How to check whether all elements in a list of lists are strings

Question

Suppose I have a list of list of strings like this:

l=[['a','kl_hg', 'FOO'],['b', 'kl_c', 'po']]

Now I would like to use an if command as follows (in pseudo-code!):

if allElementsOf(l).isString():
#do something

From this question I learned how to check a single variable whether it is a string. For a single list I could therefore do:

dummyL = ['a','kl_hg', 'FOO']
if all(isinstance(s, basestring) for s in dummyL):
#do something

and for my actual list l I could then do:

if all(isinstance(el, basestring) for sl in l for el in sl):
#do something

Is that the way to do it or is there a faster solution since that takes some time for huge lists of lists?

How could you possibly beat `O(m*n)`? You need to check each element in each list in the list! `all` already gives you short-cutting (i.e. the first element it finds that *isn't* a string will end the loop). — jonrsharpe, Jul 14 '15 at 17:10
For large lists `chain.from_iterable(l)` might be a bit faster but the complexity is not going to change — Padraic Cunningham, Jul 14 '15 at 17:13
@PadraicCunningham: Thanks for the suggestion. I guess the solution you ha din mind looks like the one danihp posted?! — Cleb, Jul 14 '15 at 17:22
sorry @PadraicCunningham, I have seen your comment after post the answer! — dani herrera, Jul 14 '15 at 17:24
@jonrsharpe: I often thought this and then found a much more efficient solution. :) — Cleb, Jul 14 '15 at 17:25
@Cleb, yes, it is the fastest way in python to flatten large lists, that is about as good as you are going to get , your worst case is always going to be `O(n*m)` and there is not really much you can do about that. — Padraic Cunningham, Jul 14 '15 at 17:25
@Cleb it might have lower fixed costs, but the algorithmic complexity is the same. — jonrsharpe, Jul 14 '15 at 17:32

score 3 · Accepted Answer · edited May 23 '17 at 12:01

3

Your approach is right, any flatting list short cut seems slowest. A fastest way may be use itertools:

import itertools
l=[['a','kl_hg', 'FOO'],['b', 'kl_c', 'po']]
if all( isinstance(x, basestring) for x in  itertools.chain.from_iterable(l) ):
    ...

edited May 23 '17 at 12:01

Community

1
1

answered Jul 14 '15 at 17:14

dani herrera

48,760
8
117
177

1

Nice, that works fine and seems faster than my approach. I upvote it and might accept it later (just in case something even better shows up). Thanks! – Cleb Jul 14 '15 at 17:19

felipsmartins · Answer 2 · 2015-07-16T01:59:11.180

3

It's odd how anyone has told any() built-in function:

seq = [['a','kl_hg', 'FOO'], ['b', 'kl_c', 'po', 13]]

def all_string(_iterable):    
    return not any([not isinstance(n, basestring) for i in _iterable 
                        for n in i])

all_string(seq) # returns False

The advantage when using any() function is that it does not evaluate the whole sequence, it returns when the first True value is found - In contrast to all().

edited Jul 16 '15 at 01:59

answered Jul 14 '15 at 17:55

felipsmartins

13,269
4
48
56

Works fine as well! Thanks for the alternative solution, I upvote your answer as well. – Cleb Jul 15 '15 at 08:19

Andru Luvisi · Answer 3 · 2015-07-14T19:50:31.977

1

You probably want to use recursion to solve this in the general case, for any level of nesting. For example:

def all_strings(thing):
    if isinstance(thing, str):
        return True
    elif isinstance(thing, list):
        for subthing in thing:
            if not all_strings(subthing):
                return False
        return True
    else:
        return False

>>> print all_strings('foo')
True
>>> print all_strings(['foo'])
True
>>> print all_strings(['foo',['foo']])
True
>>> print all_strings(['foo',[1, 'foo']])
False
>>>

edited Jul 14 '15 at 19:50

answered Jul 14 '15 at 17:16

Andru Luvisi

24,367
6
53
66

Don't remember why I did not upvote this one; just came back to this old question and your solution works perfectly fine. Upvoted it now. – Cleb Oct 31 '16 at 16:06

score 0 · Answer 4 · answered Jul 15 '15 at 18:15

0

You can use this:

for x in l:
    for a in range(3):
        if type((x[a])) == str:
            print(x[a], ' is a string')

answered Jul 15 '15 at 18:15

BattleDrum

798
7
13

How to check whether all elements in a list of lists are strings

4 Answers4