Using Python's list index() method on a list of tuples or objects?

Question

Python's list type has an index() method that takes one parameter and returns the index of the first item in the list matching the parameter. For instance:

>>> some_list = ["apple", "pear", "banana", "grape"]
>>> some_list.index("pear")
1
>>> some_list.index("grape")
3

Is there a graceful (idiomatic) way to extend this to lists of complex objects, like tuples? Ideally, I'd like to be able to do something like this:

>>> tuple_list = [("pineapple", 5), ("cherry", 7), ("kumquat", 3), ("plum", 11)]
>>> some_list.getIndexOfTuple(1, 7)
1
>>> some_list.getIndexOfTuple(0, "kumquat")
2

getIndexOfTuple() is just a hypothetical method that accepts a sub-index and a value, and then returns the index of the list item with the given value at that sub-index. I hope

Is there some way to achieve that general result, using list comprehensions or lambas or something "in-line" like that? I think I could write my own class and method, but I don't want to reinvent the wheel if Python already has a way to do it.

Answer 1

How about this?

>>> tuple_list = [("pineapple", 5), ("cherry", 7), ("kumquat", 3), ("plum", 11)]
>>> [x for x, y in enumerate(tuple_list) if y[1] == 7]
[1]
>>> [x for x, y in enumerate(tuple_list) if y[0] == 'kumquat']
[2]

As pointed out in the comments, this would get all matches. To just get the first one, you can do:

>>> [y[0] for y in tuple_list].index('kumquat')
2

There is a good discussion in the comments as to the speed difference between all the solutions posted. I may be a little biased but I would personally stick to a one-liner as the speed we're talking about is pretty insignificant versus creating functions and importing modules for this problem, but if you are planning on doing this to a very large amount of elements you might want to look at the other answers provided, as they are faster than what I provided.

Answer 2

One possibility is to use the itemgetter function from the operator module:

import operator

f = operator.itemgetter(0)
print map(f, tuple_list).index("cherry") # yields 1

The call to itemgetter returns a function that will do the equivalent of foo[0] for anything passed to it. Using map, you then apply that function to each tuple, extracting the info into a new list, on which you then call index as normal.

map(f, tuple_list)

is equivalent to:

[f(tuple_list[0]), f(tuple_list[1]), ...etc]

which in turn is equivalent to:

[tuple_list[0][0], tuple_list[1][0], tuple_list[2][0]]

which gives:

["pineapple", "cherry", ...etc]

Answer 3

You can do this with a list comprehension and index()

tuple_list = [("pineapple", 5), ("cherry", 7), ("kumquat", 3), ("plum", 11)]
[x[0] for x in tuple_list].index("kumquat")
2
[x[1] for x in tuple_list].index(7)
1

Answer 4

Those list comprehensions are messy after a while.

I like this Pythonic approach:

from operator import itemgetter

def collect(l, index):
   return map(itemgetter(index), l)

# And now you can write this:
collect(tuple_list,0).index("cherry")   # = 1
collect(tuple_list,1).index("3")        # = 2

If you need your code to be all super performant:

# Stops iterating through the list as soon as it finds the value
def getIndexOfTuple(l, index, value):
    for pos,t in enumerate(l):
        if t[index] == value:
            return pos

    # Matches behavior of list.index
    raise ValueError("list.index(x): x not in list")

getIndexOfTuple(tuple_list, 0, "cherry")   # = 1