Index of item in a list of lists of lists (Python)

Question

I am trying to find the index of a letter in the following list of lists of lists:

For example:

>>> alphabet = [[["A","B","C"],["D","E","F"],["G","H","I"]],[["J","K","L"],["M","N","O"],["P","Q","R"]],[["S","T","U"],["V","W","X"],["Y","Z","_"]]]
>>> find("H",alphabet)
(0,2,1)

What is the most Pythonic way of doing this?

@HarryE-W -- That really depends on what you're actually using the lists for ... In your example, a flat list seems like the way to go: alphabet = ["A","B","C","D",...] — mgilson
– mgilson, Commented Jan 2, 2013 at 15:02
@mgilson - I am just using it as a way of storing the alphabet, but I need to retrieve the index of a letter in the format I outlined in the example above. I am effectively putting the alphabet into a 3x3x3 cube and retrieving the x,y, and z index. — Harry E-W
– Harry E-W, Commented Jan 2, 2013 at 15:04
@HarryE-W: wouldn't it be easier to calculate the address from the ordinal? H is the 8th letter, calculate it's position into the cube from there? — Martijn Pieters
– Martijn Pieters, Commented Jan 2, 2013 at 15:08
@HarryE-W -- In that case, you can probably do it analytically -- which is basically the same thing that Martijn beat me to suggesting by a couple seconds ... — mgilson
– mgilson, Commented Jan 2, 2013 at 15:08

Gareth Latty · Accepted Answer · 2013-01-02 15:47:32Z

If you really want a solution that deals with this to any depth, this is the kind of thing you are looking for (as a simple recursive function):

def find_recursive(needle, haystack):
    for index, item in enumerate(haystack):
        if not isinstance(item, str):
            try:
                path = find_recursive(needle, item)
                if path is not None:
                    return (index, ) + path
            except TypeError:
                pass
        if needle == item:
            return index,
    return None

Edit: Just remembered, in 2.x, you want basestring to allow for unicode strings too - this solution is fine for 3.x users.

DSM · Accepted Answer · 2013-01-02 15:12:33Z

You could simply change the data structure and use a dict:

>>> import itertools
>>> import string
>>> lets = string.ascii_uppercase
>>> where = dict(zip(lets, itertools.product(range(3), repeat=3)))
>>> where
{'A': (0, 0, 0), 'C': (0, 0, 2), 'B': (0, 0, 1), 'E': (0, 1, 1), 'D': (0, 1, 0), 'G': (0, 2, 0), 'F': (0, 1, 2), 'I': (0, 2, 2), 'H': (0, 2, 1), 'K': (1, 0, 1), 'J': (1, 0, 0), 'M': (1, 1, 0), 'L': (1, 0, 2), 'O': (1, 1, 2), 'N': (1, 1, 1), 'Q': (1, 2, 1), 'P': (1, 2, 0), 'S': (2, 0, 0), 'R': (1, 2, 2), 'U': (2, 0, 2), 'T': (2, 0, 1), 'W': (2, 1, 1), 'V': (2, 1, 0), 'Y': (2, 2, 0), 'X': (2, 1, 2), 'Z': (2, 2, 1)}
>>> where["H"]
(0, 2, 1)

but note that I don't double the locations of the U to pad, and so

>>> where["U"]
(2, 0, 2)

Ashwini Chaudhary · Accepted Answer · 2013-01-02 15:10:23Z

3

In [9]: def find(val,lis):
        ind=[(j,i,k) for j,x in enumerate(lis) for i,y in enumerate(x) \
                                             for k,z in enumerate(y) if z==val]
        return ind[0] if ind else None
   ...: 

In [10]: find("H",alphabet)
Out[10]: (0, 2, 1)

In [14]: find("M",alphabet)
Out[14]: (1, 1, 0)

answered Jan 2, 2013 at 15:10

Ashwini Chaudhary

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1 Comment

jdotjdot Over a year ago

Would be even better if you could write it to take care of an indefinite number of nestings.

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