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请将此问题移至 Code Review -area .它更适合那里，因为我知道下面的代码是垃圾，我想要重要的反馈来完成重写。我几乎是在重新发明轮子。
# 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']
[更新]
我会在这里解释一下理论，也许这样可以简化问题。上面的代码尝试用下面的转移矩阵 A 构造马尔可夫链。您可以想象为掷硬币的模式 100 对应于它。
>>> 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. ]])
问题中的average 8 成为矩阵N=(I-Q)^-1 中第一行值的总和其中Q在上面。
>>> (I-Q)**-1
matrix([[ 2., 4., 2.],
[ 0., 4., 2.],
[ 0., 2., 2.]])
>>> numpy.sum(((I-Q)**-1)[0])
8.0
现在，您可能会看到这个显然只是模式匹配的问题变成了一个马尔可夫链。我看不出为什么你不能用类似于矩阵或矩阵的东西来代替困惑的 for-while-if 条件。我不知道如何实现它们，但迭代器可能是一种研究方法，特别是对于需要分解的更多状态。
但是Numpy出现了一个问题，-Inf和NaN是干什么用的？从上面的 (I-Q)**-1 矩阵中检查它们应该收敛到的值。 N 来自 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]])