Boolean Model
It is a simple retrieval model based on set theory and boolean algebra. Queries are designed as boolean expressions which have precise semantics. The retrieval strategy is based on binary decision criterion. The boolean model considers that index terms are present or absent in a document.
Problem:
Consider 5 documents with a vocabulary of 6 terms
document 1 = ‘ term1 term3 ‘
document 2 = ‘ term 2 term4 term6 ‘
document 3 = ‘ term1 term2 term3 term4 term5 ‘
document 4 = ‘ term1 term3 term6 ‘
document 5 = ‘ term3 term4 ‘
Our documents in a boolean model
| term1 | term2 | term3 | term4 | term5 | term6 | |
| document1 | 1 | 0 | 1 | 0 | 0 | 0 |
| document2 | 0 | 1 | 0 | 1 | 0 | 1 |
| document3 | 1 | 1 | 1 | 1 | 1 | 0 |
| document4 | 1 | 0 | 1 | 0 | 0 | 1 |
| document5 | 0 | 0 | 1 | 1 | 0 | 0 |
Consider the query: Find the document consisting of term1 and term3 and not term2 ( term1 ∧ term3 ∧ ¬ term2)
| term1 | ¬term2 | term3 | term4 | term5 | term6 | |
| document1 | 1 | 1 | 1 | 0 | 0 | 0 |
| document2 | 0 | 0 | 0 | 1 | 0 | 1 |
| document3 | 1 | 0 | 1 | 1 | 1 | 0 |
| document4 | 1 | 1 | 1 | 0 | 0 | 1 |
| document5 | 0 | 1 | 1 | 1 | 0 | 0 |
document 1 : 1 ∧ 1∧ 1 = 1
document 2 : 0 ∧ 0 ∧ 0 = 0
document 3 : 1 ∧ 1 ∧ 0 = 0
document 4 : 1 ∧ 1 ∧ 1 = 1
document 5 : 0 ∧ 1 ∧ 1 = 0
Based on the above computation document1 and document4 are relevant to the given query
The CSV file which is given as input:
documents , term1, term2, term3
document1, ice cream, mango, litchi
document2 , hockey, cricket, sport
document3, litchi, mango, chocolate
document4 , nice, good, cute
Code: Python code showing the implementation of the boolean model for document retrieval
Python
import pandas# module to read the contents of the file from a csv filefrom contextlib import redirect_stdout# module to redirect the output to a text fileterms = []# list to store the terms present in the documentskeys = []# list to store the names of the documentsvec_Dic = {}# dictionary to store the name of the document and the boolean vector as listdicti = {}# dictionary to store the name of the document and the terms present in it as a# vectordummy_List = []# list for performing some operations and clearing themdef filter(documents, rows, cols): '''function to read and separate the name of the documents and the terms present in it to a separate list from the data frame and also create a dictionary which has the name of the document as key and the terms present in it as the list of strings which is the value of the key''' for i in range(rows): for j in range(cols): # traversal through the data frame if(j == 0): # first column has the name of the document in the csv file keys.append(documents.loc[i].iat[j]) else: dummy_List.append(documents.loc[i].iat[j]) # dummy list to update the terms in the dictionary if documents.loc[i].iat[j] not in terms: # add the terms to the list if it is not present else continue terms.append(documents.loc[i].iat[j]) copy = dummy_List.copy() # copying the dummy list to a different list dicti.update({documents.loc[i].iat[0]: copy}) # adding the key value pair to a dictionary dummy_List.clear() # clearing the dummy listdef bool_Representation(dicti, rows, cols): '''In this function we get a boolean representation of the terms present in the documents in the form of lists, later we create a dictionary which contains the name of the documents as key and value as the list of boolean values representing the terms present in the document''' terms.sort() # we sort the elements in the alphabetical order for the convience, the order # of the term does not make any difference for i in (dicti): # for every document in the dictionary we check for each string present in # the list for j in terms: # if the string is present in the list we append 1 else we append 0 if j in dicti[i]: dummy_List.append(1) else: dummy_List.append(0) # appending 1 or 0 for obtaining the boolean representation copy = dummy_List.copy() # copying the dummy list to a different list vec_Dic.update({i: copy}) # adding the key value pair to a dictionary dummy_List.clear() # clearing the dummy listdef query_Vector(query): '''In this function we represent the query in the form of boolean vector''' qvect = [] # query vector which is returned at the end of the function for i in terms: # if the word present in the list of terms is also present in the query # then append 1 else append 0 if i in query: qvect.append(1) else: qvect.append(0) return qvect # return the query vector which is obtained in the boolean formdef prediction(q_Vect): '''In this function we make the prediction regarding which document is related to the given query by performing the boolean operations''' dictionary = {} listi = [] count = 0 # initialisation of the dictionary , list and a variable which is further # required for performing the computation term_Len = len(terms) # number of terms present in the term list for i in vec_Dic: # for every document in the dictionary containing the terms present in it # the form of boolean vector for t in range(term_Len): if(q_Vect[t] == vec_Dic[i][t]): # if the words present in the query is also present in the # document or if the words present in the query is also absent in # the document count += 1 # increase the value of count variable by one # the condition in which words present in document and absent in #query , present in query and absent in document is not considered dictionary.update({i: count}) # dictionary updation here the name of the document is the key and the # count variable computed earlier is the value count = 0 # reinitialisaion of count variable to 0 for i in dictionary: listi.append(dictionary[i]) # here we append the count value to list listi = sorted(listi, reverse=True) # we sort the list in the descending order which is needed to rank the #documents according to the relevance ans = ' ' # variable to store the name of the document which is most relevant with open('output.txt', 'w') as f: with redirect_stdout(f): # to redirect the output to a text file print("ranking of the documents") for count, i in enumerate(listi): key = check(dictionary, i) # Function call to get the key when the value is known if count == 0: ans = key # to store the name of the document which is most relevant print(key, "rank is", count+1) # print the name of the document along with its rank dictionary.pop(key) # remove the key from the dictionary after printing print(ans, "is the most relevant document for the given query") # to print the name of the document which is most relevantdef check(dictionary, val): '''Function to return the key when the value is known''' for key, value in dictionary.items(): if(val == value): # if the given value is same as the value present in the dictionary # return the key return keydef main(): documents = pandas.read_csv(r'documents.csv') # to read the data from the csv file as a dataframe rows = len(documents) # to get the number of rows cols = len(documents.columns) # to get the number of columns filter(documents, rows, cols) # function call to read and separate the name of the documents and the terms # present in it to a separate list from the data frame and also create a # dictionary which has the name of the document as key and the terms present in # it as the list of strings which is the value of the key bool_Representation(dicti, rows, cols) # In this function we get a boolean representation of the terms present in the # documents in the form of lists, later we create a dictionary which contains # the name of the documents as key and value as the list of boolean values #representing the terms present in the document print("Enter query") query = input() # to get the query input from the user, the below input is given for obtaining # the output as in output.txt file # hockey is a national sport query = query.split(' ') # splitting the query as a list of strings q_Vect = query_Vector(query) # function call to represent the query in the form of boolean vector prediction(q_Vect) # Function call to make the prediction regarding which document is related to # the given query by performing the boolean operationsmain() |
Output: The output obtained in the text file when the query is “hockey is a national sport”
ranking of the documents
document2 rank is 1
document1 rank is 2
document3 rank is 3
document4 rank is 4
document2 is the most relevant document for the given query
VECTOR SPACE MODEL:
Code: Python code showing the implementation of the vector space model for document retrieval
Python
# implementation of vector space model for document retrievalimport pandas# module to read the contents of the file from a csv filefrom contextlib import redirect_stdout# module to redirect the output to a text fileimport math# module to perform mathematical functionsterms = []# list to store the terms present in the documentskeys = []# list to store the names of the documentsvec_Dic = {}# dictionary to store the name of the document and the weight as listdicti = {}# dictionary to store the name of the document and the terms present in it as a# vectordummy_List = []# list for performing some operations and clearing themterm_Freq = {}# dictionary to store the term and the number of times of its occurrence in the# documentsidf = {}# dictionary to store the term and the inverse document frequencyweight = {}# dictionary to store the term and the weight which is the product of term# frequency and inverse document frequencydef filter(documents, rows, cols): '''function to read and separate the name of the documents and the terms present in it to a separate list from the data frame and also create a dictionary which has the name of the document as key and the terms present in it as the list of strings which is the value of the key''' for i in range(rows): for j in range(cols): # traversal through the data frame if(j == 0): # first column has the name of the document in the csv file keys.append(documents.loc[i].iat[j]) else: dummy_List.append(documents.loc[i].iat[j]) # dummy list to update the terms in the dictionary if documents.loc[i].iat[j] not in terms: # add the terms to the list if it is not present else continue terms.append(documents.loc[i].iat[j]) copy = dummy_List.copy() # copying the dummy list to a different list dicti.update({documents.loc[i].iat[0]: copy}) # adding the key value pair to a dictionary dummy_List.clear() # clearing the dummy listdef compute_Weight(doc_Count, cols): '''Function to compute the weight for each of the terms in the document. Here the weight is calculated with the help of term frequency and inverse document frequency''' for i in terms: # initially adding all the elements into the dictionary and initialising # the values as zero if i not in term_Freq: term_Freq.update({i: 0}) for key, value in dicti.items(): # to get the number of occurrence of each terms for k in value: if k in term_Freq: term_Freq[k] += 1 # value incremented by one if the term is found in the documents idf = term_Freq.copy() # copying the term frequency dictionary to a dictionary named idf which is # further neede for computation for i in term_Freq: term_Freq[i] = term_Freq[i]/cols # term frequency is number of occurrence divided by total number of # documents for i in idf: if idf[i] != doc_Count: idf[i] = math.log2(cols / idf[i]) # inverse document frequency log of total number of documents divided # by number of occurrence of the terms else: idf[i] = 0 # this is to avoid the zero division error for i in idf: weight.update({i: idf[i]*term_Freq[i]}) # weight is the product of term frequency and the inverse document # frequency for i in dicti: for j in dicti[i]: dummy_List.append(weight[j]) copy = dummy_List.copy() vec_Dic.update({i: copy}) dummy_List.clear() # above operations performed to get the dictionary of weighted vector # for each of the documentsdef get_Weight_For_Query(query): '''function to get the weight for each terms present in the query, here we consider the term frequency as the weight of the terms''' query_Freq = {} # initialisation of the dictionary with query terms as key and its weight as # the values for i in terms: # initially adding all the elements into the dictionary and initialising # the values as zero if i not in query_Freq: query_Freq.update({i: 0}) for val in query: # to get the number of occurrence of each terms if val in query_Freq: query_Freq[val] += 1 # value incremented by one if the term is found in the documents for i in query_Freq: query_Freq[i] = query_Freq[i] / len(query) # term frequency obtained by dividing the number of occurrence of terms by # total number of terms in the query return query_Freq # return the dictionary in which the key is the term and the value is the # weightdef similarity_Computation(query_Weight): ''' Function to calculate the similarity measure in which the weight of the query and the document is multiplied in the numerator and the weight is squared and squareroot is taken the weights of the query and document''' numerator = 0 denomi1 = 0 denomi2 = 0 # initialisation of the variables with zero which is needed for computation similarity = {} # initialisation of dictionary which has the name of document as key and the # similarity measure as value for document in dicti: for terms in dicti[document]: # cosine similarity is calculated numerator += weight[terms] * query_Weight[terms] denomi1 += weight[terms] * weight[terms] denomi2 += query_Weight[terms] * query_Weight[terms] # the summation values of the weight is calculated and later they are # divided if denomi1 != 0 and denomi2 != 0: # to avoid the zero division error simi = numerator / (math.sqrt(denomi1) * math.sqrt(denomi2)) similarity.update({document: simi}) #dictionary is updated numerator = 0 denomi2 = 0 denomi1 = 0 # reinitialisation of the variables to zero return (similarity) # the dictionary containing similarity measure as the valuedef prediction(similarity, doc_count): '''Function to predict the document which is relevant to the query ''' with open('output.txt', 'w') as f: with redirect_stdout(f): # to redirect the output to a text file ans = max(similarity, key=similarity.get) print(ans, "is the most relevant document") # to print the name of the document which is most relevant print("ranking of the documents") for i in range(doc_count): ans = max(similarity, key=lambda x: similarity[x]) print(ans, "rank is", i+1) # to print the document name and its rank similarity.pop(ans)def main(): documents = pandas.read_csv(r'documents.csv') # to read the data from the csv file as a dataframe rows = len(documents) # to get the number of rows cols = len(documents.columns) # to get the number of columns filter(documents, rows, cols) # function call to read and separate the name of the documents and the terms # present in it to a separate list from the data frame and also create a # dictionary which has the name of the document as key and the terms present # in it as the list of strings which is the value of the key compute_Weight(rows, cols) # Function to compute the weight for each of the terms in the document. # Here the weight is calculated with the help of term frequency and inverse # document frequency print("Enter the query") query = input() # to get the query input from the user, the below input is given for obtaining # the output as in output.txt file # one three three query = query.split(' ') # splitting the query as a list of strings query_Weight = get_Weight_For_Query(query) # function call to get the weight for each terms present in the query, here we # consider the term frequency as the weight of the terms''' similarity = similarity_Computation(query_Weight) # Function call to calculate the similarity measure in which the weight of the # query and the document is multiplied in the numerator and the weight is # squared and squareroot is taken the weights of the query and document prediction(similarity, rows) # Function call to predict the document which is relevant to the querymain() |
Output: The output obtained in the text file when the query is “one three three”
ranking of the documents
document3 rank is 1
document2 rank is 2
document1 rank is 3
document4 rank is 4
document3 is the most relevant document
