需要帮助的家伙。需要对列表进行向量化，即用某个计算出的数字替换子列表的每个元素。子列表每个元素的公式https://i.stack.imgur.com/73xj9.png该列表具有以下结构：list = [[document0],[document1],[document2],[document3].......]每个文档由术语组成，例如document1 = ['i','love','you']
我写了我的计算函数，但它不能正常工作:(
def tfc(slova):
import math as m
meshokslov1=[{}for i in range(len(slova))]
meshokslov2=[{}for i in range(len(slova))]
SummaKvadratov=0
for i in range(len(slova)):
for j in range(len(slova[i])):
n=0
for q in slova:
if q.count(slova[i][j])!=0:
n+=1
if slova[i][j] in meshokslov1:
continue
else:
meshokslov1[i][slova[i][j]]=slova[i].count(slova[i][j])*m.log10(len(slova)/n)
SummaKvadratov+=(slova[i].count(slova[i][j])*m.log10(len(slova)/n))**2
for i in range(len(slova)):
for j in range(len(slova[i])):
if slova[i][j] in meshokslov2:
continue
else:
meshokslov2[i][slova[i][j]]=meshokslov1[i][slova[i][j]]/(SummaKvadratov**0.5)
return meshokslov2
这是一个解决方案，遵循针对您的问题的自上而下的设计：
import math
def frequency(term, document):
return document.count(term) / len(document)
def totalNumOfDocuments(inList):
return len(inList)
def numOfDocumentsForTerm(inList, term):
return len([doc for doc in inList if term in doc])
def TFCWeighting(inList):
tfc = []
N = totalNumOfDocuments(inList)
for document in inList:
temp = []
for term in document:
#numerator
freq = frequency(term, document)
n = numOfDocumentsForTerm(inList, term)
logarithm = math.log(N/n)
numerator = freq * logarithm
#denominator
summation = sum([(frequency(t, document) * math.log(N / numOfDocumentsForTerm(inList, term))) ** 2 for t in document])
denominator = math.sqrt(summation)
temp.append(round(numerator / denominator,3))
tfc.append(temp)
return tfc
l1=[['can','help','you'],['thank','you'],['help','help','help','help','help','help']]
l2=[['I','help'],['will','you']]
print(TFCWeighting(l1))
print(TFCWeighting(l2))
输出：
[[0.577, 0.577, 0.577], [0.707, 0.707], [0.408, 0.408, 0.408, 0.408, 0.408, 0.408]]
[[0.707, 0.707], [0.707, 0.707]]
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我来说两句