如何枚举（）Python中的元组列表？

I've got some code like this:

我有一些像这样的代码：

letters = [('a', 'A'), ('b', 'B')]
i = 0
for (lowercase, uppercase) in letters:
    print "Letter #%d is %s/%s" % (i, lowercase, uppercase)
    i += 1

I've been told that there's an enumerate() function that can take care of the "i" variable for me:

我被告知有一个enumerate（）函数可以为我处理“i”变量：

for i, l in enumerate(['a', 'b', 'c']):
    print "%d: %s" % (i, l)

However, I can't figure out how to combine the two: How do I use enumerate when the list in question is made of tuples? Do i have to do this?

但是，我无法弄清楚如何将两者结合起来：当有问题的列表是由元组组成时，如何使用枚举？我必须这样做吗？

letters = [('a', 'A'), ('b', 'B')]
for i, tuple in enumerate(letters):
    (lowercase, uppercase) = tuple
    print "Letter #%d is %s/%s" % (i, lowercase, uppercase)

Or is there a more elegant way?

还是有更优雅的方式？

3 个解决方案

#1

This is a neat way to do it:

这是一个巧妙的方法：

letters = [('a', 'A'), ('b', 'B')]
for i, (lowercase, uppercase) in enumerate(letters):
    print "Letter #%d is %s/%s" % (i, lowercase, uppercase)

#2

This is how I'd do it:

我就是这样做的：

import itertools

letters = [('a', 'A'), ('b', 'B')]
for i, lower, upper in zip(itertools.count(),*zip(*letters)):
    print "Letter #%d is %s/%s" % (i, lower, upper)

EDIT: unpacking becomes redundant. This is a more compact way, which might work or not depending on your use case:

编辑：拆包变得多余。这是一种更紧凑的方式，根据您的使用情况可能有效或无效：

import itertools

letters = [('a', 'A'), ('b', 'B')]
for i in zip(itertools.count(),*zip(*letters)):
    print "Letter #%d is %s/%s" % i

#3

You can do this way too:

你也可以这样做：

letters = [('a', 'A'), ('b', 'B')]
for i, letter in enumerate(letters):
    print "Letter #%d is %s/%s" % (i, letter[0], letter[1])

#1