How to iterate through a string and find the duplicated values
My question is that I can't seem to find the right method for iterating through subjects2 and pick out the duplicated strings. Below is my method:
nosubjects =
subjects2 = ["hi","hi","bi","ki","si","bi","li"]
for i in subjects2:
if subjects2.count(i)==2:
nosubjects.extend(i)
print(nosubjects)
But when I print it out it appears like this:
['hi','hi']
['h', 'i', 'h', 'i','b', 'i']
['hi', 'i', 'h', 'i', 'b', 'i', 'b', 'i']
Please help thanks!
python arrays
edited Jan 19 at 6:27
sccoding
224115
asked Jan 19 at 5:54
QuestionQuestion
82
add a comment |
My question is that I can't seem to find the right method for iterating through subjects2 and pick out the duplicated strings. Below is my method:
nosubjects =
subjects2 = ["hi","hi","bi","ki","si","bi","li"]
for i in subjects2:
if subjects2.count(i)==2:
nosubjects.extend(i)
print(nosubjects)
But when I print it out it appears like this:
['hi','hi']
['h', 'i', 'h', 'i','b', 'i']
['hi', 'i', 'h', 'i', 'b', 'i', 'b', 'i']
Please help thanks!
python arrays
edited Jan 19 at 6:27
sccoding
224115
asked Jan 19 at 5:54
QuestionQuestion
82
add a comment |
My question is that I can't seem to find the right method for iterating through subjects2 and pick out the duplicated strings. Below is my method:
nosubjects =
subjects2 = ["hi","hi","bi","ki","si","bi","li"]
for i in subjects2:
if subjects2.count(i)==2:
nosubjects.extend(i)
print(nosubjects)
But when I print it out it appears like this:
['hi','hi']
['h', 'i', 'h', 'i','b', 'i']
['hi', 'i', 'h', 'i', 'b', 'i', 'b', 'i']
Please help thanks!
python arrays
edited Jan 19 at 6:27
sccoding
224115
asked Jan 19 at 5:54
QuestionQuestion
82
My question is that I can't seem to find the right method for iterating through subjects2 and pick out the duplicated strings. Below is my method:
nosubjects =
subjects2 = ["hi","hi","bi","ki","si","bi","li"]
for i in subjects2:
if subjects2.count(i)==2:
nosubjects.extend(i)
print(nosubjects)
But when I print it out it appears like this:
['hi','hi']
['h', 'i', 'h', 'i','b', 'i']
['hi', 'i', 'h', 'i', 'b', 'i', 'b', 'i']
Please help thanks!
python arrays
python arrays
edited Jan 19 at 6:27
sccoding
224115
asked Jan 19 at 5:54
QuestionQuestion
82
edited Jan 19 at 6:27
sccoding
224115
asked Jan 19 at 5:54
QuestionQuestion
82
edited Jan 19 at 6:27
sccoding
224115
edited Jan 19 at 6:27
sccoding
224115
edited Jan 19 at 6:27
sccoding
224115
224115
asked Jan 19 at 5:54
QuestionQuestion
82
asked Jan 19 at 5:54
QuestionQuestion
82
asked Jan 19 at 5:54
QuestionQuestion
82
82
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Problems in your code:
- You are trying to check the count of each element in the list, due to which duplicated elements will be checked multiple times.
- You are printing
nosubjectsinside the if condition which will cause it to be printed multiple times
Use sets. First to get unique set of elements in the list, then you can check if the count of each element in the set exceeds 1 in the original list.
nosubjects =
subjects2 = ['hi','hi','bi','ki','si','bi','li']
for i in set(subjects2):
if subjects2.count(i)>=2:
nosubjects.append(i)
print(nosubjects)
Using list comprehension:
subjects2 = ['hi','hi','bi','ki','si','bi','li']
nosubjects = [i for i in set(subjects2) if subjects2.count(i) >=2]
print(nosubjects)
answered Jan 19 at 5:57
JayJay
14.4k2163120
Can the downvoter, please leave a comment, so I can improve my answer.
– Jay
Jan 19 at 6:03
add a comment |
Use collections.Counter to get count of each element and take only those whose count exceeds 1:
from collections import Counter
subjects2 = ['hi', 'hi', 'bi', 'ki', 'si', 'bi', 'li']
nosubjects = [x for x, i in Counter(subjects2).items() if i > 1]
print(nosubjects)
# ['hi', 'bi']
answered Jan 19 at 6:04
AustinAustin
9,9763828
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Problems in your code:
- You are trying to check the count of each element in the list, due to which duplicated elements will be checked multiple times.
- You are printing
nosubjectsinside the if condition which will cause it to be printed multiple times
Use sets. First to get unique set of elements in the list, then you can check if the count of each element in the set exceeds 1 in the original list.
nosubjects =
subjects2 = ['hi','hi','bi','ki','si','bi','li']
for i in set(subjects2):
if subjects2.count(i)>=2:
nosubjects.append(i)
print(nosubjects)
Using list comprehension:
subjects2 = ['hi','hi','bi','ki','si','bi','li']
nosubjects = [i for i in set(subjects2) if subjects2.count(i) >=2]
print(nosubjects)
answered Jan 19 at 5:57
JayJay
14.4k2163120
Can the downvoter, please leave a comment, so I can improve my answer.
– Jay
Jan 19 at 6:03
add a comment |
Problems in your code:
- You are trying to check the count of each element in the list, due to which duplicated elements will be checked multiple times.
- You are printing
nosubjectsinside the if condition which will cause it to be printed multiple times
Use sets. First to get unique set of elements in the list, then you can check if the count of each element in the set exceeds 1 in the original list.
nosubjects =
subjects2 = ['hi','hi','bi','ki','si','bi','li']
for i in set(subjects2):
if subjects2.count(i)>=2:
nosubjects.append(i)
print(nosubjects)
Using list comprehension:
subjects2 = ['hi','hi','bi','ki','si','bi','li']
nosubjects = [i for i in set(subjects2) if subjects2.count(i) >=2]
print(nosubjects)
answered Jan 19 at 5:57
JayJay
14.4k2163120
Can the downvoter, please leave a comment, so I can improve my answer.
– Jay
Jan 19 at 6:03
add a comment |
Problems in your code:
- You are trying to check the count of each element in the list, due to which duplicated elements will be checked multiple times.
- You are printing
nosubjectsinside the if condition which will cause it to be printed multiple times
Use sets. First to get unique set of elements in the list, then you can check if the count of each element in the set exceeds 1 in the original list.
nosubjects =
subjects2 = ['hi','hi','bi','ki','si','bi','li']
for i in set(subjects2):
if subjects2.count(i)>=2:
nosubjects.append(i)
print(nosubjects)
Using list comprehension:
subjects2 = ['hi','hi','bi','ki','si','bi','li']
nosubjects = [i for i in set(subjects2) if subjects2.count(i) >=2]
print(nosubjects)
answered Jan 19 at 5:57
JayJay
14.4k2163120
Problems in your code:
- You are trying to check the count of each element in the list, due to which duplicated elements will be checked multiple times.
- You are printing
nosubjectsinside the if condition which will cause it to be printed multiple times
Use sets. First to get unique set of elements in the list, then you can check if the count of each element in the set exceeds 1 in the original list.
nosubjects =
subjects2 = ['hi','hi','bi','ki','si','bi','li']
for i in set(subjects2):
if subjects2.count(i)>=2:
nosubjects.append(i)
print(nosubjects)
Using list comprehension:
subjects2 = ['hi','hi','bi','ki','si','bi','li']
nosubjects = [i for i in set(subjects2) if subjects2.count(i) >=2]
print(nosubjects)
answered Jan 19 at 5:57
JayJay
14.4k2163120
edited Jan 19 at 6:05
answered Jan 19 at 5:57
JayJay
14.4k2163120
answered Jan 19 at 5:57
JayJay
14.4k2163120
answered Jan 19 at 5:57
JayJay
14.4k2163120
14.4k2163120
Can the downvoter, please leave a comment, so I can improve my answer.
– Jay
Jan 19 at 6:03
add a comment |
Can the downvoter, please leave a comment, so I can improve my answer.
– Jay
Jan 19 at 6:03
Can the downvoter, please leave a comment, so I can improve my answer.
– Jay
Jan 19 at 6:03
Can the downvoter, please leave a comment, so I can improve my answer.
– Jay
Jan 19 at 6:03
add a comment |
Use collections.Counter to get count of each element and take only those whose count exceeds 1:
from collections import Counter
subjects2 = ['hi', 'hi', 'bi', 'ki', 'si', 'bi', 'li']
nosubjects = [x for x, i in Counter(subjects2).items() if i > 1]
print(nosubjects)
# ['hi', 'bi']
answered Jan 19 at 6:04
AustinAustin
9,9763828
add a comment |
Use collections.Counter to get count of each element and take only those whose count exceeds 1:
from collections import Counter
subjects2 = ['hi', 'hi', 'bi', 'ki', 'si', 'bi', 'li']
nosubjects = [x for x, i in Counter(subjects2).items() if i > 1]
print(nosubjects)
# ['hi', 'bi']
answered Jan 19 at 6:04
AustinAustin
9,9763828
add a comment |
Use collections.Counter to get count of each element and take only those whose count exceeds 1:
from collections import Counter
subjects2 = ['hi', 'hi', 'bi', 'ki', 'si', 'bi', 'li']
nosubjects = [x for x, i in Counter(subjects2).items() if i > 1]
print(nosubjects)
# ['hi', 'bi']
answered Jan 19 at 6:04
AustinAustin
9,9763828
Use collections.Counter to get count of each element and take only those whose count exceeds 1:
from collections import Counter
subjects2 = ['hi', 'hi', 'bi', 'ki', 'si', 'bi', 'li']
nosubjects = [x for x, i in Counter(subjects2).items() if i > 1]
print(nosubjects)
# ['hi', 'bi']
answered Jan 19 at 6:04
AustinAustin
9,9763828
answered Jan 19 at 6:04
AustinAustin
9,9763828
answered Jan 19 at 6:04
AustinAustin
9,9763828
answered Jan 19 at 6:04
AustinAustin
9,9763828
9,9763828
add a comment |
add a comment |
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