CODEWITHSUNDEEP
pythonlist
miik
miik

Reputation: 663

list of ints into a list of tuples python

[1, 109, 2, 109, 2, 130, 2, 131, 2, 132, 3, 28, 3, 127] I have this array and I want to turn it into a list of tuples, each tuple has 2 values in. so this list becomes [(1, 109), (2, 109), (2, 130), (2, 131), (2, 132), (3, 28), (3, 127)]

Upvotes: 6

Views: 559

Answers (4)

eandersson
eandersson

Reputation: 26352

You can use zip combined with slicing to create a new list of tuples.

my_new_list = zip(my_list[0::2], my_list[1::2])

This would generate a new list with the following output

[(1, 109), (2, 109), (2, 130), (2, 131), (2, 132), (3, 28), (3, 127)]

The process behind this is quite simple. We first split the existing list into two new lists using slicing.

print my_list[0::2] # [1, 2, 2, 2, 2, 3, 3]
print my_list[1::2] # [109, 109, 130, 131, 132, 28, 127]

Then use zip to combine these two lists into one list of tuples.

print zip(my_list[0::2], my_list[1::2])

Upvotes: 12

dansalmo
dansalmo

Reputation: 11686

Inspired by ovgolovin's answer:

>>> it=iter(L)
>>> [(el, next(it)) for el in it]
[(1, 109), (2, 109), (2, 130), (2, 131), (2, 132), (3, 28), (3, 127)]

May be a little easier to see how it works. Can also reverse the tuple order:

>>> it=iter(L)
>>> [(next(it), el) for el in it]
[(109, 1), (109, 2), (130, 2), (131, 2), (132, 2), (28, 3), (127, 3)]

Upvotes: 0

ovgolovin
ovgolovin

Reputation: 13410

>>> it = iter(L)
>>> zip(*[it]*2)
[(1, 109), (2, 109), (2, 130), (2, 131), (2, 132), (3, 28), (3, 127)]

Explanation

it = iter(L) creates iterator on the initial list

[it]*2 crates the list consisting of the same iterator twice.

* used at the first place in the parameter is used to unpack the parameters, so that zip(*[it]*2) turns into zip(it,it). Though you can use zip(it,it), zip(*[it]*2) is a more general construction, which can be extended to any number of values in the resultant tuple.

The core of this approach is that zip on each iteration tries to get one value from each argument. But it turns out that all the arguments are effectively the same iterator, so it yields every time a new value.

Upvotes: 9

BrtH
BrtH

Reputation: 2664

Another approach:

lst = [1, 109, 2, 109, 2, 130, 2, 131, 2, 132, 3, 28, 3, 127]

print [(lst[i], lst[i + 1]) for i in xrange(0, len(lst), 2)]
#[(1, 109), (2, 109), (2, 130), (2, 131), (2, 132), (3, 28), (3, 127)]

Upvotes: 2

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