Python using enumerate inside list comprehension

2019-01-07 02:43发布

Lets suppose I have a list like this:

mylist = ["a","b","c","d"]

To get the values printed along with their index I can use Python's enumerate function like this

>>> for i,j in enumerate(mylist):
...     print i,j
...
0 a
1 b
2 c
3 d
>>>

Now, when I try to use it inside a list comprehension it gives me this error

>>> [i,j for i,j in enumerate(mylist)]
  File "<stdin>", line 1
    [i,j for i,j in enumerate(mylist)]
           ^
SyntaxError: invalid syntax

So, my question is: what is the correct way of using enumerate inside list comprehension?

7条回答
Summer. ? 凉城
2楼-- · 2019-01-07 03:20

Try this:

[(i, j) for i, j in enumerate(mylist)]

You need to put i,j inside a tuple for the list comprehension to work. Alternatively, given that enumerate() already returns a tuple, you can return it directly without unpacking it first:

[pair for pair in enumerate(mylist)]

Either way, the result that gets returned is as expected:

> [(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd')]
查看更多
再贱就再见
3楼-- · 2019-01-07 03:21

Just to be really clear, this has nothing to do with enumerate and everything to do with list comprehension syntax.

This list comprehension returns a list of tuples:

[(i,j) for i in range(3) for j in 'abc']

this a list of dicts:

[{i:j} for i in range(3) for j in 'abc']

a list of lists:

[[i,j] for i in range(3) for j in 'abc']

a syntax error:

[i,j for i in range(3) for j in 'abc']

Which is inconsistent (IMHO) and confusing with dictionary comprehensions syntax:

>>> {i:j for i,j in enumerate('abcdef')}
{0: 'a', 1: 'b', 2: 'c', 3: 'd', 4: 'e', 5: 'f'}

And a set of tuples:

>>> {(i,j) for i,j in enumerate('abcdef')}
set([(0, 'a'), (4, 'e'), (1, 'b'), (2, 'c'), (5, 'f'), (3, 'd')])

As Óscar López stated, you can just pass the enumerate tuple directly:

>>> [t for t in enumerate('abcdef') ] 
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd'), (4, 'e'), (5, 'f')]
查看更多
放荡不羁爱自由
4楼-- · 2019-01-07 03:24

If you're using long lists, it appears the list comprehension's faster, not to mention more readable.

~$ python -mtimeit -s"mylist = ['a','b','c','d']" "list(enumerate(mylist))"
1000000 loops, best of 3: 1.61 usec per loop
~$ python -mtimeit -s"mylist = ['a','b','c','d']" "[(i, j) for i, j in enumerate(mylist)]"
1000000 loops, best of 3: 0.978 usec per loop
~$ python -mtimeit -s"mylist = ['a','b','c','d']" "[t for t in enumerate(mylist)]"
1000000 loops, best of 3: 0.767 usec per loop
查看更多
冷血范
5楼-- · 2019-01-07 03:24

All great answer guys. I know the question here is specific to enumeration but how about something like this, just another perspective

from itertools import izip, count
a = ["5", "6", "1", "2"]
tupleList = list( izip( count(), a ) )
print(tupleList)

It becomes more powerful, if one has to iterate multiple lists in parallel in terms of performance. Just a thought

a = ["5", "6", "1", "2"]
b = ["a", "b", "c", "d"]
tupleList = list( izip( count(), a, b ) )
print(tupleList)
查看更多
Deceive 欺骗
6楼-- · 2019-01-07 03:32

Or, if you don't insist on using a list comprehension:

>>> mylist = ["a","b","c","d"]
>>> list(enumerate(mylist))
[(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd')]
查看更多
爱情/是我丢掉的垃圾
7楼-- · 2019-01-07 03:33

Be explicit about the tuples.

[(i, j) for (i, j) in enumerate(mylist)]
查看更多
登录 后发表回答