Is there a way in Python to index a list of containers (tuples, lists, dictionaries) by an element of a container?

Question

I have been poking around for a recipe / example to index a list of tuples without taking a modification of the decorate, sort, undecorate approach.

For example:

l=[(a,b,c),(x,c,b),(z,c,b),(z,c,d),(a,d,d),(x,d,c) . . .]

The approach I have been using is to build a dictionary using defaultdict of the second element

from collections import defaultdict

tdict=defaultdict(int)

for myTuple in l:
    tdict[myTuple[1]]+=1

Then I have to build a list consisting of only the second item in the tuple for each item in the list. While there are a number of ways to get there a simple approach is to:

tempList=[myTuple[1] for myTuple in l]

and then generate an index of each item in tdict

indexDict=defaultdict(dict)
for key in tdict:
    indexDict[key]['index']=tempList.index(key)

Clearly this does not seem very Pythonic. I have been trying to find examples or insights thinking that I should be able to use something magical to get the index directly. No such luck so far.

Note, I understand that I can take my approach a little more directly and not generating tdict.

output could be a dictionary with the index

indexDict={'b':{'index':0},'c':{'index':1},'d':{'index':4},. . .}

After learning a lot from Nadia's responses I think the answer is no.

While her response works I think it is more complicated than needed. I would simply

 def build_index(someList):
    indexDict={}
    for item in enumerate(someList):
        if item[1][1] not in indexDict:
           indexDict[item[1][1]]=item[0]
    return indexDict

Answer 1

If i think this is what you're asking...

l = ['asd', 'asdxzc']
d = {}

for i, x in enumerate(l):
    d[x] = {'index': i}

Answer 2

This will generate the result you want

dict((myTuple[1], index) for index, myTuple in enumerate(l))

>>> l = [(1, 2, 3), (4, 5, 6), (1, 4, 6)]
>>> dict((myTuple[1], index) for index, myTuple in enumerate(l))
{2: 0, 4: 2, 5: 1}

And if you insist on using a dictionary to represent the index:

dict((myTuple[1], {'index': index}) for index, myTuple in enumerate(l))

The result will be:

{2: {'index': 0}, 4: {'index': 2}, 5: {'index': 1}}

---

EDIT If you want to handle key collision then you'll have to extend the solution like this:

def build_index(l):
    indexes = [(myTuple[1], index) for index, myTuple in enumerate(l)]
    d = {}
    for e, index in indexes:
        d[e] = min(index, d.get(e, index))
    return d

>>> l = [(1, 2, 3), (4, 5, 6), (1, 4, 6), (2, 4, 6)]
>>> build_index(l)
{2: 0, 4: 2, 5: 1}

---

EDIT 2

And a more generalized and compact solution (in a similar definition to sorted)

def index(l, key):
    d = {}
    for index, myTuple in enumerate(l):
        d[key(myTuple)] = min(index, d.get(key(myTuple), index))
    return d

>>> index(l, lambda a: a[1])
{2: 0, 4: 2, 5: 1}

So the answer to your question is yes: There is a way in Python to index a list of containers (tuples, lists, dictionaries) by an element of a container without preprocessing. But your request of storing the result in a dictionary makes it impossible to be a one liner. But there is no preprocessing here. The list is iterated only once.