CODEWITHSUNDEEP
pythonnested-lists
Sam Jarman
Sam Jarman

Reputation: 7377

Python: Return 2 ints for index in 2D lists given item

I've been tinkering in python this week and I got stuck on something.
If I had a 2D list like this:

myList = [[1,2],[3,4],[5,6]]

and I did this

>>>myList.index([3,4])

it would return

1

However, I want the index of something in side one of the lists, like this

    >>>myList.index(3)

and it would return 

1, 0

Is there anything that can do this?

Cheers

Upvotes: 16

Views: 38046

Answers (8)

ZWW
ZWW

Reputation: 1

Caution: The first answer up there:

def index_2d(myList, v):
    for i, x in enumerate(myList):
        if v in x:
            return (i, x.index(v))

will cause a lot of errors as long as there are any equal/identical elements, for any v in each x. Because x.index(v) only returns the first (leftmost) index of any element that has the same value of v, in x, NOT the actual index of the current v. That is, this only works if all elements in each row are unique.

Instead, try this to avoid non-unique values:

import numpy
arr_2D = numpy.array(...)
                  
for x, row in enumerate(arr_2D.tolist()):
    for y, cell in enumerate(row):
    cell_pos = (x, y)

Upvotes: 0

Ghislain Bugnicourt
Ghislain Bugnicourt

Reputation: 334

Based on kevpie's answer I managed to get a 2D list containing the coordinates of all occurences

myList = [[0,1],[1,1],[0,0],[1,0]]
coordsList = [[x, y] for x, li in enumerate(myList) for y, val in enumerate(li) if val==1]

Now coordsList contains all indexes for value 1 in myList :

[[0, 1], [1, 0], [1, 1], [3, 0]]

Upvotes: 0

Rishabh Pathak
Rishabh Pathak

Reputation: 1

Try this! this worked for me :)

def ret_pos(mylist,val_to_find):
    for i in (len(mylist)):
        for j in (len(i)):
            if mylist[i][j]== val_to_find:
                postn=[i,j]
    return(postn);

Upvotes: 0

kevpie
kevpie

Reputation: 26108

If you are doing many lookups you could create a mapping.

>>> myList = [[1,2],[3,4],[5,6]]
>>> d = dict( (j,(x, y)) for x, i in enumerate(myList) for y, j in enumerate(i) )
>>> d
{1: (0, 0), 2: (0, 1), 3: (1, 0), 4: (1, 1), 5: (2, 0), 6: (2, 1)}
>>> d[3]
(1, 0)

Upvotes: 5

jfs
jfs

Reputation: 414745

Using simple genexpr:

def index2d(list2d, value):
    return next((i, j) for i, lst in enumerate(list2d) 
                for j, x in enumerate(lst) if x == value)

Example

print index2d([[1,2],[3,4],[5,6]], 3)
# -> (1, 0)

Upvotes: 2

dawg
dawg

Reputation: 104082

def td(l,tgt):
    rtr=[]
    for sub in l:
        if tgt in sub:
            rtr.append(    (l.index(sub),sub.index(tgt))    )

    return rtr        


myList = [[1,2],[3,4],[5,6]]

print td(myList,3)

This will return more than one instance of the sub list, if any.

Upvotes: 0

Mark Byers
Mark Byers

Reputation: 838974

Try this:

def index_2d(myList, v):
    for i, x in enumerate(myList):
        if v in x:
            return (i, x.index(v))

Usage:

>>> index_2d(myList, 3)
(1, 0)

Upvotes: 26

Omnifarious
Omnifarious

Reputation: 56078

There is nothing that does this already, unless it's in numpy, which I don't know much about. This means you'll have to write code that does it. And that means questions like "What does [[1, 2], [2, 3], [3, 4]].index(3) return?" are important.

Upvotes: 1

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