統一列表項成對

Question

給出一個列表

[1,2,3,4,5,6] 

與2n元素

如何獲取列表

[1+2,3+4,5+6] 

與n元素？

Answer 1

從itertools recipes section：

def grouper(n, iterable, fillvalue=None): 
    "Collect data into fixed-length chunks or blocks" 
    # grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx 
    args = [iter(iterable)] * n 
    return izip_longest(fillvalue=fillvalue, *args) 

然後調用與：

for paired in grouper(2, inputlist): 
    # paired is a tuple of two elements from the inputlist at a time. 

grouper返回迭代;如果你有有一個列表，只需消耗迭代到一個新的列表：

newlist = list(grouper(2, inputlist)) 

Answer 2

a = [1,2,3,4,5,6] 
b = [a[i]+a[i+1] for i in xrange(0,len(a),2)] 

Answer 3

li = [1,2,10,20,100,200,2000,3000,2,2,3,3,5,5] 
print li 

it = iter(li) 
if len(li)%2==0: 
    print [x+it.next() for x in it] 

給

[1, 2, 10, 20, 100, 200, 2000, 3000, 2, 2, 3, 3, 5, 5] 
[3, 30, 300, 5000, 4, 6, 10]