CODEWITHSUNDEEP
pythonpython-3.x
Rankinstudio
Rankinstudio

Reputation: 612

List of tuples remove duplicates by first index

I'm having a hard time with this one. I have a list of tuples but am only interested in removing duplicates based on the first item of the tuples.

I need to remove the duplicates from the individual lists by only looking at the first item in the tuple.

I need to compare the lists for duplicates where unique is in uniqueFound and only keep those from unique that are not in uniqueFound, also only interested in the first item of the tuple. The other numbers and values should not be considered. Here is what I came up with, and it's still passing N1 to the final list when it shouldn't. There has to be an easier way to do this?

#unique found
uniqueFound = [('N1', '20.3', 1, 58, 14),
         ('N2', '20.1', 1, 44, 14),
         ('N3', '21', 1, 23, 14),
         ('N1', '21', 1, 23, 14),
         ('N5', '33', 1, 34, 14)
         ]
unique = [('N0', '23', 2, 22, 1),
         ('N6', '33.1', 1, 25, 11),
         ('N3', '21', 1, 33, 12),
         ('N1', '44', 1, 2, 44),
         ('N7', '33.2', 4, 22, 12),
         ('N7', '3.2', 2, 21, 132)
         ]

#Remove duplicates
uniqueFound = dict((x[0], x) for x in uniqueFound).values()
unique = dict((x[0], x) for x in unique).values()

fmatches=[]
uniquefinal = []

for x in uniqueFound:
    for y in unique:
        if x[0] == y[0]:
            fmatches.append(y)
        else:
            continue
fmatches = dict((x[0], x) for x in fmatches).values()
for x in unique:
    for y in fmatches:
        if x[0] == y[0]:
            continue
        else:
            uniquefinal.append(x)
duplicatesrem = dict((x[0], x) for x in uniquefinal).values()
uniqueFinal = list(duplicatesrem)
print(uniqueFinal)

which yields

[('N0', '23', 2, 22, 1), ('N6', '33.1', 1, 25, 11), ('N3', '21', 1, 33, 12), ('N1', '44', 1, 2, 44), ('N7', '3.2', 2, 21, 132)]

What I'm going for (in unique but not in uniqueFound):

[('N0', '23', 2, 22, 1), ('N6', '33.1', 1, 25, 11), ('N7', '3.2', 2, 21, 132)]

Thanks for your help.

Upvotes: 1

Views: 128

Answers (3)

Mykola Zotko
Mykola Zotko

Reputation: 17864

Another solution that utilizes dictionaries:

found = {i[0] for i in uniqueFound}
dct = {k: v for k, *v in unique if k not in found}
[(k, *v) for k, v in dct.items()]
# [('N0', '23', 2, 22, 1), ('N6', '33.1', 1, 25, 11), ('N7', '3.2', 2, 23, 1)]

Upvotes: 1

RichieV
RichieV

Reputation: 5183

drop = {t[0] for t in uniqueFound} # set comprehension

# build a conditional list comprehension
uniqueFinal = [e for e in unique if e[0] not in drop]

Upvotes: 0

Jan Stránský
Jan Stránský

Reputation: 1691

uniqueFound = [('N1', '20.3', 1, 58, 14),
         ('N2', '20.1', 1, 44, 14),
         ('N3', '21', 1, 23, 14),
         ('N1', '21', 1, 23, 14),
         ('N5', '33', 1, 34, 14)
         ]
unique = [('N0', '23', 2, 22, 1),
         ('N6', '33.1', 1, 25, 11),
         ('N3', '21', 1, 33, 12),
         ('N1', '44', 1, 2, 44),
         ('N7', '33.2', 4, 22, 12),
         ('N7', '3.2', 2, 23, 1)
         ]

aux1 = set(e[0] for e in uniqueFound)
aux2 = set()
result = []
for e in unique:
    e0 = e[0]
    if e0 in aux1: continue
    if e0 in aux2: continue
    aux2.add(e0)
    result.append(e)
print(result)

aux1 is an auxiliary set containing "keys" from uniqueFound.

aux2 is an auxiliary set containing already used "keys"

if a "key" (e0) is found in aux1 or aux2, do nothing. Otherwise add "key" to aux2 and the element to the result

Upvotes: 2

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