Remove duplicates in tuple without using set() [closed]

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x = (1,1,-13,8,5,0,4,-1,-4)

a = filter(lambda i,j: i == j, x)

print(tuple(a))

I am getting an error instead of the correct answer (1,-13,8,5,0,4,-1,-4). What is my error?

Answer 1

x = (1,1,-13,8,5,0,4,-1,-4)
seen = []
answer = []
for elem in x:
    if elem not in seen:
        seen.append(elem)
        answer.append(elem)
print(tuple(answer))

Ouput:

(1, -13, 8, 5, 0, 4, -1, -4)

Answer 2

x = (1, 1, -13, 8, 5, 0, 4, -1, -4)

print(tuple([item for index, item in enumerate(x) if item not in x[:index]]))

Output:

(1, -13, 8, 5, 0, 4, -1, -4)

Answer 3

filter will iterate through x and pass each and every element to the lamdba function. But, it will be passing only one element at a time. So, the lambda function cannot accept two elements (unless the last one has a default value).

Apart from that, there are so many solutions without using set. For example, you can use collections.OrderedDict, like this

x = (1, 1, -13, 8, 5, 0, 4, -1, -4)
from collections import OrderedDict
print tuple(OrderedDict.fromkeys(x))
# (1, -13, 8, 5, 0, 4, -1, -4)

if the order of elements doesn't matter, you can use a normal dictionary itself, like this

print tuple({}.fromkeys(x))
# (0, 1, 4, 5, 8, -13, -4, -1)

Or you can use a temporary seen list, like this

x = (1, 1, -13, 8, 5, 0, 4, -1, -4)
seen, result = [], tuple()
for item in x:
    if item not in seen:
        seen.append(item)
        result += (item, )
print result
# (1, -13, 8, 5, 0, 4, -1, -4)

Answer 4

Assuming you may have duplicates anywhere in the list and not only consecutive repetitions, filter won't help you much.

You can use reduce with a custom function:

reduce(lambda acc, e: acc if e in acc else acc + (e, ), x, ())

Also if you wanted to remove successive repetitions only, it's pretty easy too:

reduce(lambda acc, e: acc if e in acc[-1:] else acc + (e, ), x, ())

Or hand-crafted code

rv = []
for i in x:
    if i not in rv:  # any repetition
    if i not in rv[-1:]  # only successive repetitions
        rv.append(i)
result = tuple(rv)

Just for the kicks, here's are a couple of answers that use filter:

map(lambda i: x[i], filter(lambda i: x[i] not in x[:i], range(len(x)))

[sub[0] for sub in filter(lambda sub: sub[0] != sub[1], [x[i:i+2] for i in range(len(x)-1)])]