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A simple way to convert a list of lists to a list of tuples is to start with an empty list. Then iterate over each list in the nested list in a simple for loop, convert it to a tuple using the tuple() function, and append it to the list of tuples.
Method #1 : Using map() + split() + tuple() This task can be achieved using the combination of these functions. The map function can be used to link the logic to each string, split function is used to split the inner contents of list to different tuple attributes and tuple function performs the task of forming a tuple.
Python list method list() takes sequence types and converts them to lists. This is used to convert a given tuple into list. Note − Tuple are very similar to lists with only difference that element values of a tuple can not be changed and tuple elements are put between parentheses instead of square bracket.
To convert a tuple to list in Python, use the list() method. The list() is a built-in Python method that takes a tuple as an argument and returns the list. The list() takes sequence types and converts them to lists.
>>> 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)
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).
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]))
Using slicing?
L = [1, "term1", 2, "term2", 3, "term3"]
L = zip(L[::2], L[1::2])
print L
[(L[i], L[i+1]) for i in xrange(0, len(L), 2)]
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')]
>>>
>>> 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')]
>>> L = [1, "term1", 3, "term2", 4, "term3", 5, "termN"]
>>> map(None,*[iter(L)]*2)
[(1, 'term1'), (3, 'term2'), (4, 'term3'), (5, 'termN')]
>>>
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)]
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