python - 如何确定单词的概率？

Question

我有两个文件。 Doc1 格式如下:

TOPIC:  0 5892.0
site 0.0371690427699
Internet 0.0261371350984
online 0.0229124236253
web 0.0218940936864
say 0.0159538357094

TOPIC:  1 12366.0
web 0.150331554262
site 0.0517548115801
say 0.0451237263464
Internet 0.0153647096879
online 0.0135856380398

...以相同的模式依此类推，直到第 99 题。

Doc2 的格式是:

0 0.566667 0 0.0333333 0 0 0 0.133333 ..........

依此类推... 每个主题的每个值总共有 100 个值。

现在，我必须找到每个单词的加权平均概率，即:

P(w) = alpha.P(w1)+ alpha.P(w2)+...... +alpha.P(wn)

where alpha = value in the nth position corresponding to the nth topic. 

也就是说对于“say”这个词，概率应该是

P(say) = 0*0.0159 + 0.5666*0.045+....... 

同样，对于每个单词，我都必须计算概率。

For  multiplication, if the word is taken from topic 0, then the 0th value from the doc2 must be considered and so on.

我只用下面的代码计算了单词的出现次数，但从未取过它们的值。所以，我很困惑。

 with open(doc2, "r") as f:
    with open(doc3, "w") as f1:

         words = " ".join(line.strip() for line in f)
         d = defaultdict(int)
         for word in words.split():  
              d[word] += 1
              for key, value in d.iteritems() :
                  f1.write(key+ ' ' + str(value) + ' ')
              print '\n'

我的输出应该是这样的:

 say = "prob of this word calculated by above formula"
 site = "
 internet = " 

等等。

我做错了什么？

Answer 1

假设您忽略了 TOPIC 行，使用 defaultdict 对值进行分组，然后在最后进行计算:

from collections import defaultdict
from itertools import groupby, imap

d = defaultdict(list)
with open("doc1") as f,open("doc2") as f2:
    values = map(float, f2.read().split()) 
    for line in f:
        if line.strip() and not line.startswith("TOPIC"):
            name, val = line.split()
            d[name].append(float(val))

for k,v in d.items():
    print("Prob for {} is {}".format(k ,sum(i*j for i, j in zip(v,values)) ))

另一种方法是边走边计算，每次点击新部分时都会增加计数，即带有 TOPIC 的行以通过索引从值中获取正确的值:

from collections import defaultdict
d = defaultdict(float)
from itertools import  imap

with open("doc1") as f,open("doc2") as f2:
    # create list of all floats from doc2
    values = imap(float, f2.read().split())
    for line in f:
        # if we have a new TOPIC increase the ind to get corresponding ndex from values
        if line.startswith("TOPIC"):
            ind = next(values)
            continue
        # ignore empty lines
        if line.strip():
            # get word and float and multiply the val by corresponding values value
            name, val = line.split()
            d[name] += float(val) * values[ind]

for k,v in d.items():
    print("Prob for {} is {}".format(k ,v) )

在 doc2 中使用两个 doc1 内容和 0 0.566667 0 0.0333333 0 输出以下内容:

Prob for web is 0.085187930859
Prob for say is 0.0255701266375
Prob for online is 0.0076985327511
Prob for site is 0.0293277438137
Prob for Internet is 0.00870667394471

您还可以使用 itertools groupby:

from collections import defaultdict
d = defaultdict(float)
from itertools import groupby, imap

with open("doc1") as f,open("doc2") as f2:
    values = imap(float, f2.read().split())
    # lambda x: not(x.strip()) will split into groups on the empty lines
    for ind, (k, v) in enumerate(groupby(f, key=lambda x: not(x.strip()))):
        if not k:
            topic = next(v) 
            #  get matching float from values
            f = next(values)
            # iterate over the group 
            for s in v:
                name, val = s.split()
                d[name] += (float(val) * f)
for k,v in d.iteritems():
    print("Prob for {} is {}".format(k,v))

对于 python3 所有 itertools imaps 应该改为 map，它在 python3 中也返回一个迭代器。