2018-10-17
Oct 17 In-Class Exercise Thread.
Hey Everyone,
Please post your solutions to the Oct 17 In-class Exercise to his thread.
Best,
Chris
(Edited: 2018-10-17) Hey Everyone,
Please post your solutions to the Oct 17 In-class Exercise to his thread.
Best,
Chris
ScoreBM25(q, d) = 5* log( 64) * (k1+1) * 1/( 1+ k1) = 5* 6 = 30
if l_d = 2 * l_average
ScoreBM25(q, d) = 5 * log (64) * (k1+1) * 1/ ( 1+ k1* (1+b)) = 30 * 2.2 / ( 1+ 1.2*1.75) = 21.3
(Edited: 2018-10-17) ScoreBM25(q, d) = 5* log( 64) * (k1+1) * 1/( 1+ k1) = 5* 6 = 30
if l_d = 2 * l_average
ScoreBM25(q, d) = 5 * log (64) * (k1+1) * 1/ ( 1+ k1* (1+b)) = 30 * 2.2 / ( 1+ 1.2*1.75) = 21.3
for term1 of query = log(N/Nt) * (1*(k+1))/1+k1(1-b+b(ld/lavg))
log(N/Nt) * (K+1)/(K1+1)
6
for 5 docs = 6 * 5 = 30
Nt = N/64
second part = 5 * 6 * (k1+1)/(k1+1+k1*b) = 30*0.75 = 21.3
(Edited: 2018-10-17) for term1 of query = log(N/Nt) * (1*(k+1))/1+k1(1-b+b(ld/lavg))
= log(N/Nt) * (K+1)/(K1+1)
= 6
for 5 docs = 6 * 5 = 30
Nt = N/64
second part = 5 * 6 * (k1+1)/(k1+1+k1*b) = 30*0.75 = 21.3
((resource:20181017_152813.jpg|Resource Description for 20181017_152813.jpg))
1)
IDF = log(64/1)=6
TF = 5*(k1+1)/(1+k1)=5
TF-IDF=30
2)
TF= 5*(k1+1)/(1+k1*(1-b+2b))
TF-IDF=30(k1+1)/(k1+k1*b+1) =30(1.2+1)/(1.2+1.2*0.75+1) = 21.29
(Edited: 2018-10-17) 1)
IDF = log(64/1)=6
TF = 5*(k1+1)/(1+k1)=5
TF-IDF=30
2)
TF= 5*(k1+1)/(1+k1*(1-b+2b))
TF-IDF=30(k1+1)/(k1+k1*b+1) =30(1.2+1)/(1.2+1.2*0.75+1) = 21.29
`N/N_t = 64, k = 1.2, b = 0.75`
`mbox{Score_BM25} (q, d) = log_2 64 * (5 * (1.2+1)/(1.2+1)) = 6 * 5 * (22)/(22) = 30`
`mbox{Score_BM25} (q, d_2) = log_2 64 * (5 * (1.2+1)/(1+1.75*1.2)) = 6 * 5 * (22)/(31) = 21.29`
(Edited: 2018-10-17) @BT@N/N_t = 64, k = 1.2, b = 0.75@BT@
@BT@mbox{Score_BM25} (q, d) = log_2 64 * (5 * (1.2+1)/(1.2+1)) = 6 * 5 * (22)/(22) = 30@BT@
@BT@mbox{Score_BM25} (q, d_2) = log_2 64 * (5 * (1.2+1)/(1+1.75*1.2)) = 6 * 5 * (22)/(31) = 21.29@BT@
1. Score_BM25(q, d) = `5 * log(64) * (1 * (1.2+1)) / (1 + 1.2((1-0.75) + 0.75*1)) = 5 * 6 * 2.2/2.2 = 30`
2. Score_BM25(q, d) = `5 * log(64) * (1 * (1.2+1)) / (1 + 1.2((1-0.75) + 0.75*2)) = 5 * 6 * 2.2/(1 + 1.2 * 1.75) = 30 * 2.2/(3.1) = 21.29`
(Edited: 2018-10-17) 1. Score_BM25(q, d) = @BT@5 * log(64) * (1 * (1.2+1)) / (1 + 1.2((1-0.75) + 0.75*1)) = 5 * 6 * 2.2/2.2 = 30@BT@
2. Score_BM25(q, d) = @BT@5 * log(64) * (1 * (1.2+1)) / (1 + 1.2((1-0.75) + 0.75*2)) = 5 * 6 * 2.2/(1 + 1.2 * 1.75) = 30 * 2.2/(3.1) = 21.29@BT@
1. log(N/Nt) = log(64) = 6
(Edited: 2018-10-17)
ScoreBM25(q,d) = 6*5 = 30
2. ScoreBM25(q,d) = (6*5((k1+1)/(1+k1(1+b))) = 6*(5*(2.2/3.1)) = 21.29
1. log(N/Nt) = log(64) = 6
ScoreBM25(q,d) = 6*5 = 30
2. ScoreBM25(q,d) = (6*5((k1+1)/(1+k1(1+b))) = 6*(5*(2.2/3.1)) = 21.29
a) IDF = `log(N/(N/64))` = log 64 = 6; TF = `((1 * (k1 + 1))/ (1 + (k1 * (1-b + b( 1)))))` = 1
score = 5 terms * (IDF*TF) = 5 * 6 *1 = 30
b) score = 5 * 6 * (2.2 */ (2.2+0.9)) = 5 * 6 * 0.71 = 21.29
(Edited: 2018-10-17) a) IDF = @BT@log(N/(N/64))@BT@ = log 64 = 6; TF = @BT@((1 * (k1 + 1))/ (1 + (k1 * (1-b + b( 1)))))@BT@ = 1
score = 5 terms * (IDF*TF) = 5 * 6 *1 = 30
b) score = 5 * 6 * (2.2 */ (2.2+0.9)) = 5 * 6 * 0.71 = 21.29
((resource:IMG_20181017_153418__01.jpg|Resource Description for IMG_20181017_153418__01.jpg))
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