我需要计算一个二元组的概率,因为在列表中存在相应的unigrams。例如,期望的结果在以下列表中,'pretty girl', 'pretty', 'girl'
都存在。因此,概率是使用列表P
,(0.0017) % (0.003 * 0.002) = 5.999999999999987e-06
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,因此我无法继续计算概率。我试图用这个代码做的是在列表中选择bigrams并找到相应的unigrams。然后我打算在P
中匹配他们的概率。
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]]
答案 0 :(得分:0)
这有效
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])