How to group a list into inclusive sequential n-tu

问题:

Possible Duplicate:
Iterate a list as pair (current, next) in Python

I have a list like this:

list = [A, B, C, D, E, F, G]

How can I group this to get the following Python output

[(A,B), (B,C), (C, D), (D,E), (E,F), (F,G)]

So the values are grouped by the secound value but the order is preserved...

回答1:

Try using zip():

zip(lst, lst[1:])

Also, you shouldn't use the name list, as you will override the built-in list type.

Example:

>>> lst = ['A', 'B', 'C', 'D', 'E', 'F', 'G']
>>> zip(lst, lst[1:])
[('A', 'B'), ('B', 'C'), ('C', 'D'), ('D', 'E'), ('E', 'F'), ('F', 'G')]

For a version that will work with generators or other one-pass iterables, you can use the pairwise recipe from the itertools docs:

from itertools import tee, izip

def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return izip(a, b)

回答2:

>>> a = [1,2,3,4,5]
>>> zip(a, a[1:])
<zip object at 0x7fe6e905ab90>
>>> list(zip(a, a[1:]))
[(1, 2), (2, 3), (3, 4), (4, 5)]

回答3:

use a list comprehension that iterates from 0 to length - 1

print [(a[i],a[i+1]) for i in range(len(a)-1)]

or even better , just use zip

print zip (a,a[1:])