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

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.

z = list(zip(*tuple_list))
z[1][z[0].index('persimon')]

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

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]

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.

No body suggest lambdas?

Y try this and works. I come to this post searching answer. I don't found that I like, but I feel a insingth :P

    l #[['rana', 1, 1], ['pato', 1, 1], ['perro', 1, 1]]

    map(lambda x:x[0], l).index("pato") #1 

Edit to add examples:

   l=[['rana', 1, 1], ['pato', 2, 1], ['perro', 1, 1], ['pato', 2, 2], ['pato', 2, 2]]

extract all items by condition: filter(lambda x:x[0]=="pato", l) #[['pato', 2, 1], ['pato', 2, 2], ['pato', 2, 2]]

extract all items by condition with index:

    >>> filter(lambda x:x[1][0]=="pato", enumerate(l))
    [(1, ['pato', 2, 1]), (3, ['pato', 2, 2]), (4, ['pato', 2, 2])]
    >>> map(lambda x:x[1],_)
    [['pato', 2, 1], ['pato', 2, 2], ['pato', 2, 2]]

Note:_ variable only works in interactive interpreter y normal text file _ need explicti assign, ie _=filter(lambda x:x[1][0]=="pato", enumerate(l))

Python's list.index(x) returns index of the first occurrence of x in the list. So we can pass objects returned by list compression to get their index.

>>> tuple_list = [("pineapple", 5), ("cherry", 7), ("kumquat", 3), ("plum", 11)]
>>> [tuple_list.index(t) for t in tuple_list if t[1] == 7]
[1]
>>> [tuple_list.index(t) for t in tuple_list if t[0] == 'kumquat']
[2]

With the same line, we can also get the list of index in case there are multiple matched elements.

>>> tuple_list = [("pineapple", 5), ("cherry", 7), ("kumquat", 3), ("plum", 11), ("banana", 7)]
>>> [tuple_list.index(t) for t in tuple_list if t[1] == 7]
[1, 4]

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','ddd')]

# 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

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)

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

Inspired by this question, I found this quite elegant:

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

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