如何在嵌套列表中找到元素的索引？

Question

I have the following lists:

a = ["a", "b", "c", "d", "e", "f"]
b = [a[5], a[4], a[3]]

If I use b.index("f") I get 0. What I want it to output, however, is 5. How can I get the index of "f" in list a through the list b?

Answer 1

You can't, because the elements in a are strings which don't "know" where they are in the list. Hence when you index them out of the list (eg a[5] ), the strings can't tell you where they came from in the list.

---

I'm not sure what your aim is through creating this new list, but you could just store the indexes of the elements rather than the elements themselves in b .

eg

b = [5, 4, 3]

which would allow you to then create a function which would "get the index of [an element] in list a through list b :

def get_ind_in_a_thru_b(e):
    i = a.index(e)
    if i in b: return i
    else: raise ValueError

which works as if there were some magical method to get the index of elements in b as if they were in a :

>>> get_ind_in_a_thru_b('f')
5
>>> get_ind_in_a_thru_b('g')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 2, in get_ind_in_a_thru_b
ValueError: 'g' is not in list
>>> get_ind_in_a_thru_b('a')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 4, in get_ind_in_a_thru_b
ValueError

note how even though 'a' is in the list a , since it is not in the list b , it is not returned

Answer 2

This isn't possible. Elements of b are resolved to strings and lose all knowledge of their indices in a . You can write a small class to store the value and index:

from operator import attrgetter

a = ["a", "b", "c", "d", "e", "f"]

class GetItemPos():
    def __init__(self, L, idx):
        self.idx = idx
        self.var = L[idx]

b = [GetItemPos(a, 5), GetItemPos(a, 4), GetItemPos(a, 3)]

indices = list(map(attrgetter('idx'), b))  # [5, 4, 3]
values = list(map(attrgetter('var'), b))   # ['f', 'e', 'd']

Answer 3

In this way, b would be ["f", "e", "d"] and the indexes of that elements are 0,1,2. You can create b in this way, though:

b = [a.index(a[5]), a.index(a[4]), a.index(a[3])]

Otherwise, if you want indexes and values you can use a dictionary:

b = {a[3]: a.index(a[3]), a[4]: a.index(a[4]),a[3]: a.index(a[3])}