Python, get index from list of lists

Question

I have a list of lists of strings, like this:

l = [['apple','banana','kiwi'],['chair','table','spoon']]

Given a string, I want its index in l. Experimenting with numpy, this is what I ended up with:

import numpy as np
l = [['apple','banana','kiwi'],['chair','table','spoon']]
def ind(s):
    i = [i for i in range(len(l)) if np.argwhere(np.array(l[i]) == s)][0]
    j = np.argwhere(np.array(l[i]) == s)[0][0]
    return i, j
s = ['apple','banana','kiwi','chair','table','spoon']
for val in s:
    try:
        print val, ind(val)
    except IndexError:
        print 'oops'

This fails for apple and chair, getting an indexerror. Also, this just looks bad to me. Is there some better approch to doing this?

Manny D · Accepted Answer · 2013-04-08 19:08:09Z

4

Returns a list of tuples containing (outer list index, inner list index), designed such that the item you're looking for can be in multiple inner lists:

l = [['apple','banana','kiwi'],['chair','table','spoon']]
def findItem(theList, item):
   return [(ind, theList[ind].index(item)) for ind in xrange(len(theList)) if item in theList[ind]]

findItem(l, 'apple') # [(0, 0)]
findItem(l, 'spoon') # [(1, 2)]

answered Apr 8, 2013 at 19:08

Manny D

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2 Comments

Carsten Over a year ago

For Python 3 you'll want to use range()

Carsten Over a year ago

Please note that for multiple occurrences of the string you're searching, you'll only find the first entry. I added an answer that finds all occurrences.

mtadd · Accepted Answer · 2013-04-08 19:17:43Z

1

If you want to use numpy, you don't need to roll your own:

import numpy as np
l = np.array([['apple','banana','kiwi'],['chair','table','spoon']])
s = ['apple','banana','kiwi','chair','table','spoon']

for a in s:
    arg = np.argwhere(l==a)
    print a, arg, tuple(arg[0]) if len(arg) else None

answered Apr 8, 2013 at 19:17

mtadd

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Comments

Rushy Panchal · Accepted Answer · 2013-04-08 19:01:05Z

0

l = [['apple','banana','kiwi'],['chair','table','spoon']]
def search(lst, item):
    for i in range(len(lst)):
        part = lst[i]
        for j in range(len(part)):
            if part[j] == item: return (i, j)
    return None

answered Apr 8, 2013 at 19:01

Rushy Panchal

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Comments

mgilson · Accepted Answer · 2013-04-08 19:02:02Z

I'd create a dictionary to map the items to their indices:

>>> import numpy as np
>>> l = [['apple','banana','kiwi'],['chair','table','spoon']]
>>> a = np.array(l,dtype=object)
>>> a
array([[apple, banana, kiwi],
       [chair, table, spoon]], dtype=object)
>>> d = {s:idx for (idx),s in np.ndenumerate(a)}
>>> d['apple']
(0, 0)
>>> d['chair']
(1, 0)

numpy + ndenumerate is nice for creating the index, but it's definitely not necessary. Of course, this is going to be most efficient if you can create the index once and then reuse it for subsequent searches.

zwol · Accepted Answer · 2013-04-08 19:03:10Z

0

One way is to make use of enumerate:

l = [['apple','banana','kiwi'],['chair','table','spoon']]
s = ['apple','banana','kiwi','chair','table','spoon']

for a in s:
    for i, ll in enumerate(l):
        for j, b in enumerate(ll):
            if a == b:
                print a, i, j

answered Apr 8, 2013 at 19:03

zwol

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Comments

KP. · Accepted Answer · 2013-04-08 19:25:28Z

In your line that computes i, you already have the answer if you apply argwhere to the entire list, rather than each sublist. There is no need to search again for j.

def ind(s):
    match = np.argwhere(np.array(l == s))
    if match:
        i, j = match[0]
    else:
        return -1, -1

This is will return the indeces of the first occurence of the string you're searching for.

Also, you might consider how this method is impacted as the complexity of the problem increases. This method will iterate over every element of your list, so the runtime cost increases as the list becomes bigger. So, if the number of test strings you're trying to find in the list also increases, you might want to think about using a dictionary to create a lookup table once, then subsequent searches for test strings are cheaper.

def make_lookup(search_list):
    lookup_table = {}
    for i, sublist in enumerate(list):
        for j, word in enumerate(sublist):
            lookup_table[word] = (i, j)
    return lookup_table

lookup_table = make_lookup(l)

def ind(s):
    if s in lookup_table:
        return lookup_table[s]
    else:
        return -1, -1

Mohammed Agboola · Accepted Answer · 2017-02-10 09:33:04Z

0

To get index of list of list in python:

theList = [[1,2,3], [4,5,6], [7,8,9]]
for i in range(len(theList)):
    if 5 in theList(i):
        print("[{0}][{1}]".format(i, theList[i].index(5))) #[1][1]

edited Feb 10, 2017 at 9:33

answered Feb 10, 2017 at 8:57

Mohammed Agboola

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1 Comment

bobsacameno Over a year ago

in the 3rd line should be theList[i] instead of theList(i)

Carsten · Accepted Answer · 2020-07-21 19:15:17Z

0

This solution will find all occurrences of the string you're searching for:

l = [['apple','banana','kiwi','apple'],['chair','table','spoon']]

def findItem(theList, item):
       return [(i, j) for i, line in enumerate(theList)
               for j, char in enumerate(line) if char == item]

findItem(l, 'apple') # [(0, 0), (0, 3)]
findItem(l, 'spoon') # [(1, 2)]

edited Jul 21, 2020 at 19:15

answered Jul 21, 2020 at 8:29

Carsten

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