选择列表中的特殊元素并计算条件概率

Question

我需要计算一个二元语法的概率，因为在一个列表中存在它对应的一元语法。例如，期望的结果是，在下面的列表中，'pretty girl', 'pretty', 'girl'都存在。因此，通过使用列表P、(0.0017) % (0.003 * 0.002) = 5.999999999999987e-06中的值，概率为

S = ['girl', 'pretty', 'pretty girl', 'our', 'our world', 'wide', 'word', 'yes', 'yike', 'yummy']

P = [0.003, 0.002, 0.0017, 0.003, 0.006, 0.004, 0.002, 0.012, 0.006, 0.003]

我有以下代码。它似乎没有给我结果，因此我不能继续计算概率。我试图用这段代码做的是在列表中选择二元语法，并找到它们对应的单音。然后我计划在P中匹配它们的概率。

In [60]: import re
In [61]: M = []
In [62]: for i in range(len(S)):
             s_split = S[i].split()
             s_split_len = len(S[i].split())
             if s_split_len == 2:
                 m = []
                 a = re.match(s_split[0], S[i])
                 b = re.match(s_split[1], S[i])
                 m.append(a)
                 m.append(b)
                 M.append(m)
                 print M

[[<_sre.SRE_Match object at 0x10447b988>, None], [<_sre.SRE_Match object at 0x10447b8b8>, None], [<_sre.SRE_Match object at 0x10447b920>, None], [<_sre.SRE_Match object at 0x10447b9f0>, None], [<_sre.SRE_Match object at 0x10447bac0>, None], [<_sre.SRE_Match object at 0x10447bb90>, None], [<_sre.SRE_Match object at 0x10447bbf8>, None], [<_sre.SRE_Match object at 0x10447bc60>, None], [<_sre.SRE_Match object at 0x10447bcc8>, None], [<_sre.SRE_Match object at 0x10447bd30>, None], [<_sre.SRE_Match object at 0x10447bd98>, None], [<_sre.SRE_Match object at 0x10447be00>, None], [<_sre.SRE_Match object at 0x10447be68>, None], [<_sre.SRE_Match object at 0x10447bed0>, None], [<_sre.SRE_Match object at 0x10447bf38>, None], [<_sre.SRE_Match object at 0x1044a8030>, None], [<_sre.SRE_Match object at 0x1044a8098>, None], [<_sre.SRE_Match object at 0x1044a8100>, None], [<_sre.SRE_Match object at 0x1044a8168>, None], [<_sre.SRE_Match object at 0x1044a81d0>, None], [<_sre.SRE_Match object at 0x1044a8238>, None], [<_sre.SRE_Match object at 0x1044a82a0>, None], [<_sre.SRE_Match object at 0x1044a8308>, None], [<_sre.SRE_Match object at 0x1044a8370>, None], [<_sre.SRE_Match object at 0x1044a83d8>, None], [<_sre.SRE_Match object at 0x1044a8440>, None], [<_sre.SRE_Match object at 0x1044a84a8>, None], [<_sre.SRE_Match object at 0x1044a8510>, None], [<_sre.SRE_Match object at 0x1044a8578>, None], [<_sre.SRE_Match object at 0x1044a85e0>, None], [<_sre.SRE_Match object at 0x1044a8648>, None], [<_sre.SRE_Match object at 0x1044a86b0>, None], [<_sre.SRE_Match object at 0x1044a8718>, None]]
[[<_sre.SRE_Match object at 0x10447b988>, None], [<_sre.SRE_Match object at 0x10447b8b8>, None], [<_sre.SRE_Match object at 0x10447b920>, None], [<_sre.SRE_Match object at 0x10447b9f0>, None], [<_sre.SRE_Match object at 0x10447bac0>, None], [<_sre.SRE_Match object at 0x10447bb90>, None], [<_sre.SRE_Match object at 0x10447bbf8>, None], [<_sre.SRE_Match object at 0x10447bc60>, None], [<_sre.SRE_Match object at 0x10447bcc8>, None], [<_sre.SRE_Match object at 0x10447bd30>, None], [<_sre.SRE_Match object at 0x10447bd98>, None], [<_sre.SRE_Match object at 0x10447be00>, None], [<_sre.SRE_Match object at 0x10447be68>, None], [<_sre.SRE_Match object at 0x10447bed0>, None], [<_sre.SRE_Match object at 0x10447bf38>, None], [<_sre.SRE_Match object at 0x1044a8030>, None], [<_sre.SRE_Match object at 0x1044a8098>, None], [<_sre.SRE_Match object at 0x1044a8100>, None], [<_sre.SRE_Match object at 0x1044a8168>, None], [<_sre.SRE_Match object at 0x1044a81d0>, None], [<_sre.SRE_Match object at 0x1044a8238>, None], [<_sre.SRE_Match object at 0x1044a82a0>, None], [<_sre.SRE_Match object at 0x1044a8308>, None], [<_sre.SRE_Match object at 0x1044a8370>, None], [<_sre.SRE_Match object at 0x1044a83d8>, None], [<_sre.SRE_Match object at 0x1044a8440>, None], [<_sre.SRE_Match object at 0x1044a84a8>, None], [<_sre.SRE_Match object at 0x1044a8510>, None], [<_sre.SRE_Match object at 0x1044a8578>, None], [<_sre.SRE_Match object at 0x1044a85e0>, None], [<_sre.SRE_Match object at 0x1044a8648>, None], [<_sre.SRE_Match object at 0x1044a86b0>, None], [<_sre.SRE_Match object at 0x1044a8718>, None], [<_sre.SRE_Match object at 0x1044a8780>, None]]

Answer 1

这是可行的

S = ['girl', 'pretty', 'pretty girl', 'our', 'our world', 'wide', 'world', 'yes', 'yike', 'yummy']

P = [0.003, 0.002, 0.0017, 0.003, 0.006, 0.004, 0.002, 0.012, 0.006, 0.003]


for i in range(len(S)):
    s_split = S[i].split()
    s_split_len = len(S[i].split())
    if s_split_len == 2:
        a = S.index(S[i])
        b = S.index(S[i].split()[0])
        c = S.index(S[i].split()[1])
        if a != None:
            if b != None:
                co = [a, b, c]
                probs = P[co[0]], P[co[1]], P[co[2]]
                print S[i], probs[0] % (probs[1] * probs[2])