在python中一次迭代列表的两个值
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iterating over two values of a list at a time in python
提问by Rasmus Damgaard Nielsen
I have a set like (669256.02, 6117662.09, 669258.61, 6117664.39, 669258.05, 6117665.08) which I need to iterate over, like
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，例如
for x,y in (669256.02, 6117662.09, 669258.61, 6117664.39, 669258.05, 6117665.08)
print (x,y)
which would print
哪个会打印
669256.02 6117662.09
669258.61 6117664.39
669258.05 6117665.08
im on Python 3.3 btw
我在 Python 3.3 上顺便说一句
采纳答案by Ashwini Chaudhary
You can use an iterator:
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
回答by jamylak
>>> 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
回答by Lllama
The grouperexample in the itertoolsrecipes section should help you here:
http://docs.python.org/library/itertools.html#itertools-recipes
食谱部分中的grouper示例itertools应该可以帮助您：http:
//docs.python.org/library/itertools.html#itertools-recipes
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)
You would then it use like this:
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)
