首先,这是家庭作业,所以我想请一些建议。我正在编写一个生成加权倒排索引的程序。加权倒排索引是以单词为关键字的字典;该值是列表列表,列表中的每个项目都包含文档编号,以及该文档在文档中出现的次数。
例如,
{"a": [[1, 2],[2,1]]}
The word "a" appears twice in document 1 and once in document 2.
我正在练习两个小文件。
FILE1.TXT:
Where should I go
When I want to have
A smoke,
A pancake,
and a nap.
FILE2.TXT:
I do not know
Where my pancake is
I want to take a nap.
这是我的程序代码:
def cleanData(myFile):
file = open(myFile, "r")
data = file.read()
wordList = []
#All numbers and end-of-sentence punctuation
#replaced with the empty string
#No replacement of apostrophes
formattedData = data.strip().lower().replace(",","")\
.replace(".","").replace("!","").replace("?","")\
.replace(";","").replace(":","").replace('"',"")\
.replace("1","").replace("2","").replace("3","")\
.replace("4","").replace("5","").replace("6","")\
.replace("7","").replace("8","").replace("9","")\
.replace("0","")
words = formattedData.split() #creates a list of all words in the document
for word in words:
wordList.append(word) #adds each word in a document to the word list
return wordList
def main():
fullDict = {}
files = ["file1.txt", "file2.txt"]
docNumber = 1
for file in files:
wordList = cleanData(file)
for word in wordList:
if word not in fullDict:
fullDict[word] = []
fileList = [docNumber, 1]
fullDict[word].append(fileList)
else:
listOfValues = list(fullDict.values())
for x in range(len(listOfValues)):
if docNumber == listOfValues[x][0]:
listOfValues[x][1] +=1
fullDict[word] = listOfValues
break
fileList = [docNumber,1]
fullDict[word].append(fileList)
docNumber +=1
return fullDict
我要做的是生成这样的东西:
{"a": [[1,3],[2,1]], "nap": [[1,1],[2,1]]}
我得到的是:
{"a": [[1,1],[1,1],[1,1],[2,1]], "nap": [[1,1],[2,1]]}
它记录所有文档中每个单词的所有出现次数,但它会分别记录重复次数。我无法弄清楚这一点。任何帮助,将不胜感激!先感谢您。 :)
答案 0 :(得分:2)
您的代码中存在两个主要问题。
问题1
listOfValues = list(fullDict.values())
for x in range(len(listOfValues)):
if docNumber == listOfValues[x][0]:
在这里,您可以获取字典的所有值,而不管当前的单词,并递增计数,但是您应该在与当前单词对应的列表中递增计数。所以,你应该把它改成
listOfValues = fullDict[word]
问题2
fileList = [docNumber,1]
fullDict[word].append(fileList)
除了增加所有单词的计数外,您总是向fullDict
添加新值。但是,只有当docNumber
中没有listOfValues
时,您才应该添加它。因此,您可以将else
与for
循环一起使用,就像这样
for word in wordList:
if word not in fullDict:
....
else:
listOfValues = fullDict[word]
for x in range(len(listOfValues)):
....
else:
fileList = [docNumber, 1]
fullDict[word].append(fileList)
进行这两项更改后,我得到以下输出
{'a': [[1, 3], [2, 1]],
'and': [[1, 1]],
'do': [[2, 1]],
'go': [[1, 1]],
'have': [[1, 1]],
'i': [[1, 2], [2, 2]],
'is': [[2, 1]],
'know': [[2, 1]],
'my': [[2, 1]],
'nap': [[1, 1], [2, 1]],
'not': [[2, 1]],
'pancake': [[1, 1], [2, 1]],
'should': [[1, 1]],
'smoke': [[1, 1]],
'take': [[2, 1]],
'to': [[1, 1], [2, 1]],
'want': [[1, 1], [2, 1]],
'when': [[1, 1]],
'where': [[1, 1], [2, 1]]}
很少有改进代码的建议。
您可以使用字典,而不是使用列表来存储文档编号和计数。这会让你的生活更轻松。
您可以使用collections.Counter
。
您可以使用简单的正则表达式,而不是使用多次替换,例如
formattedData = re.sub(r'[.!?;:"0-9]', '', data.strip().lower())
如果我要清理cleanData
,我会这样做
import re
def cleanData(myFile):
with open(myFile, "r") as input_file:
data = input_file.read()
return re.sub(r'[.!?;:"0-9]', '', data.strip().lower()).split()
在main
循环中,您可以使用Brad Budlong建议的改进,例如
def main():
fullDict = {}
files = ["file1.txt", "file2.txt"]
for docNumber, currentFile in enumerate(files, 1):
for word in cleanData(currentFile):
if word not in fullDict:
fullDict[word] = [[docNumber, 1]]
else:
for x in fullDict[word]:
if docNumber == x[0]:
x[1] += 1
break
else:
fullDict[word].append([docNumber, 1])
return fullDict
答案 1 :(得分:1)
我首选的for循环实现不使用len和range函数进行迭代。由于这些都是可变列表,因此您不需要知道索引,只需要拥有每个列表,然后可以在没有索引的情况下进行修改。我用以下内容替换了for循环,并获得与thefourtheye相同的输出。
for word in wordList:
if word not in fullDict:
fullDict[word] = [[docNumber, 1]]
else:
for val in fullDict[word]:
if val[0] == docNumber:
val[1] += 1
break
else:
fullDict[word].append([docNumber, 1])