## Efficient strings containing each other

I have two sets of strings (A and B), and I want to know all pairs of strings a in A and b in B where a is a substring of b. The first step of coding this was the following: for a in A: for b in B: if a in b: print (a,b) However, I wanted to know-- is there a more efficient way to do this with regular expressions (e.g. instead of checking if a in b:, check if the regexp '.*' + a + '.*': matches 'b'. I thought that maybe using something like this would let me cache the Knuth-Morris-Pratt failure function for all a. Also, using a list comprehension for the inner for b in B: loop will likely give a pretty big speedup (and a nested list comprehension may be even better). I'm not very interested in making a giant leap in the asymptotic runtime of the algorithm (e.g. using a suffix tree or anything else complex and clever). I'm more concerned with the constant (I just need to do this for several pairs of A and B sets, and I don't want it to run all week). Do you know any tricks or have any generic advice to do this more quickly? Thanks a lot for any insight you can share! Edit: Using the advice of @ninjagecko and @Sven Marnach, I built a quick prefix table of 10-mers: import collections prefix_table = collections.defaultdict(set) for k, b in enumerate(B): for i in xrange(len(prot_seq)-10): j = i+10+1 prefix_table[b[i:j]].add(k) for a in A: if len(a) >= 10: for k in prefix_table[a[:10]]: # check if a is in b # (missing_edges is necessary, but not sufficient) if a in B[k]: print (a,b) else: for k in xrange(len(prots_and_seqs)): # a is too small to use the table; check if # a is in any b if a in B[k]: print (a, b)

## Solutions/Answers:

### Answer 1:

Of course you can easily write this as a list comprehension:

```
[(a, b) for a in A for b in B if a in b]
```

This might slightly speed up the loop, but don’t expect too much. I doubt using regular expressions will help in any way with this one.

**Edit**: Here are some timings:

```
import itertools
import timeit
import re
import collections
with open("/usr/share/dict/british-english") as f:
A = [s.strip() for s in itertools.islice(f, 28000, 30000)]
B = [s.strip() for s in itertools.islice(f, 23000, 25000)]
def f():
result = []
for a in A:
for b in B:
if a in b:
result.append((a, b))
return result
def g():
return [(a, b) for a in A for b in B if a in b]
def h():
res = [re.compile(re.escape(a)) for a in A]
return [(a, b) for a in res for b in B if a.search(b)]
def ninjagecko():
d = collections.defaultdict(set)
for k, b in enumerate(B):
for i, j in itertools.combinations(range(len(b) + 1), 2):
d[b[i:j]].add(k)
return [(a, B[k]) for a in A for k in d[a]]
print "Nested loop", timeit.repeat(f, number=1)
print "List comprehension", timeit.repeat(g, number=1)
print "Regular expressions", timeit.repeat(h, number=1)
print "ninjagecko", timeit.repeat(ninjagecko, number=1)
```

Results:

```
Nested loop [0.3641810417175293, 0.36279606819152832, 0.36295199394226074]
List comprehension [0.362030029296875, 0.36148500442504883, 0.36158299446105957]
Regular expressions [1.6498990058898926, 1.6494300365447998, 1.6480278968811035]
ninjagecko [0.06402897834777832, 0.063711881637573242, 0.06389307975769043]
```

**Edit 2:** Added a variant of the alogrithm suggested by ninjagecko to the timings. You can see it is much better than all the brute force approaches.

**Edit 3:** Used sets instead of lists to eliminate the duplicates. (I did not update the timings — they remained essentially unchanged.)

### Answer 2:

Let’s assume your words are bounded at a reasonable size (let’s say 10 letters). Do the following to achieve linear(!) time complexity, that is, `O(A+B)`

:

- Initialize a hashtable or trie
- For each string b in B:
- For every substring of that string
- Add the substring to the hashtable/trie (this is no worse than
`55*O(B)`

=`O(B)`

), with metadata of which string it belonged to

- Add the substring to the hashtable/trie (this is no worse than

- For every substring of that string
- For each string a in A:
- Do an
`O(1)`

query to your hashtable/trie to find all B-strings it is in, yield those

- Do an

(As of writing this answer, no response yet if OP’s “words” are bounded. If they are unbounded, this solution still applies, but there is a dependency of `O(maxwordsize^2)`

, though actually it’s nicer in practice since not all words are the same size, so it might be as nice as `O(averagewordsize^2)`

with the right distribution. For example if all the words were of size 20, the problem size would grow by a factor of 4 more than if they were of size 10. But if sufficiently few words were increased from size 10->20, then the complexity wouldn’t change much.)

**edit:** https://stackoverflow.com/q/8289199/711085 is actually a theoretically better answer. I was looking at the linked Wikipedia page before that answer was posted, and was thinking “linear in the string size is not what you want”, and only later realized it’s exactly what you want. Your intuition to build a regexp `(Aword1|Aword2|Aword3|...)`

is correct since the finite-automaton which is generated behind the scenes will perform matching quickly IF it supports simultaneous overlapping matches, which not all regexp engines might. Ultimately what you should use depends on if you plan to reuse the As or Bs, or if this is just a one-time thing. The above technique is much easier to implement but only works if your words are bounded (and introduces a DoS vulnerability if you don’t reject words above a certain size limit), but may be what you are looking for if you don’t want the Aho-Corasick string matching finite automaton or similar, or it is unavailable as a library.

### Answer 3:

A very fast way to search for a lot of strings is to make use of a **finite automaton** (so you were not that far with the guess of regexp), namely the Aho Corasick string matching machine, which is used in tools like *grep*, *virus scanners* and the like.

First it compiles the strings you want to search for (in your case the words in A) into a finite-state automaton with failure function (see the paper from ’75 if you are interested in details). This automaton then reads the input string(s) and outputs all found search strings (probably you want to modify it a bit, so that it outputs the string in which the search string was found aswell).

This method has the advantage that it searches all search strings at the same time and thus needs to look at every character of the input string(s) only once (*linear complexity*)!

There are implementations of the aho corasick pattern matcher at pypi, but i haven’t tested them, so I can’t say anything about performance, usability or correctness of these implementations.

**EDIT**: I tried this implementation of the Aho-Corasick automaton and it is indeed the fastest of the suggested methods so far, and also easy to use:

```
import pyahocorasick
def aho(A, B):
t = pyahocorasick.Trie();
for a in A:
t.add_word(a, a)
t.make_automaton()
return [(s,b) for b in B for (i,res) in t.iter(b) for s in res]
```

One thing I observed though, was when testing this implementation with @SvenMarnachs script it yielded *slightly less results* than the other methods and I am not sure why. I wrote a mail to the creator, maybe he will figure it out.

### Answer 4:

There are specialized index structures for this, see for example

http://en.wikipedia.org/wiki/Suffix_tree

You’d build a suffix-tree or something similar for B, then use A to query it.

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