PYTHON

python remove duplicates from list

mylist = ["a", "b", "b", "c", "a"]
mylist = sorted(set(mylist))
print(mylist)

remove duplicates from list python preserve order

list(dict.fromkeys(items))

python remove duplicates from list

# remove duplicate from given_list using list comprehension
res = []
[res.append(x) for x in given_list if x not in res]

python remove duplicates from a list

# HOW TO REMOVE DUPLICATES FROM A LIST:
# 1) CREATE A LIST
my_list = [1, 2, 3, 4, 5, 5, 5, 1]
# 2) CONVERT IT TO A SET AND THEN BACK INTO A LIST
my_list = list(set(my_list))
# 3) DONE! 
print(my_list) #WILL PRINT: [1, 2, 3, 4, 5]

how to make python remove the duplicates in list

  mylist = ["a", "b", "a", "c", "c"]
mylist = list(dict.fromkeys(mylist))

  print(mylist) 

remove duplicates python

mylist = ["a", "b", "a", "c", "c"]
mylist = list(dict.fromkeys(mylist))

remove duplicates function python

def remove_dupiclates(list_):
	new_list = []
	for a in list_:
    	if a not in new_list:
        	new_list.append(a)
	return new_list

python remove duplicates from list

bad_list = ["hi", 1, 2, 2, 3, 5, "hi", "hi"]
good_list = list(set(bad_list))

pytthon remove duplicates from list

ar = [1,2,1,2,1,3,2]
ar = list(sorted(set(ar)))
print(ar)

How to remove duplicates from a Python List

olist= ["a", "a", "b", "c", "d", "e", "f"] 

tlist = []
for x in range(0, len(olist)):
    intlist = False    
    for y in range(0, len(tlist)):
        if(olist[x] == tlist[y]):
            intlist = True
    
    if(intlist == False):
        tlist.append(olist[x])

print(tlist)
#['a', 'b', 'c', 'd', 'e', 'f', 1, 2]

remove duplicates from a list of lists python

k = [[1, 2], [4], [5, 6, 2], [1, 2], [3], [4]]
new_k = []
for elem in k:
    if elem not in new_k:
        new_k.append(elem)
k = new_k
print k
# prints [[1, 2], [4], [5, 6, 2], [3]]

python remove duplicates from list

# Python 3 code to demonstrate 
# removing duplicated from list 
# using naive methods 
  
# initializing list
test_list = [1, 3, 5, 6, 3, 5, 6, 1]
print ("The original list is : " +  str(test_list))
  
# using naive method
# to remove duplicated 
# from list 
res = []
for i in test_list:
    if i not in res:
        res.append(i)
  
# printing list after removal 
print ("The list after removing duplicates : " + str(res))

sort and remove duplicates list python

myList = sorted(set(myList))

how to eliminate duplicate values in list python

def dedupe(items):
    seen = set()
    for item in items:
        if item not in seen:
            yield item
            seen.add(item) 

a = [1, 5, 2, 1, 9, 1, 5, 10]
list(dedupe(a))# [1, 5, 2, 9, 10]

python remove duplicates

word = input().split()

for i in word:
  if word.count(i) > 1:
    word.remove(i)

python remove duplicates from list

''' we can convert the list to set and then back to list'''
a=[1,1,2,3,4,5,6,6,7]
'''b=(list(set(a))) # will have only unique elemenets'''

python remove duplicates

if mylist:
    mylist.sort()
    last = mylist[-1]
    for i in range(len(mylist)-2, -1, -1):
        if last == mylist[i]:
            del mylist[i]
        else:
            last = mylist[i]
# Quicker if all elements are hashables:
mylist = list(set(mylist))

python removing duplicate item

list_1 = [1, 2, 1, 4, 6]
list_2 = [7, 8, 2, 1]

print(list(set(list_1) ^ set(list_2)))

python remove duplicates from list

# get unique items in list aa with order maintained (python 3.7 and up)
list(dict.fromkeys(aa)) 

remove duplicates from list python

>>> list(dict.fromkeys('abracadabra'))
['a', 'b', 'r', 'c', 'd']

python list remove duplicates keep order

import pandas as pd

my_list = [0, 1, 2, 3, 4, 1, 2, 3, 5]

>>> pd.Series(my_list).drop_duplicates().tolist()
# Output:
# [0, 1, 2, 3, 4, 5]

remove duplicates from list

 ArrayList<Object> withDuplicateValues;
 HashSet<Object> withUniqueValue = new HashSet<>(withDuplicateValues);
 
 withDuplicateValues.clear();
 withDuplicateValues.addAll(withUniqueValue);

remove duplicate item on a list

    final stores = storeListFilterForSearch.map((e) => e.storeId).toSet();
    storeListFilterForSearch.retainWhere((x) => stores.remove(x.storeId));