how to get [(1, 2, 3, 4), (5, 6, 7, 8)] in my code using python

this is my code :

a = [(1,2),(5,6)]
b = [(3,4),(7,8)]

print zip(a,b)

and it show :

[((1, 2), (3, 4)), ((5开发者_开发技巧, 6), (7, 8))]

but i want get :

[(1, 2, 3, 4), (5, 6, 7, 8)]

so what can i do ,

thanks

Try this:

[aa+bb for aa,bb in zip(a,b)]

>>> a = [(1,2),(5,6)]
>>> b = [(3,4),(7,8)]
>>> [x+y for x,y in zip(a,b)]
[(1, 2, 3, 4), (5, 6, 7, 8)]
>>> 

you can use the following list comprehension:

[x + y for (i, x) in enumerate(a) for (j, y) in enumerate(b) if i == j]

Result:

[(1, 2, 3, 4), (5, 6, 7, 8)]

Update #2

You can also use this list comprehension for your specific input:

[a[0] + b[0], a[1] + b[1]]

which is obviously faster than the zip version.

from timeit import Timer

s1 = """\
a = [(1, 2), (5, 6)]
b = [(3, 4), (7, 8)]

[x + y for (i, x) in enumerate(a) for (j, y) in enumerate(b) if i == j]
"""

s2 = """\
a = [(1, 2), (5, 6)]
b = [(3, 4), (7, 8)]

[x + y for x, y in zip(a,b)]
"""

s3 = """\
a = [(1, 2), (5, 6)]
b = [(3, 4), (7, 8)]

[a[0] + b[0], a[1] + b[1]]
"""
t1 = Timer(s1)
t2 = Timer(s2)
t3 = Timer(s3)

print ("non-zip: {0} | zip: {1} | list-concat: {2}".format(t1.timeit(), t2.timeit(), t3.timeit()))

Results (Python 2.6.6 on Linux x86-64):

non-zip: 2.17631602287 | zip: 1.20438694954 | list-concat: 0.658749103546