Getting list index by matching strings list of lists

use in but in the sublists, not in the global list of lists. Combined with enumerate to get the index of the matching list(s)

p = [['a','b','c'],['x','y','z']]

idx = [i for i,sublist in enumerate(p) if 'a' in sublist]

that gives you a list of indexes. Here we have only one element:

>>> idx
[0]

You can use the same technique to compute the index of a in the sublist BTW (with idx = [i for i,sublist in enumerate(p) if 'a' in sublist])

To get a sole value or None, just iterate once on a generator comprehension:

idx = next((i for i,sublist in enumerate(p) if 'a' in sublist),None)

If you set the maximum nesting depth to 2 you can use the following code to get both indices.

p = [['a','b','c'],['x','y','z']]

def get_indices(p, search_for='a'):
    return [(i, sublist.index(search_for)) for i,sublist in enumerate(p) if search_for in sublist]

print('a', get_indices(p, search_for='a'))
print('b', get_indices(p, search_for='b'))
print('c', get_indices(p, search_for='c'))
print('x', get_indices(p, search_for='x'))
print('y', get_indices(p, search_for='y'))
print('z', get_indices(p, search_for='z'))
print('k', get_indices(p, search_for='k'))

output:

a [(0, 0)]
b [(0, 1)]
c [(0, 2)]
x [(1, 0)]
y [(1, 1)]
z [(1, 2)]
k []

Try this:

list_count = 0
for x in list:
   list_count += 1
   if 'a' in x:
       print ("Index in sublist" %(x.index('a')))
       print("Index of outer list" %(list_count))