Python：将 for 循环与列表相关联答案

【问题标题】：Python: Associate for loop with a listPython：将 for 循环与列表相关联
【发布时间】：2013-12-15 21:44:22
【问题描述】：

我找不到任何解决方法。在我的真实示例中，这将用于将颜色蒙版与对象相关联。

我认为最好的解释方式是举个例子：

objects = ['pencil','pen','keyboard','table','phone']
colors  = ['red','green','blue']

for n, i in enumerate(objects):
    print n,i #0 pencil

    # print the index of colors

我需要的结果：

#0 pencil    # 0 red
#1 pen       # 1 green
#2 keyboard  # 2 blue
#3 table     # 0 red
#4 phone     # 1 green

所以，总会有 3 种颜色与对象相关联，如何在 python 的 for 循环中得到这个结果？每次 n （迭代器）大于颜色列表的长度时，我如何告诉它返回并打印第一个索引等等？

【问题讨论】：

标签： python list loops for-loop

【解决方案1】：

使用%:

for n, obj in enumerate(objects):
    print n, obj, colors[n % len(colors)]

或zip()和itertools.cycle()：

from itertools import cycle

for n, (obj, color) in enumerate(zip(objects, cycle(colors))):
    print n, obj, color

演示：

>>> objects = ['pencil','pen','keyboard','table','phone']
>>> colors  = ['red','green','blue']
>>> for n, obj in enumerate(objects):
...     print n, obj, colors[n % len(colors)]
... 
0 pencil red
1 pen green
2 keyboard blue
3 table red
4 phone green
>>> from itertools import cycle
>>> for n, (obj, color) in enumerate(zip(objects, cycle(colors))):
...     print n, obj, color
... 
0 pencil red
1 pen green
2 keyboard blue
3 table red
4 phone green

【讨论】：

【解决方案2】：

>>> from itertools import count, izip, cycle
>>> objects = ['pencil','pen','keyboard','table','phone']
>>> colors  = ['red','green','blue']
>>> for i, obj, color in izip(count(), objects, cycle(colors)):
...     print i, obj, color
... 
0 pencil red
1 pen green
2 keyboard blue
3 table red
4 phone green

或

>>> for i, obj, j, color in izip(count(), objects, cycle(range(len(colors))), cycle(colors)):
...     print i, obj, j, color
... 
0 pencil 0 red
1 pen 1 green
2 keyboard 2 blue
3 table 0 red
4 phone 1 green

【讨论】：