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:

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

I would place this as a comment to Triptych, but I can't comment yet due to lack of rating:

Using the enumerator method to match on sub-indices in a list of tuples. e.g.

li = [(1,2,3,4), (11,22,33,44), (111,222,333,444), ('a','b','c','d'),
        ('aa','bb','cc','dd'), ('aaa','bbb','ccc','ffffd')]

# want pos of item having [22,44] in positions 1 and 3:

def getIndexOfTupleWithIndices(li, indices, vals):

    # if index is a tuple of subindices to match against:
    for pos,k in enumerate(li):
        match = True
        for i in indices:
            if k[i] != vals[i]:
                match = False
                break;
        if (match):
            return pos

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

idx = [1,3]
vals = [22,44]
print getIndexOfTupleWithIndices(li,idx,vals)    # = 1
idx = [0,1]
vals = ['a','b']
print getIndexOfTupleWithIndices(li,idx,vals)    # = 3
idx = [2,1]
vals = ['cc','bb']
print getIndexOfTupleWithIndices(li,idx,vals)    # = 4

Answer 4

ok, it might be a mistake in vals(j), the correction is:

def getIndex(li,indices,vals):
for pos,k in enumerate(lista):
    match = True
    for i in indices:
        if k[i] != vals[indices.index(i)]:
            match = False
            break
    if(match):
        return pos

Answer 5

tuple_list = [("pineapple", 5), ("cherry", 7), ("kumquat", 3), ("plum", 11)]

def eachtuple(tupple, pos1, val):
    for e in tupple:
        if e == val:
            return True

for e in tuple_list:
    if eachtuple(e, 1, 7) is True:
        print tuple_list.index(e)

for e in tuple_list:
    if eachtuple(e, 0, "kumquat") is True:
        print tuple_list.index(e)

Answer 6

I guess the following is not the best way to do it (speed and elegance concerns) but well, it could help :

from collections import OrderedDict as od
t = [('pineapple', 5), ('cherry', 7), ('kumquat', 3), ('plum', 11)]
list(od(t).keys()).index('kumquat')
2
list(od(t).values()).index(7)
7
# bonus :
od(t)['kumquat']
3

list of tuples with 2 members can be converted to ordered dict directly, data structures are actually the same, so we can use dict method on the fly.