Elisha512 - 6 months ago 50

Python Question

I am writing a python script where I have multiple strings.

For example:

`x = "brownasdfoersjumps"`

y = "foxsxzxasis12sa[[#brown"

z = "thissasbrownxc-34a@s;"

In all these three strings, they have one sub string in common which is

`brown`

`dict = { [commonly occuring substring] => [total number of occurences in the strings provided] }`

What would be the best way of doing that. Considering that I will have more than 200 strings each time. What would be an easy/efficient way of doing it?

Any suggestions will be highly appreciated.

Answer

This is a relatively optimised naïve algorithm. You first transform each sequence into a set of all its ngrams. Then you intersect all sets and find the longest ngram in the intersection.

```
from functools import partial, reduce
from itertools import chain
from typing import Iterator
def ngram(seq: str, n: int) -> Iterator[str]:
return (seq[i: i+n] for i in range(0, len(seq)-n+1))
def allngram(seq: str) -> Iterator[str]:
lengths = range(len(seq))
ngrams = map(partial(ngram, seq), lengths)
return set(chain.from_iterable(ngrams))
sequences = ["brownasdfoersjumps",
"foxsxzxasis12sa[[#brown",
"thissasbrownxc-34a@s;"]
seqs_ngrams = map(allngram, sequences)
intersection = reduce(set.intersection, seqs_ngrams)
longest = max(intersection, key=len) # -> brown
```

While this might get you through short sequences, this algorithm is extremely inefficient on long sequences. If your sequences are long, you can add a heuristic to limit the largest possible ngram length (i.e. the longest possible common substring). One obvious value for such a heuristic may be the shortest sequence's length.

```
def allngram(seq: str, maxn=None) -> Iterator[str]:
lengths = range(maxn) if maxn else range(len(seq))
ngrams = map(partial(ngram, seq), lengths)
return set(chain.from_iterable(ngrams))
sequences = ["brownasdfoersjumps",
"foxsxzxasis12sa[[#brown",
"thissasbrownxc-34a@s;"]
maxn = min(map(len, sequences))
seqs_ngrams = map(partial(allngram, maxn=maxn), sequences)
intersection = reduce(set.intersection, seqs_ngrams)
longest = max(intersection, key=len) # -> brown
```

This may still take too long (or make your machine run out of RAM), so you might want to read about some optimal algorithms (see the link I left in my comment to your question).