CODEWITHSUNDEEP
pythonstringlistnumpy
Binyamin Even
Binyamin Even

Reputation: 3382

return element in list refering to index

Assuming I have a list of words:

l = ['example', 'to', 'a', 'list', 'of', 'words']

And I get an index i, let's say 10.

What I need is to return the element in l that contains the i-th char,

So in the example of 10, as the 10-th element (zero-based) is the l from the word list - what I need to return is the word list.

Iv'e been trying to think of a simple way to do this, and I didn't find something elegant.

Any help will be appreciated!

Upvotes: 1

Views: 132

Answers (6)

superb rain
superb rain

Reputation: 5531

Repeating each word for each of its letters, and then just picking the i-th word:

[s for s in l for _ in s][i]

Or more efficiently with a generator and itertools.islice:

next(islice((s for s in l for _ in s), i, None))

Upvotes: 0

Carlos Leite
Carlos Leite

Reputation: 302

I got that wrong ... the @ -----Keeo it simple ...

In [1]: l = ['example', 'to', 'a', 'list', 'of', 'words']

In [2]: s = "".join(l)

In [3]: s
Out[3]: 'exampletoalistofwords'

In [4]: s[10]
Out[4]: 'l'

I believe, even in 1 line of code it is still readable (pythonic)

 "".join( ['example', 'to', 'a', 'list', 'of', 'words'])[10]

Upvotes: -1

Jab
Jab

Reputation: 27515

You can use next and the walrus operator to keep track of your counts too. Basically keep subtracting the length of every string from i and then once i is less than 0, that's the string:

l = ['example', 'to', 'a', 'list', 'of', 'words']
i = 10

result = next(s for s in l if (i:= i-len(s)) < 0)

Result:

'list'

Upvotes: 3

Andrej Kesely
Andrej Kesely

Reputation: 195553

from itertools import count

l = ['example', 'to', 'a', 'list', 'of', 'words']

c = count()
print(next(s for s in l for _, i in zip(s, c) if i == 10))

Prints:

list

Another solution (using bisect module):

from bisect import bisect
from itertools import accumulate

l = ['example', 'to', 'a', 'list', 'of', 'words']

lengths = [*accumulate(map(len, l))]
print(l[bisect(lengths, 10)])

Upvotes: 1

Jack
Jack

Reputation: 569

i = 10

for word in l:
    i -= len(word)
    if i < 0:
        break

# the value of word is 'list'

and if you want it in a function

def at_index(l, i):
    for word in l:
        i -= len(word)
        if i < 0:
            return word
    return None

Upvotes: 3

Jack Fleeting
Jack Fleeting

Reputation: 24940

Something like this?

ind = 10
ll = 0
for item in l:
    ll+=len(item)
    if ll>=ind+1:
        print(item)
        break

Upvotes: 0

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