Python中的多元组到双组元组？

Question

What is the nicest way of splitting this:

tuple = ('a', 'b', 'c', 'd', 'e', 'f', 'g', 'h')

into this:

tuples = [('a', 'b'), ('c', 'd'), ('e', 'f'), ('g', 'h')]

Assuming that the input always has an even number of values.

Answer 1

zip() is your friend:

t = ('a', 'b', 'c', 'd', 'e', 'f', 'g', 'h')
zip(t[::2], t[1::2])

Answer 2

[(tuple[a], tuple[a+1]) for a in range(0,len(tuple),2)]

Answer 3

Or, using itertools (see the recipe for grouper ):

from itertools import izip
def group2(iterable):
   args = [iter(iterable)] * 2
   return izip(*args)

tuples = [ab for ab in group2(tuple)]

Answer 4

I present this code based on Peter Hoffmann's answer as a response to dfa's comment .

It is guaranteed to work whether or not your tuple has an even number of elements.

[(tup[i], tup[i+1]) for i in range(0, (len(tup)/2)*2, 2)]

The (len(tup)/2)*2 range parameter calculates the highest even number less or equal to the length of the tuple so it is guaranteed to work whether or not the tuple has an even number of elements.

The result of the method is going to be a list. This can be converted to tuples using the tuple() function.

Sample:

def inPairs(tup):
    return [(tup[i], tup[i+1]) for i in range(0, (len(tup)/2)*2, 2)]

# odd number of elements
print("Odd Set")
odd = range(5)
print(odd)
po = inPairs(odd)
print(po)

# even number of elements
print("Even Set")
even = range(4)
print(even)
pe = inPairs(even)
print(pe)

Output

Odd Set
[0, 1, 2, 3, 4]
[(0, 1), (2, 3)]
Even Set
[0, 1, 2, 3]
[(0, 1), (2, 3)]

Answer 5

Here's a general recipe for any-size chunk, if it might not always be 2:

def chunk(seq, n):
    return [seq[i:i+n] for i in range(0, len(seq), n)]

chunks= chunk(tuples, 2)

Or, if you enjoy iterators:

def iterchunk(iterable, n):
    it= iter(iterable)
    while True:
        chunk= []
        try:
            for i in range(n):
                chunk.append(it.next())
        except StopIteration:
            break
        finally:
            if len(chunk)!=0:
                yield tuple(chunk)