我在Python中有两个迭代,我想成对地遍历它们:
foo = (1, 2, 3)
bar = (4, 5, 6)
for (f, b) in some_iterator(foo, bar):
print "f: ", f, "; b: ", b
它应该导致:
f: 1; b: 4
f: 2; b: 5
f: 3; b: 6
一种方法是迭代索引:
for i in xrange(len(foo)):
print "f: ", foo[i], "; b: ", b[i]
但这对我来说似乎有点不合时宜.有没有更好的方法呢？
什么是Pythonic方法来实现以下目标？
# Original lists:
list_a = [1, 2, 3, 4]
list_b = [5, 6, 7, 8]
# List of tuples from 'list_a' and 'list_b':
list_c = [(1,5), (2,6), (3,7), (4,8)]
每个成员list_c都是一个元组,其第一个成员来自list_a,而第二个来自list_b.
l1 = [4, 6, 8]
l2 = [a, b, c]
结果= [(4,a),(6,b),(8,c)]
我怎么做？