Removing duplicates from list of lists in Python

Question

Can anyone suggest a good solution to remove duplicates from nested lists if wanting to evaluate duplicates based on first element of each nested list?

The main list looks like this:

L = [['14', '65', 76], ['2', '5', 6], ['7', '12', 33], ['14', '22', 46]]

If there is another list with the same element at first position [k][0] that had already occurred, then I'd like to remove that list and get this result:

L = [['14', '65', 76], ['2', '5', 6], ['7', '12', 33]]

Can you suggest an algorithm to achieve this goal?

Answer 1

Do you care about preserving order / which duplicate is removed? If not, then:

dict((x[0], x) for x in L).values()

will do it. If you want to preserve order, and want to keep the first one you find then:

def unique_items(L):
 found = set()
 for item in L:
 if item[0] not in found:
 yield item
 found.add(item[0])
print list(unique_items(L))

• Jiaaro: your conversion to a dict was so much more elegant than mind that I stole it :)
• xuiqzy: Doesn't the first one also preserve order because dicts preserve order since Python 3.7 and the keys are inserted in the order that the comprehension produces them?

Answer 2

use a dict instead like so:

L = {'14': ['65', 76], '2': ['5', 6], '7': ['12', 33]}
L['14'] = ['22', 46]

if you are receiving the first list from some external source, convert it like so:

L = [['14', '65', 76], ['2', '5', 6], ['7', '12', 33], ['14', '22', 46]]
L_dict = dict((x[0], x[1:]) for x in L)

Answer 3

Use Pandas :

import pandas as pd
L = [['14', '65', 76], ['2', '5', 6], ['7', '12', 33], ['14', '22', 46],['7','a','b']]
df = pd.DataFrame(L)
df = df.drop_duplicates()
L_no_duplicates = df.values.tolist()

If you want to drop duplicates in specific columns only use instead:

df = df.drop_duplicates([1,2])

Answer 4

i am not sure what you meant by "another list", so i assume you are saying those lists inside L

a=[]
L = [['14', '65', 76], ['2', '5', 6], ['7', '12', 33], ['14', '22', 46],['7','a','b']]
for item in L:
 if not item[0] in a:
 a.append(item[0])
 print item

• RichieHindle: This would be more efficient if you used a set for 'a' - you're O(N^2) using a list like that, and amortised O(N) using a set.
• ghostdog74: that has not come to mind, thanks for the info. nevertheless, that code works in older Python version that doesn't come with set. ;)

Answer 5

If the order does not matter, code below

print [ [k] + v for (k, v) in dict( [ [a[0], a[1:]] for a in reversed(L) ] ).items() ]

gives

[['2', '5', '6'], ['14', '65', '76'], ['7', '12', '33']]

Answer 6

def Remove(duplicate):
 final_list = []
 for num in duplicate:
 if num not in final_list:
 final_list.append(num)
 return final_list
duplicate = [2, 4, 10, 20, 5, 2, 20, 4]
print(Remove(duplicate))

• SLDem: Plase provide some comments about your code and changes you made with the original code.