How to find similar patterns in lists/arrays of strings - string

Question

I am looking for ways to find matching patterns in lists or arrays of strings, specifically in .NET, but algorithms or logic from other languages would be helpful.
Say I have 3 arrays (or in this specific case List(Of String))
Array1
"Do"
"Re"
"Mi"
"Fa"
"So"
"La"
"Ti"
Array2
"Mi"
"Fa"
"Jim"
"Bob"
"So"
Array3
"Jim"
"Bob"
"So"
"La"
"Ti"
I want to report on the occurrences of the matches of
("Mi", "Fa") In Arrays (1,2)
("So") In Arrays (1,2,3)
("Jim", "Bob", "So") in Arrays (2,3)
("So", "La", "Ti") in Arrays (1, 3)
...and any others.
I am using this to troubleshoot an issue, not to make a commercial product of it specifically, and would rather not do it by hand (there are 110 lists of about 100-200 items).
Are there any algorithms, existing code, or ideas that will help me accomplish finding the results described?

Answer 1

The simplest way to code would be to build a Dictionary then loop through each item in each array. For each item do this:
Check if the item is in the dictonary if so add the list to the array.
If the item is not in the dictionary add it and the list.
Since as you said this is non-production code performance doesn't matter so this approach should work fine.

Answer 2

Here's a solution using SuffixTree module to locate subsequences:
#!/usr/bin/env python
from SuffixTree import SubstringDict
from collections import defaultdict
from itertools import groupby
from operator import itemgetter
import sys
def main(stdout=sys.stdout):
"""
>>> import StringIO
>>> s = StringIO.StringIO()
>>> main(stdout=s)
>>> print s.getvalue()
[['Mi', 'Fa']] In Arrays (1, 2)
[['So', 'La', 'Ti']] In Arrays (1, 3)
[['Jim', 'Bob', 'So']] In Arrays (2, 3)
[['So']] In Arrays (1, 2, 3)
<BLANKLINE>
"""
# array of arrays of strings
arr = [
["Do", "Re", "Mi", "Fa", "So", "La", "Ti",],
["Mi", "Fa", "Jim", "Bob", "So",],
["Jim", "Bob", "So", "La", "Ti",],
]
#### # 28 seconds (27 seconds without lesser substrs inspection (see below))
#### N, M = 100, 100
#### import random
#### arr = [[random.randrange(100) for _ in range(M)] for _ in range(N)]
# convert to ASCII alphabet (for SubstringDict)
letter2item = {}
item2letter = {}
c = 1
for item in (i for a in arr for i in a):
if item not in item2letter:
c += 1
if c == 128:
raise ValueError("too many unique items; "
"use a less restrictive alphabet for SuffixTree")
letter = chr(c)
letter2item[letter] = item
item2letter[item] = letter
arr_ascii = [''.join(item2letter[item] for item in a) for a in arr]
# populate substring dict (based on SuffixTree)
substring_dict = SubstringDict()
for i, s in enumerate(arr_ascii):
substring_dict[s] = i+1
# enumerate all substrings, save those that occur more than once
substr2indices = {}
indices2substr = defaultdict(list)
for str_ in arr_ascii:
for start in range(len(str_)):
for size in reversed(range(1, len(str_) - start + 1)):
substr = str_[start:start + size]
if substr not in substr2indices:
indices = substring_dict[substr] # O(n) SuffixTree
if len(indices) > 1:
substr2indices[substr] = indices
indices2substr[tuple(indices)].append(substr)
#### # inspect all lesser substrs
#### # it could diminish size of indices2substr[ind] list
#### # but it has no effect for input 100x100x100 (see above)
#### for i in reversed(range(len(substr))):
#### s = substr[:i]
#### if s in substr2indices: continue
#### ind = substring_dict[s]
#### if len(ind) > len(indices):
#### substr2indices[s] = ind
#### indices2substr[tuple(ind)].append(s)
#### indices = ind
#### else:
#### assert set(ind) == set(indices), (ind, indices)
#### substr2indices[s] = None
#### break # all sizes inspected, move to next `start`
for indices, substrs in indices2substr.iteritems():
# remove substrs that are substrs of other substrs
substrs = sorted(substrs, key=len) # sort by size
substrs = [p for i, p in enumerate(substrs)
if not any(p in q for q in substrs[i+1:])]
# convert letters to items and print
items = [map(letter2item.get, substr) for substr in substrs]
print >>stdout, "%s In Arrays %s" % (items, indices)
if __name__=="__main__":
# test
import doctest; doctest.testmod()
# measure performance
import timeit
t = timeit.Timer(stmt='main(stdout=s)',
setup='from __main__ import main; from cStringIO import StringIO as S; s = S()')
N = 1000
milliseconds = min(t.repeat(repeat=3, number=N))
print("%.3g milliseconds" % (1e3*milliseconds/N))
It takes about 30 seconds to process 100 lists of 100 items each. SubstringDict in the above code might be emulated by grep -F -f.
Old solution:
In Python (save it to 'group_patterns.py' file):
#!/usr/bin/env python
from collections import defaultdict
from itertools import groupby
def issubseq(p, q):
"""Return whether `p` is a subsequence of `q`."""
return any(p == q[i:i + len(p)] for i in range(len(q) - len(p) + 1))
arr = (("Do", "Re", "Mi", "Fa", "So", "La", "Ti",),
("Mi", "Fa", "Jim", "Bob", "So",),
("Jim", "Bob", "So", "La", "Ti",))
# store all patterns that occure at least twice
d = defaultdict(list) # a map: pattern -> indexes of arrays it's within
for i, a in enumerate(arr[:-1]):
for j, q in enumerate(arr[i+1:]):
for k in range(len(a)):
for size in range(1, len(a)+1-k):
p = a[k:k + size] # a pattern
if issubseq(p, q): # `p` occures at least twice
d[p] += [i+1, i+2+j]
# group patterns by arrays they are within
inarrays = lambda pair: sorted(set(pair[1]))
for key, group in groupby(sorted(d.iteritems(), key=inarrays), key=inarrays):
patterns = sorted((pair[0] for pair in group), key=len) # sort by size
# remove patterns that are subsequences of other patterns
patterns = [p for i, p in enumerate(patterns)
if not any(issubseq(p, q) for q in patterns[i+1:])]
print "%s In Arrays %s" % (patterns, key)
The following command:
$ python group_patterns.py
prints:
[('Mi', 'Fa')] In Arrays [1, 2]
[('So',)] In Arrays [1, 2, 3]
[('So', 'La', 'Ti')] In Arrays [1, 3]
[('Jim', 'Bob', 'So')] In Arrays [2, 3]
The solution is terribly inefficient.

Answer 3

As others have mentioned the function you want is Intersect. If you are using .NET 3.0 consider using LINQ's Intersect function.
See the following post for more information
Consider using LinqPAD to experiment.
www.linqpad.net

Answer 4

I hacked the program below in about 10 minutes of Perl. It's not perfect, it uses a global variable, and it just prints out the counts of every element seen by the program in each list, but it's a good approximation to what you want to do that's super-easy to code.
Do you actually want all combinations of all subsets of the elements common to each array? You could enumerate all of the elements in a smarter way if you wanted, but if you just wanted all elements that exist at least once in each array you could use the Unix command "grep -v 0" on the output below and that would show you the intersection of all elements common to all arrays. Your question is missing a little bit of detail, so I can't perfectly implement something that solves your problem.
If you do more data analysis than programming, scripting can be very useful for asking questions from textual data like this. If you don't know how to code in a scripting language like this, I would spend a month or two reading about how to code in Perl, Python or Ruby. They can be wonderful for one-off hacks such as this, especially in cases when you don't really know what you want. The time and brain cost of writing a program like this is really low, so that (if you're fast) you can write and re-write it several times while still exploring the definition of your question.
#!/usr/bin/perl -w
use strict;
my #Array1 = ( "Do", "Re", "Mi", "Fa", "So", "La", "Ti");
my #Array2 = ( "Mi", "Fa", "Jim", "Bob", "So" );
my #Array3 = ( "Jim", "Bob", "So", "La", "Ti" );
my %counts;
sub count_array {
my $array = shift;
my $name = shift;
foreach my $e (#$array) {
$counts{$e}{$name}++;
}
}
count_array( \#Array1, "Array1" );
count_array( \#Array2, "Array2" );
count_array( \#Array3, "Array3" );
my #names = qw/ Array1 Array2 Array3 /;
print join ' ', ('element',#names);
print "\n";
my #unique_names = keys %counts;
foreach my $unique_name (#unique_names) {
my #counts = map {
if ( exists $counts{$unique_name}{$_} ) {
$counts{$unique_name}{$_};
} else {
0;
}
}
#names;
print join ' ', ($unique_name,#counts);
print "\n";
}
The program's output is:
element Array1 Array2 Array3
Ti 1 0 1
La 1 0 1
So 1 1 1
Mi 1 1 0
Fa 1 1 0
Do 1 0 0
Bob 0 1 1
Jim 0 1 1
Re 1 0 0

Answer 5

Looks like you want to use an intersection function on sets of data. Intersection picks out elements that are common in both (or more) sets.
The problem with this viewpoint is that sets cannot contain more than one of each element, i.e. no more than one Jim per set, also it cannot recognize several elements in a row counting as a pattern, you can however modify a comparison function to look further to see just that.
There mey be functions like intersect that works on bags (which is kind of like sets, but tolerate identical elements).
These functions should be standard in most languages or pretty easy to write yourself.

Answer 6

I'm sure there's a MUCH more elegant way, but...
Since this isn't production code, why not just hack it and convert each array into a delimited string, then search each string for the pattern you want? i.e.
private void button1_Click(object sender, EventArgs e)
{
string[] array1 = { "do", "re", "mi", "fa", "so" };
string[] array2 = { "mi", "fa", "jim", "bob", "so" };
string[] pattern1 = { "mi", "fa" };
MessageBox.Show(FindPatternInArray(array1, pattern1).ToString());
MessageBox.Show(FindPatternInArray(array2, pattern1).ToString());
}
private bool FindPatternInArray(string[] AArray, string[] APattern)
{
return string.Join("~", AArray).IndexOf(string.Join("~", APattern)) >= 0;
}

Answer 7

First, start by counting each item.
You make a temp list : "Do" = 1, "Mi" = 2, "So" = 3, etc.
you can remove from the temp list all the ones that match = 1 (ex: "Do").
The temp list contains the list of non-unique items (save it somewhere).
Now, you try to make lists of two from one in the temp list, and a following in the original lists.
"So" + "La" = 2, "Bob" + "So" = 2, etc.
Remove the ones with = 1.
You have the lists of couple that appears at least twice (save it somewhere).
Now, try to make lists of 3 items, by taking a couple from the temp list, and take the following from the original lists.
("Mi", "Fa") + "So" = 1, ("Mi", "Fa") + "Jim" = 1, ("So", "La") + "Ti" = 2
Remove the ones with = 1.
You have the lists of 3 items that appears at least twice (save it).
And you continue like that until the temp list is empty.
At the end, you take all the saved lists and you merge them.
This algorithm is not optimal (I think we can do better with suitable data structures), but it is easy to implement :)

Answer 8

Suppose a password consisted of a string of nine characters from the English alphabet (26 characters). If each possible password could be tested in a millisecond, how long would it take to test all possible passwords?