Simplifying for-if messes with better structure? [closed]

Please, move this question to Code Review -area. It is better suited there because I know the code below is junk and I want critical feedback to complete rewrite. I am pretty much reinventing the wheel.

# Description: you are given a bitwise pattern and a string
# you need to find the number of times the pattern matches in the string.
# The pattern is determined by markov chain.
# For simplicity, suppose the ones and zeros as unbiased coin flipping
# that stops as it hits the pattern, below.
#
# Any one liner or simple pythonic solution?

import random

def matchIt(yourString, yourPattern):
        """find the number of times yourPattern occurs in yourString"""

        count = 0
        matchTimes = 0

        # How can you simplify the for-if structures?
        # THIS IS AN EXAMPLE HOW NOT TO DO IT, hence Code-Smell-label
        # please, read clarifications in [Update]

        for coin in yourString:
            #return to base
            if  count == len(pattern):
                    matchTimes = matchTimes + 1
                    count = 0

            #special case to return to 2, there could be more this type of conditions
            #so this type of if-conditionals are screaming for a havoc
            if count == 2 and pattern[count] == 1:
                    count = count - 1

            #the work horse
            #it could be simpler by breaking the intial string of lenght 'l'
            #to blocks of pattern-length, the number of them is 'l - len(pattern)-1'
            if coin == pattern[count]:
                    count=count+1

        average = len(yourString)/matchTimes

        return [average, matchTimes]



# Generates the list
myString =[]
for x in range(10000):
    myString= myString + [int(random.random()*2)]

pattern = [1,0,0]
result = matchIt(myString, pattern)

print("The sample had "+str(result[1])+" matches and its size was "+str(len(myString))+".n" +
        "So it took "+str(result[0])+" steps in average.n" +
        "RESULT: "+str([a for a in "FAILURE" if result[0] != 8]))


# Sample Output
# 
# The sample had 1656 matches and its size was 10000.
# So it took 6 steps in average.
# RESULT: ['F', 'A', 'I', 'L', 'U', 'R', 'E']

[Update]

I will explain here a bit about theory, perhaps, the problem can be simplified that way. The above code try to construct the markov chain with transition matrix A below. The pattern 100 that you can imagine as coin flipping corresponds to it.

>>> Q=numpy.matrix('0.5 0.5 0; 0 0.5 0.5; 0 0.5 0')     
>>> I=numpy.identity(3)
>>> I
array([[ 1.,  0.,  0.],
       [ 0.,  1.,  0.],
       [ 0.,  0.,  1.]])
>>> Q
matrix([[ 0.5,  0.5,  0. ],
        [ 0. ,  0.5,  0.5],
        [ 0. ,  0.5,  0. ]])
>>> A=numpy.matrix('0.5 0.5 0 0; 0 0.5 0.5 0; 0 0.5 0 0.5; 0 0 0 1')
>>> A
matrix([[ 0.5,  0.5,  0. ,  0. ],
        [ 0. ,  0.5,  0.5,  0. ],
        [ 0. ,  0.5,  0. ,  0.5],
        [ 0. ,  0. ,  0. ,  1. ]])

The average 8 in the question becomes the sum of values on the first row in the matrix N=(I-Q)^-1 where Q above.

>>> (I-Q)**-1
matrix([[ 2.,  4.,  2.],
        [ 0.,  4.,  2.],
        [ 0.,  2.,  2.]])
>>> numpy.sum(((I-Q)**-1)[0])
8.0

Now, you can probably see that this apparently-only-pattern-matching-problem becomes a markov chain. I cannot see a reason why you could not substitute the messy for-while-if conditions with something similar to matrices or matrices. I don’t know how to implement them but iterators could be a way to go, researching, particularly with more states where you need to decompose.

But a problem emerged with Numpy, what are the things -Inf and NaN for? Check the values to which they should converge, above, from (I-Q)**-1 matrix. The N is from N=I+Q+Q^2+Q^3+...=frac{I-Q^{n}}{I-Q}.

>>> (I-Q**99)/(I-Q)
matrix([[  2.00000000e+00,   1.80853571e-09,             -Inf],
        [             NaN,   2.00000000e+00,   6.90799171e-10],
        [             NaN,   6.90799171e-10,   1.00000000e+00]])
>>> (I-Q**10)/(I-Q)
matrix([[ 1.99804688,  0.27929688,        -Inf],
        [        NaN,  1.82617188,  0.10742188],
        [        NaN,  0.10742188,  0.96679688]])

Answer

Ok – a standard(-ish) string search:

def matchIt(needle, haystack):
    """
    @param needle:   string, text to seek
    @param haystack: string, text to search in

    Return number of times needle is found in haystack,
        allowing overlapping instances.

    Example: matchIt('abab','ababababab') -> 4
    """
    lastSeenAt = -1
    timesSeen = 0
    while True:
        nextSeen = haystack.find(needle, lastSeenAt+1)
        if nextSeen==-1:
            return timesSeen
        else:
            lastSeenAt = nextSeen
            timesSeen += 1

but you want to do this to a list of numbers? No problem; we just need to make a list class with a find() method, like so:

import itertools
class FindableList(list):
    def find(self, sub, start=None, end=None):
        """
        @param sub: list, pattern to look for in self

        @param start: int, first possible start-of-list
            If not specified, start at first item

        @param: end: int, last+1 possible start-of-list
            If not specified, end such that entire self is searched

        Returns;
            Starting offset if a match is found, else -1
        """
        if start is None or start < 0:
            start = 0

        # N.B. If end is allowed to be too high,
        # zip() will silently truncate the list comparison
        # and you will probably get extra spurious matches.
        lastEnd = len(self) - len(sub) + 1
        if end is None or end > lastEnd:
            end = lastEnd

        rng = xrange if xrange else range
        iz  = itertools.izip
        isl = itertools.islice

        for pos in rng(start, end):
            if all(a==b for a,b in iz(sub, isl(self, pos, end))):
                return pos

        # no match found
        return -1

then the example looks like

matchIt([1,2,1,2], FindableList([1,2,1,2,1,2,1,2,1,2])) -> 4

and your code becomes:

# Generate a list
randIn = lambda x: int(x*random.random())
myString =[randIn(2) for i in range(10000)]

pattern = [1,0,0]
result = matchIt(pattern, myString)

print("The sample had {0} matches and its size was {1}.n".format(result, len(myString)))