迭代列表中的每?jī)蓚€(gè)元素我該怎么做for循環(huán)還是一個(gè)列表理解，這樣每次迭代就給了我兩個(gè)元素？l = [1,2,3,4,5,6]for i,k in ???:
print str(i), '+', str(k), '=', str(i+k)產(chǎn)出：1+2=3
3+4=7
5+6=11
3 回答
繁星淼淼
TA貢獻(xiàn)1775條經(jīng)驗(yàn) 獲得超11個(gè)贊
pairwise()grouped()
from itertools import izipdef pairwise(iterable): "s -> (s0, s1), (s2, s3), (s4, s5), ..." a = iter(iterable) return izip(a, a)for x, y in pairwise(l): print "%d + %d = %d" % (x, y, x + y)
from itertools import izipdef grouped(iterable, n): "s -> (s0,s1,s2,...sn-1), (sn,sn+1,sn+2,...s2n-1), (s2n,s2n+1,s2n+2,...s3n-1), ..." return izip(*[iter(iterable)]*n)for x, y in grouped(l, 2): print "%d + %d = %d" % (x, y, x + y)
izipzip()import.
N.B.pairwiseitertoolss -> (s0, s1), (s1, s2), (s2, s3), ...
吃雞游戲
TA貢獻(xiàn)1829條經(jīng)驗(yàn) 獲得超7個(gè)贊
data = [1,2,3,4,5,6]for i,k in zip(data[0::2], data[1::2]): print str(i), '+', str(k), '=', str(i+k)
data[0::2]意味著創(chuàng)建以下元素的子集集合 (index % 2 == 0)zip(x,y)從x和y集合創(chuàng)建一個(gè)元組集合，相同的索引元素。
守著一只汪
TA貢獻(xiàn)1872條經(jīng)驗(yàn) 獲得超4個(gè)贊
>>> l = [1,2,3,4,5,6]
>>> zip(l,l[1:])
[(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)]
>>> zip(l,l[1:])[::2]
[(1, 2), (3, 4), (5, 6)]
>>> [a+b for a,b in zip(l,l[1:])[::2]]
[3, 7, 11]
>>> ["%d + %d = %d" % (a,b,a+b) for a,b in zip(l,l[1:])[::2]]
['1 + 2 = 3', '3 + 4 = 7', '5 + 6 = 11']
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