CODEWITHSUNDEEP
pythonlisttuples
user3401408
user3401408

Reputation:

How to convert a list to a list of tuples?

I am newbie to Python and need to convert a list to dictionary. I know that we can convert a list of tuples to a dictionary.

This is the input list: 

L = [1,term1, 3, term2, x, term3,... z, termN]

and I want to convert this list to a list of tuples (or directly to a dictionary) like this:

[(1, term1), (3, term2), (x, term3), ...(z, termN)]

How can we do that easily in Python?

Upvotes: 67

Views: 49701

Answers (7)

Sahil Chhabra
Sahil Chhabra

Reputation: 11666

The below code will take care of both even and odd sized list :

[set(L[i:i+2]) for i in range(0, len(L),2)]

Upvotes: 2

thefourtheye
thefourtheye

Reputation: 239443

>>> L = [1, "term1", 3, "term2", 4, "term3", 5, "termN"]
# Create an iterator
>>> it = iter(L)
# zip the iterator with itself
>>> zip(it, it)
[(1, 'term1'), (3, 'term2'), (4, 'term3'), (5, 'termN')]

You want to group three items at a time?

>>> zip(it, it, it)

You want to group N items at a time?

# Create N copies of the same iterator
it = [iter(L)] * N
# Unpack the copies of the iterator, and pass them as parameters to zip
>>> zip(*it)

Upvotes: 138

perreal
perreal

Reputation: 97938

List directly into a dictionary using zip to pair consecutive even and odd elements:

m = [ 1, 2, 3, 4, 5, 6, 7, 8 ] 
d = { x : y for x, y in zip(m[::2], m[1::2]) }

or, since you are familiar with the tuple -> dict direction:

d = dict(t for t in zip(m[::2], m[1::2]))

even:

d = dict(zip(m[::2], m[1::2]))

Upvotes: 12

Nishant Nawarkhede
Nishant Nawarkhede

Reputation: 8400

Try this ,

>>> L = [1, "term1", 3, "term2", 4, "term3", 5, "termN"]
>>> it = iter(L)
>>> [(x, next(it)) for x in it ]
[(1, 'term1'), (3, 'term2'), (4, 'term3'), (5, 'termN')]
>>>

OR

>>> L = [1, "term1", 3, "term2", 4, "term3", 5, "termN"]
>>> [i for i in zip(*[iter(L)]*2)]
[(1, 'term1'), (3, 'term2'), (4, 'term3'), (5, 'termN')]

OR

>>> L = [1, "term1", 3, "term2", 4, "term3", 5, "termN"]
>>> map(None,*[iter(L)]*2)
[(1, 'term1'), (3, 'term2'), (4, 'term3'), (5, 'termN')]
>>>

Upvotes: 5

velis
velis

Reputation: 10025

Using slicing?

L = [1, "term1", 2, "term2", 3, "term3"]
L = zip(L[::2], L[1::2])

print L

Upvotes: 10

Xing Fei
Xing Fei

Reputation: 287

[(L[i], L[i+1]) for i in xrange(0, len(L), 2)]

Upvotes: 4

Pablo Francisco Pérez Hidalgo
Pablo Francisco Pérez Hidalgo

Reputation: 27692

Try with the group clustering idiom:

zip(*[iter(L)]*2)

From https://docs.python.org/2/library/functions.html:

The left-to-right evaluation order of the iterables is guaranteed. This makes possible an idiom for clustering a data series into n-length groups using zip(*[iter(s)]*n).

Upvotes: 18

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