使用 Python 中的索引迭代列表

Question

I could swear I've seen the function (or method) that takes a list, like this [3, 7, 19] and makes it into iterable list of tuples, like so: [(0,3), (1,7), (2,19)] to use it instead of:

for i in range(len(name_of_list)):
    name_of_list[i] = something

but I can't remember the name and googling "iterate list" gets nothing.

Answer 1

>>> a = [3,4,5,6]
>>> for i, val in enumerate(a):
...     print i, val
...
0 3
1 4
2 5
3 6
>>>

Answer 2

Yep, that would be the enumerate function! Or more to the point, you need to do:

list(enumerate([3,7,19]))

[(0, 3), (1, 7), (2, 19)]

Answer 3

Here's another using the zip function.

>>> a = [3, 7, 19]
>>> zip(range(len(a)), a)
[(0, 3), (1, 7), (2, 19)]

Answer 4

Here it is a solution using map function:

>>> a = [3, 7, 19]
>>> map(lambda x: (x, a[x]), range(len(a)))
[(0, 3), (1, 7), (2, 19)]

And a solution using list comprehensions:

>>> a = [3,7,19]
>>> [(x, a[x]) for x in range(len(a))]
[(0, 3), (1, 7), (2, 19)]

Answer 5

python enumerate function will be satisfied your requirements

result = list(enumerate([1,3,7,12]))
print result

output

[(0, 1), (1, 3), (2, 7),(3,12)]

Answer 6

If you have multiple lists, you can do this combining enumerate and zip :

list1 = [1, 2, 3, 4, 5]
list2 = [10, 20, 30, 40, 50]
list3 = [100, 200, 300, 400, 500]
for i, (l1, l2, l3) in enumerate(zip(list1, list2, list3)):
    print(i, l1, l2, l3)

Output:

 0 1 10 100 1 2 20 200 2 3 30 300 3 4 40 400 4 5 50 500

Note that parenthesis is required after i . Otherwise you get the error: ValueError: need more than 2 values to unpack

Answer 7

You are looking python enumerate function

Do as below

something = list(enumerate([1,3,7,12])) print something

output [(0, 1), (1, 3), (2, 7),(3,12)]