python - 在python中一次迭代列表的两个值

Question

我有一个像 (669256.02, 6117662.09, 669258.61, 6117664.39, 669258.05, 6117665.08) 这样的集合，我需要对其进行迭代，比如

    for x,y in (669256.02, 6117662.09, 669258.61, 6117664.39, 669258.05, 6117665.08)
        print (x,y)

这将打印

    669256.02 6117662.09
    669258.61 6117664.39
    669258.05 6117665.08

我在 Python 3.3 顺便说一句

Answer 1

您可以使用迭代器：

>>> lis = (669256.02, 6117662.09, 669258.61, 6117664.39, 669258.05, 6117665.08)
>>> it = iter(lis)
>>> for x in it:
...     print (x, next(it))
...     
669256.02 6117662.09
669258.61 6117664.39
669258.05 6117665.08

Answer 2

>>> nums = (669256.02, 6117662.09, 669258.61, 6117664.39, 669258.05, 6117665.08)
>>> for x, y in zip(*[iter(nums)]*2):
        print(x, y)


669256.02 6117662.09
669258.61 6117664.39
669258.05 6117665.08

Answer 3

食谱部分 中的grouper示例应该在这里对您有所帮助：http: //docs.python.org/library/itertools.html#itertools-recipesitertools

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

然后你会像这样使用它：

for x, y in grouper(my_set, 2, 0.0):  # Use 0.0 to pad with a float
    print(x, y)