythonic solution to this
input = [2, 3, 4, 5, 3, 6, 4, 1]
first_dupe = None
for elem in input:
if input.count(elem) > 1:
first_dupe = elem
break
flg = True
new_list =
for x in input:
if x != first_dupe or flg is True:
new_list.append(x)
if x == first_dupe:
flg = False
print(new_list)
python list
asked Jan 3 at 2:45
ShanilShanil
5717
I need to remove the duplicate occurrence of the 1st element in the list which is duplicate (present more than once) while preserving the order of the input list. For eg: for the input of
in = [2, 3, 4, 5, 3, 6, 4, 1]
output should be
out = [2, 3, 4, 5, 6, 4, 1]
I have tried below and is giving correct result , I just wanted to check with the community if there is a better or more pythonic solution to this
input = [2, 3, 4, 5, 3, 6, 4, 1]
first_dupe = None
for elem in input:
if input.count(elem) > 1:
first_dupe = elem
break
flg = True
new_list =
for x in input:
if x != first_dupe or flg is True:
new_list.append(x)
if x == first_dupe:
flg = False
print(new_list)
python list
python list
asked Jan 3 at 2:45
ShanilShanil
5717
asked Jan 3 at 2:45
ShanilShanil
5717
asked Jan 3 at 2:45
ShanilShanil
5717
asked Jan 3 at 2:45
ShanilShanil
5717
asked Jan 3 at 2:45
ShanilShanil
5717
5717
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
If you just want to remove the first duplicate, you can create set, append elements to a new list if they are in the set while removing each element from the set as well. When an element is not seen, append the rest of the list.
This is fairly efficient if the duplicate is seen early, but has O(1) if the element is seen late.
x = [2, 3, 4, 5, 3, 6, 4, 1]
s = set(x)
out =
for i,z in enumerate(x):
if z in s:
out.append(z)
s.remove(z)
else:
break
out += x[i+1:]
out
# returns:
[2, 3, 4, 5, 6, 4, 1]
answered Jan 3 at 2:56
JamesJames
13.8k11633
add a comment |
You could keep track of what you have already used and check if the value is in there.
lst = [2, 3, 4, 5, 3, 6, 4, 1]
used = set()
for i, x in enumerate(lst):
if x in used:
lst.pop(i)
break
used.add(x)
print(lst)
Also, don't give variables, or anything else, the same name as a keyword in python. input is already a built-in function.
answered Jan 3 at 2:53
All KnowerAll Knower
413214
It's better forusedto beset.
– dyukha
Jan 3 at 2:55
This is better than the original solution, but: 1. Iterating usingfor i in range(len())is not pythonic. 2.usedcould probably be aset.
– Tomothy32
Jan 3 at 2:55
Thanks, I guess the best alternative would beenumerate?
– All Knower
Jan 3 at 3:22
add a comment |
This is my version of Chiheb Nexus' answer.
def find_dup(lst):
seen = set()
it = iter(lst)
for item in it:
if item not in seen:
seen.add(item)
yield item
else:
yield from it
lst = [2, 3, 4, 5, 3, 6, 4, 1]
list(find_dup(lst))
[2, 3, 4, 5, 6, 4, 1]
answered Jan 3 at 5:03
John La RooyJohn La Rooy
215k41279431
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you just want to remove the first duplicate, you can create set, append elements to a new list if they are in the set while removing each element from the set as well. When an element is not seen, append the rest of the list.
This is fairly efficient if the duplicate is seen early, but has O(1) if the element is seen late.
x = [2, 3, 4, 5, 3, 6, 4, 1]
s = set(x)
out =
for i,z in enumerate(x):
if z in s:
out.append(z)
s.remove(z)
else:
break
out += x[i+1:]
out
# returns:
[2, 3, 4, 5, 6, 4, 1]
answered Jan 3 at 2:56
JamesJames
13.8k11633
add a comment |
If you just want to remove the first duplicate, you can create set, append elements to a new list if they are in the set while removing each element from the set as well. When an element is not seen, append the rest of the list.
This is fairly efficient if the duplicate is seen early, but has O(1) if the element is seen late.
x = [2, 3, 4, 5, 3, 6, 4, 1]
s = set(x)
out =
for i,z in enumerate(x):
if z in s:
out.append(z)
s.remove(z)
else:
break
out += x[i+1:]
out
# returns:
[2, 3, 4, 5, 6, 4, 1]
answered Jan 3 at 2:56
JamesJames
13.8k11633
add a comment |
If you just want to remove the first duplicate, you can create set, append elements to a new list if they are in the set while removing each element from the set as well. When an element is not seen, append the rest of the list.
This is fairly efficient if the duplicate is seen early, but has O(1) if the element is seen late.
x = [2, 3, 4, 5, 3, 6, 4, 1]
s = set(x)
out =
for i,z in enumerate(x):
if z in s:
out.append(z)
s.remove(z)
else:
break
out += x[i+1:]
out
# returns:
[2, 3, 4, 5, 6, 4, 1]
answered Jan 3 at 2:56
JamesJames
13.8k11633
If you just want to remove the first duplicate, you can create set, append elements to a new list if they are in the set while removing each element from the set as well. When an element is not seen, append the rest of the list.
This is fairly efficient if the duplicate is seen early, but has O(1) if the element is seen late.
x = [2, 3, 4, 5, 3, 6, 4, 1]
s = set(x)
out =
for i,z in enumerate(x):
if z in s:
out.append(z)
s.remove(z)
else:
break
out += x[i+1:]
out
# returns:
[2, 3, 4, 5, 6, 4, 1]
answered Jan 3 at 2:56
JamesJames
13.8k11633
answered Jan 3 at 2:56
JamesJames
13.8k11633
answered Jan 3 at 2:56
JamesJames
13.8k11633
answered Jan 3 at 2:56
JamesJames
13.8k11633
13.8k11633
add a comment |
add a comment |
You could keep track of what you have already used and check if the value is in there.
lst = [2, 3, 4, 5, 3, 6, 4, 1]
used = set()
for i, x in enumerate(lst):
if x in used:
lst.pop(i)
break
used.add(x)
print(lst)
Also, don't give variables, or anything else, the same name as a keyword in python. input is already a built-in function.
answered Jan 3 at 2:53
All KnowerAll Knower
413214
It's better forusedto beset.
– dyukha
Jan 3 at 2:55
This is better than the original solution, but: 1. Iterating usingfor i in range(len())is not pythonic. 2.usedcould probably be aset.
– Tomothy32
Jan 3 at 2:55
Thanks, I guess the best alternative would beenumerate?
– All Knower
Jan 3 at 3:22
add a comment |
You could keep track of what you have already used and check if the value is in there.
lst = [2, 3, 4, 5, 3, 6, 4, 1]
used = set()
for i, x in enumerate(lst):
if x in used:
lst.pop(i)
break
used.add(x)
print(lst)
Also, don't give variables, or anything else, the same name as a keyword in python. input is already a built-in function.
answered Jan 3 at 2:53
All KnowerAll Knower
413214
It's better forusedto beset.
– dyukha
Jan 3 at 2:55
This is better than the original solution, but: 1. Iterating usingfor i in range(len())is not pythonic. 2.usedcould probably be aset.
– Tomothy32
Jan 3 at 2:55
Thanks, I guess the best alternative would beenumerate?
– All Knower
Jan 3 at 3:22
add a comment |
You could keep track of what you have already used and check if the value is in there.
lst = [2, 3, 4, 5, 3, 6, 4, 1]
used = set()
for i, x in enumerate(lst):
if x in used:
lst.pop(i)
break
used.add(x)
print(lst)
Also, don't give variables, or anything else, the same name as a keyword in python. input is already a built-in function.
answered Jan 3 at 2:53
All KnowerAll Knower
413214
You could keep track of what you have already used and check if the value is in there.
lst = [2, 3, 4, 5, 3, 6, 4, 1]
used = set()
for i, x in enumerate(lst):
if x in used:
lst.pop(i)
break
used.add(x)
print(lst)
Also, don't give variables, or anything else, the same name as a keyword in python. input is already a built-in function.
answered Jan 3 at 2:53
All KnowerAll Knower
413214
edited Jan 3 at 3:42
answered Jan 3 at 2:53
All KnowerAll Knower
413214
answered Jan 3 at 2:53
All KnowerAll Knower
413214
answered Jan 3 at 2:53
All KnowerAll Knower
413214
413214
It's better forusedto beset.
– dyukha
Jan 3 at 2:55
This is better than the original solution, but: 1. Iterating usingfor i in range(len())is not pythonic. 2.usedcould probably be aset.
– Tomothy32
Jan 3 at 2:55
Thanks, I guess the best alternative would beenumerate?
– All Knower
Jan 3 at 3:22
add a comment |
It's better forusedto beset.
– dyukha
Jan 3 at 2:55
This is better than the original solution, but: 1. Iterating usingfor i in range(len())is not pythonic. 2.usedcould probably be aset.
– Tomothy32
Jan 3 at 2:55
Thanks, I guess the best alternative would beenumerate?
– All Knower
Jan 3 at 3:22
It's better for
used to be set.– dyukha
Jan 3 at 2:55
It's better for
used to be set.– dyukha
Jan 3 at 2:55
This is better than the original solution, but: 1. Iterating using
for i in range(len()) is not pythonic. 2. used could probably be a set.– Tomothy32
Jan 3 at 2:55
This is better than the original solution, but: 1. Iterating using
for i in range(len()) is not pythonic. 2. used could probably be a set.– Tomothy32
Jan 3 at 2:55
Thanks, I guess the best alternative would be
enumerate?– All Knower
Jan 3 at 3:22
Thanks, I guess the best alternative would be
enumerate?– All Knower
Jan 3 at 3:22
add a comment |
This is my version of Chiheb Nexus' answer.
def find_dup(lst):
seen = set()
it = iter(lst)
for item in it:
if item not in seen:
seen.add(item)
yield item
else:
yield from it
lst = [2, 3, 4, 5, 3, 6, 4, 1]
list(find_dup(lst))
[2, 3, 4, 5, 6, 4, 1]
answered Jan 3 at 5:03
John La RooyJohn La Rooy
215k41279431
add a comment |
This is my version of Chiheb Nexus' answer.
def find_dup(lst):
seen = set()
it = iter(lst)
for item in it:
if item not in seen:
seen.add(item)
yield item
else:
yield from it
lst = [2, 3, 4, 5, 3, 6, 4, 1]
list(find_dup(lst))
[2, 3, 4, 5, 6, 4, 1]
answered Jan 3 at 5:03
John La RooyJohn La Rooy
215k41279431
add a comment |
This is my version of Chiheb Nexus' answer.
def find_dup(lst):
seen = set()
it = iter(lst)
for item in it:
if item not in seen:
seen.add(item)
yield item
else:
yield from it
lst = [2, 3, 4, 5, 3, 6, 4, 1]
list(find_dup(lst))
[2, 3, 4, 5, 6, 4, 1]
answered Jan 3 at 5:03
John La RooyJohn La Rooy
215k41279431
This is my version of Chiheb Nexus' answer.
def find_dup(lst):
seen = set()
it = iter(lst)
for item in it:
if item not in seen:
seen.add(item)
yield item
else:
yield from it
lst = [2, 3, 4, 5, 3, 6, 4, 1]
list(find_dup(lst))
[2, 3, 4, 5, 6, 4, 1]
answered Jan 3 at 5:03
John La RooyJohn La Rooy
215k41279431
answered Jan 3 at 5:03
John La RooyJohn La Rooy
215k41279431
answered Jan 3 at 5:03
John La RooyJohn La Rooy
215k41279431
answered Jan 3 at 5:03
John La RooyJohn La Rooy
215k41279431
215k41279431
add a comment |
add a comment |
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