using index() on multidimensional lists

Question

For a one dimensional list, the index of an item is found as follows:

 a_list = [\'a\', \'b\', \'new\', \'mpilgrim\', \'new\']
 a_list.index(\'mpilgrim\')


        

Answer 1

For n-dimensional recursive search, you can try something like this:

from copy import copy
def scope(word, list, indexes = None):
    result = []
    if not indexes:
        indexes = []
    for index, item in enumerate(list):
        try:
            current_index = indexes + [index]
            result.append(current_index + [item.index(word)])
        except ValueError:
            pass

        if type(item[0]) == type([]):
            indexes.append(index)
            result.extend(scope(word, item, copy(indexes)))

    return result

And the result is:

>>> d_list = [['a', 'b', 'new', 'mpilgrim', 'new'], [['a', 'b', 'new', 'mpilgrim', 'new'], ['b', 'd', 'new', 'mpilgrim', 'new']]]
>>> word = 'mpilgrim'
>>> result = scope(word, d_list)
[[0, 3], [1, 0, 3], [1, 1, 3]]

Probably there are better ways to do it, but that is the one I figured out without getting any library.

EDIT: Actually, it was not perfect and one library must be added. It's copy. Now it's ok.

Answer 2

For two dimensional list; you can iterate over rows and using .index function for looking for item:

def find(l, elem):
    for row, i in enumerate(l):
        try:
            column = i.index(elem)
        except ValueError:
            continue
        return row, column
    return -1

tl = [[1,2,3],[4,5,6],[7,8,9]]

print(find(tl, 6)) # (1,2)
print(find(tl, 1)) # (0,0)
print(find(tl, 9)) # (2,2)
print(find(tl, 12)) # -1

Answer 3

For multidimensional arrays:

def find(needle,haystack):
  if needle == haystack: return []
  # Strings are iterable, too
  if isinstance(haystack,str) and len(haystack)<=1: return None
  try:
    for i,e in enumerate(haystack):
      r = find(needle,e)
      if r is not None: 
        r.insert(0,i)
        return r
  except TypeError:
    pass
  return None    


ml = [[1,2,3],[4,5,6],[7,8,9]]
print find(2,ml)
ml = [3,[[1,2,3],[4,5,6],[7,8,9]]]
print find(2,ml)
ml = [[["ab", "bc", "cde"]]]
print find("d",ml)

There should be a better way to avoid the try/except block, but I could not find one: In Python, how do I determine if an object is iterable?

Answer 4

I don't know of an automatic way to do it, but if

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

and you want to find the location of 3, you can do:

x = [x for x in a if 3 in x][0]

print 'The index is (%d,%d)'%(a.index(x),x.index(3))

The output is:

The index is (1,0)

Answer 5

list_2d = [[1,2],[3,4],[5,6]]

element = 1

index_row = [list_2d.index(row) for row in list_2d if element in row]

index_column = [row.index(element) for row in list_2d if element in row]

Answer 6

You can use the following sample method too:

data = [[1, 1,2],[12,4],[6]]

def m_array_index(arr, searchItem):
    for i,x in enumerate(a):
        for j,y in enumerate(x):
            if y == searchItem:
                return i,j
    return -1,-1#not found

print m_array_index(data, 6)

Or with all occurrences(sure code could be optimized - modified to work with generators and so on - but here is just a sample):

occurrences = lambda arr, val: tuple((i,j) for i,x in enumerate(arr) for j,y in enumerate(x) if y == val) or ((-1,-1))

print occurrences(data, 1) # ((0, 0), (0, 1))
print occurrences(data, 12) # ((1, 0),)
print occurrences(data, 11) # (-1, -1)