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?
3 Answers
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 (e.g. 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.
e.g.
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
Comments
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']
Comments
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])}
[a.index(e) for e in b]