CODEWITHSUNDEEP
pythonlist
Tapojyoti Mandal
Tapojyoti Mandal

Reputation: 882

Removing duplicates in list of lists

I have a list consisting of lists, and each sublist has 4 items(integers and floats) in it. My problem is that I want to remove those sublists whose index=1 and index=3 match with other sublists.

[[1, 2, 0, 50], [2, 19, 0, 25], [3, 12, 25, 0], [4, 18, 50, 50], [6, 19, 50, 67.45618854993529], [7, 4, 50, 49.49657024231138], [8, 12, 50, 41.65340802385248], [9, 12, 50, 47.80600357035001], [10, 18, 50, 47.80600357035001], [11, 18, 50, 53.222014760339356], [12, 18, 50, 55.667812693447615], [13, 12, 50, 41.65340802385248], [14, 12, 50, 47.80600357035001], [15, 13, 50, 47.80600357035001], [16, 3, 50, 49.49657024231138], [17, 3, 50, 49.49657024231138], [18, 4, 50, 49.49657024231138], [19, 5, 50, 49.49657024231138]]

For example,[7, 4, 50, 49.49657024231138] and [18, 4, 50, 49.49657024231138] have the same integers at index 1 and 3. So I want to remove one, which one doesn't matter.

I have looked at codes which allow me to do this on the basis of single index.

def unique_items(L):
found = set()
for item in L:
    if item[1] not in found:
        yield item
        found.add(item[1])

I have been using this code which allows me to remove lists but only on the basis of a single index.(I haven't really understood the code completely.But it is working.)

Hence, the problem is removing sublists only on the basis of duplicate values of index=1 and index=3 in the list of lists.

Upvotes: 0

Views: 72

Answers (3)

Saksham Varma
Saksham Varma

Reputation: 2130

This should work:

from pprint import pprint

d = {}
for sublist in lists:
    k = str(sublist[1]) + ',' + str(sublist[3])
    if k not in d:
        d[k] = sublist
pprint(d.values())

Upvotes: 0

Tim Pietzcker
Tim Pietzcker

Reputation: 336108

This is how you could make it work:

def unique_items(L):
    # Build a set to keep track of all the indices we've found so far
    found = set()  
    for item in L:
        # Now check if the 2nd and 4th index of the current item already are in the set
        if (item[1], item[3]) not in found: 
            # if it's new, then add its 2nd and 4th index as a tuple to our set
            found.add((item[1], item[3])
            # and give back the current item 
            # (I find this order more logical, but it doesn't matter much)
            yield item

Upvotes: 1

alexanderlukanin13
alexanderlukanin13

Reputation: 4715

If you need to compare (item[1], item[3]), use a tuple. Tuple is hashable type, so it can be used as a set member or dict key.

def unique_items(L):
    found = set()
    for item in L:
        key = (item[1], item[3])  # use tuple as key
        if key not in found:
            yield item
            found.add(key)

Upvotes: 3

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