如何用两个'for'列举列表理解？

I am trying to do

我想做

ls = [myfunc(a,b,i) for a in a_list for b in b_list]

but also pass in i into myfunc, which is an index starting at 0 and incrementing for each new element.

但也将i传入myfunc，这是一个从0开始的索引，并为每个新元素递增。

For example:

例如：

a_list = 'abc'
b_list = 'def'

should result in

应该导致

ls = [myfunc('a','d',0),
      myfunc('a','e',1),
      myfunc('a','f',2),
      myfunc('b','d',3),
      myfunc('b','e',4),
      ...
      myfunc('c','f',8]

I know that I can use enumerate() for just the normal case, ie.

我知道我可以使用enumerate（）来表示正常情况，即。

ls = [myfunc(a,i) for a,i in enumerate(a_list)]

But I can't figure out how to do it cleanly when there are two fors. I couldn't find this question posted previously either.

但是，当有两个人时，我无法弄清楚如何干净利落地做到这一点。我以前找不到这个问题。

3 个解决方案

#1

You are creating a Cartesian product over two lists, so use itertools.product() instead of a double for loop. This gives you a single iterable you can easily add enumerate() to:

您正在两个列表上创建笛卡尔积，因此请使用itertools.product（）而不是double for循环。这为您提供了一个可迭代的，您可以轻松地将enumerate（）添加到：

from itertools import product

ls = [myfunc(a, b, i) for i, (a, b) in enumerate(product(a_list, b_list))]

For cases where you can't use product(), you'd put the multiple loops in a generator expression, then add enumerate() to that. Say you needed to filter some values of a_list:

对于无法使用product（）的情况，您可以将多个循环放在生成器表达式中，然后将enumerate（）添加到该表达式中。假设您需要过滤a_list的某些值：

gen = (a, b for a in a_list if some_filter(a) for b in b_list)
ls = [myfunc(a, b, i) for i, (a, b) in enumerate(gen)]

Another option is to add a separate counter; itertools.count() gives you a counter object that produces a new value with next():

另一种选择是增加一个单独的柜台; itertools.count（）为您提供一个计数器对象，该对象使用next（）生成一个新值：

from itertools import count

counter = count()
ls = [myfunc(a, b, next(counter)) 
      for a in a_list if some_filter(a)
      for b in b_list]

After all, in essence enumerate(iterable, start=0) is the equivalent of zip(itertools.count(start), iterable).

毕竟，实质上枚举（iterable，start = 0）相当于zip（itertools.count（start），iterable）。

#2

You can use enumerate on the sequence of pairs.

您可以在对序列上使用枚举。

ls = [myfunc(a,b,i) for (i,(a,b)) in
      enumerate((a,b) for a in a_list for b in b_list)]

#3

For a simple nested loop, use itertools.product() as @Martijn suggested.

对于一个简单的嵌套循环，请使用itertools.product（）作为@Martijn建议。

If the expression is more complex you can use itertools.count:

如果表达式更复杂，可以使用itertools.count：

i_gen = itertools.count()
ls = [myfunc(a, b, next(i_gen)) for a in a_list for b in b_list]

#1