What is the most efficient way to search nested lists in python?

Question

I have a list that contains nested lists and I need to know the most efficient way to search within those nested lists.

e.g., if I have

[['a','b','c'],
['d','e','f']]

and I have to search the entire list above, what is the most efficient way to find 'd'?

Answer 1

>>> lis=[['a','b','c'],['d','e','f']]
>>> any('d' in x for x in lis)
True

generator expression using any

$ python -m timeit -s "lis=[['a','b','c'],['d','e','f'],[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]" "any('d' in x for x in lis)" 
1000000 loops, best of 3: 1.32 usec per loop

generator expression

$ python -m timeit -s "lis=[['a','b','c'],['d','e','f'],[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]" "'d' in (y for x in lis for y in x)"
100000 loops, best of 3: 1.56 usec per loop

list comprehension

$ python -m timeit -s "lis=[['a','b','c'],['d','e','f'],[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]" "'d' in [y for x in lis for y in x]"
100000 loops, best of 3: 3.23 usec per loop

How about if the item is near the end, or not present at all? any is faster than the list comprehension

$ python -m timeit -s "lis=[['a','b','c'],['d','e','f'],[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]"
    "'NOT THERE' in [y for x in lis for y in x]"
100000 loops, best of 3: 4.4 usec per loop

$ python -m timeit -s "lis=[['a','b','c'],['d','e','f'],[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]" 
    "any('NOT THERE' in x for x in lis)"
100000 loops, best of 3: 3.06 usec per loop

Perhaps if the list is 1000 times longer? any is still faster

$ python -m timeit -s "lis=1000*[['a','b','c'],['d','e','f'],[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]"
    "'NOT THERE' in [y for x in lis for y in x]"
100 loops, best of 3: 3.74 msec per loop
$ python -m timeit -s "lis=1000*[['a','b','c'],['d','e','f'],[1,2,3],[4,5,6],[7,8,9],[10,11,12],[13,14,15],[16,17,18]]" 
    "any('NOT THERE' in x for x in lis)"
100 loops, best of 3: 2.48 msec per loop

We know that generators take a while to set up, so the best chance for the LC to win is a very short list

$ python -m timeit -s "lis=[['a','b','c']]"
    "any('c' in x for x in lis)"
1000000 loops, best of 3: 1.12 usec per loop
$ python -m timeit -s "lis=[['a','b','c']]"
    "'c' in [y for x in lis for y in x]"
1000000 loops, best of 3: 0.611 usec per loop

And any uses less memory too

Answer 2

Using list comprehension, given:

mylist = [['a','b','c'],['d','e','f']]
'd' in [j for i in mylist for j in i]

yields:

True

and this could also be done with a generator (as shown by @AshwiniChaudhary)

Update based on comment below:

Here is the same list comprehension, but using more descriptive variable names:

'd' in [elem for sublist in mylist for elem in sublist]

The looping constructs in the list comprehension part is equivalent to

for sublist in mylist:
   for elem in sublist

and generates a list that where 'd' can be tested against with the in operator.