问题
I faced some problem with solving the next problem:
We have a list of elements (integers), and we should return a list consisting of only the non-unique elements in this list. Without changing order of the list I think the best way is to delete or remove all unique element.
Take note that I just start to learn python and would like only the simplest solutions.
Here is my code:
def checkio(data):
for i in data:
if data.count(i) == 1: #if element seen in the list just ones, we delet this el
ind = data.index(i)
del data[ind]
return data
回答1:
Your function can be made to work by iterating over the list in reverse:
def checkio(data):
for index in range(len(data) - 1, -1, -1):
if data.count(data[index]) == 1:
del data[index]
return data
print(checkio([3, 3, 5, 8, 1, 4, 5, 2, 4, 4, 3, 0]))
[3, 3, 5, 4, 5, 4, 4, 3]
print(checkio([1, 2, 3, 4]))
[]
This works, because it only deletes numbers in the section of the list that has already been iterated over.
回答2:
You can implement a OrderedCounter
, eg:
from collections import OrderedDict, Counter
class OrderedCounter(Counter, OrderedDict):
pass
data = [1, 3, 1, 2, 3, 5, 8, 1, 5, 2]
duplicates = [k for k, v in OrderedCounter(data).items() if v > 1]
# [1, 3, 2, 5]
So you count the occurrence of each value, then filter on if it has a frequency of more than one. Inheriting from OrderedDict
means the order of the original elements is preserved.
Going by comments, you want all duplicated elements reserved, so you can pre-build a set of the duplicate entries, then re-iterate your original list, eg:
from collections import Counter
data = [1, 3, 1, 2, 3, 5, 8, 1, 5, 2]
duplicates = {k for k, v in Counter(data).items() if v > 1}
result = [el for el in data if el in duplicates]
# [1, 3, 1, 2, 3, 5, 1, 5, 2]
回答3:
Try this:
>>> a=[1,2,3,3,4,5,6,6,7,8,9,2,0,0]
>>> a=[i for i in a if a.count(i)>1]
>>> a
[2, 3, 3, 6, 6, 2, 0, 0]
>>> a=[1, 2, 3, 1, 3]
>>> a=[i for i in a if a.count(i)>1]
>>> a
[1, 3, 1, 3]
>>> a=[1, 2, 3, 4, 5]
>>> a=[i for i in a if a.count(i)>1]
a
[]
回答4:
Just I used list Comprehensions.
def checkio(data):
a=[i for i in data if data.count(i)>1]
return a
print checkio([1,1,2,2,1,1,1,3,4,5,6,7,8])
回答5:
def checkio(data):
lis = []
for i in data:
if data.count(i)>1:
lis.append(i)
print(lis)
checkio([1,2,3,3,2,1])
Yeah it's a bit late to contribute to this thread but just wanted to put it there on the net for anyone else use.
回答6:
Following what you have started, iterating on the list of integers, but not counting or deleting elements, try just testing if the element has already been seen, append it to a list of duplicated elements:
def checkio(data):
elements = []
duplicates = []
for i in data:
if i not in elements:
elements.append(i)
else:
if i not in duplicates:
duplicates.append(i)
return duplicates
d = [1, 3, 1, 2, 3, 5, 8, 1, 5, 2]
print (checkio(d))
#[1, 3, 5, 2]
回答7:
numbers = [1, 1, 1, 1, 3, 4, 9, 0, 1, 1, 1]
x=set(numbers)
print(x)
You can use the set key word too to get the desired solution.
回答8:
I used an integer and bool to check every time the list was modified within a while loop.
rechecks = 1
runscan = True
while runscan == True:
for i in data:
if data.count(i) <2:
data.remove(i)
rechecks+=1
#need to double check now
if rechecks >0:
runscan = True
rechecks-=1
else:
runscan = False
return data
来源:https://stackoverflow.com/questions/26870699/delete-unique-elements-from-a-list