Solving the issue of small number multiplied by small number gives a even smaller number
$begingroup$
So for my project I have certain scores associated with a food and I have to recommend a foodie a food based on the highest score:
For example: Ice-cream: 0.00004
Hot dogs: 0.00005
Fries: 0.00009
Based on this, foodie1 will get a recommendation of eating fries, then hot-dogs, and then icecream.
But, before I do that. I need to individualize the score based on the foodie's background. So for example foodie1 likes cold foods and therefore I want to multiply the score of ice-cream with the score of cold foods to give it just a slight bump based on the preference. But, the bump shouldn't be that overpowering that instead of recommending fries first it directly jumps to recommending ice cream (rather it should make the recommendation order: fries, ice cream, hot-dogs).
Issue is that the score for cold foods is also a small number like 0.1. Currently, I am using the process of multiplication. When I multiply 0.1*0.00004 it gives 0.0000004 which does exactly the opposite of what I want it to do (aka pushes ice cream to the bottom of the recommendation list). I can't use addition because then ice-cream overpowers the list.
Is there a work-around this like normalization for example?
normal-distribution
asked Dec 5 '18 at 21:47
Sabah LalaSabah Lala
31
$endgroup$
add a comment |
$begingroup$
So for my project I have certain scores associated with a food and I have to recommend a foodie a food based on the highest score:
For example: Ice-cream: 0.00004
Hot dogs: 0.00005
Fries: 0.00009
Based on this, foodie1 will get a recommendation of eating fries, then hot-dogs, and then icecream.
But, before I do that. I need to individualize the score based on the foodie's background. So for example foodie1 likes cold foods and therefore I want to multiply the score of ice-cream with the score of cold foods to give it just a slight bump based on the preference. But, the bump shouldn't be that overpowering that instead of recommending fries first it directly jumps to recommending ice cream (rather it should make the recommendation order: fries, ice cream, hot-dogs).
Issue is that the score for cold foods is also a small number like 0.1. Currently, I am using the process of multiplication. When I multiply 0.1*0.00004 it gives 0.0000004 which does exactly the opposite of what I want it to do (aka pushes ice cream to the bottom of the recommendation list). I can't use addition because then ice-cream overpowers the list.
Is there a work-around this like normalization for example?
normal-distribution
asked Dec 5 '18 at 21:47
Sabah LalaSabah Lala
31
$endgroup$
$begingroup$
multiply with $(1+score)$ to give the bump. It's like score% increase. So you'll multiple with 1.1 instead of 0.1
$endgroup$
– karakfa
Dec 5 '18 at 21:52
$begingroup$
The number $0.1$ means "only 10% as good." If your boss says "we'll pay you 10% of your old salary next year" then you are getting a very large pay cut. As karakfa wrote, if you want "10% better", use $1.1$.
$endgroup$
– David K
Dec 5 '18 at 21:55
$begingroup$
@karakfa You mean add 1 to both scores or just the multiplier score of cold foods?
$endgroup$
– Sabah Lala
Dec 6 '18 at 15:40
$begingroup$
you use the same term for both, it gets confusing. I'll use different words with larger numbers. Assume I got recommendation for A: 4, B: 5, C: 10. Based background you know that I value A more, so you want to raise it by 30%. Now my updated values will be A4:*1.3=5.2, B:5, C:10.
$endgroup$
– karakfa
Dec 6 '18 at 15:55
add a comment |
$begingroup$
So for my project I have certain scores associated with a food and I have to recommend a foodie a food based on the highest score:
For example: Ice-cream: 0.00004
Hot dogs: 0.00005
Fries: 0.00009
Based on this, foodie1 will get a recommendation of eating fries, then hot-dogs, and then icecream.
But, before I do that. I need to individualize the score based on the foodie's background. So for example foodie1 likes cold foods and therefore I want to multiply the score of ice-cream with the score of cold foods to give it just a slight bump based on the preference. But, the bump shouldn't be that overpowering that instead of recommending fries first it directly jumps to recommending ice cream (rather it should make the recommendation order: fries, ice cream, hot-dogs).
Issue is that the score for cold foods is also a small number like 0.1. Currently, I am using the process of multiplication. When I multiply 0.1*0.00004 it gives 0.0000004 which does exactly the opposite of what I want it to do (aka pushes ice cream to the bottom of the recommendation list). I can't use addition because then ice-cream overpowers the list.
Is there a work-around this like normalization for example?
normal-distribution
asked Dec 5 '18 at 21:47
Sabah LalaSabah Lala
31
$endgroup$
So for my project I have certain scores associated with a food and I have to recommend a foodie a food based on the highest score:
For example: Ice-cream: 0.00004
Hot dogs: 0.00005
Fries: 0.00009
Based on this, foodie1 will get a recommendation of eating fries, then hot-dogs, and then icecream.
But, before I do that. I need to individualize the score based on the foodie's background. So for example foodie1 likes cold foods and therefore I want to multiply the score of ice-cream with the score of cold foods to give it just a slight bump based on the preference. But, the bump shouldn't be that overpowering that instead of recommending fries first it directly jumps to recommending ice cream (rather it should make the recommendation order: fries, ice cream, hot-dogs).
Issue is that the score for cold foods is also a small number like 0.1. Currently, I am using the process of multiplication. When I multiply 0.1*0.00004 it gives 0.0000004 which does exactly the opposite of what I want it to do (aka pushes ice cream to the bottom of the recommendation list). I can't use addition because then ice-cream overpowers the list.
Is there a work-around this like normalization for example?
normal-distribution
normal-distribution
asked Dec 5 '18 at 21:47
Sabah LalaSabah Lala
31
asked Dec 5 '18 at 21:47
Sabah LalaSabah Lala
31
asked Dec 5 '18 at 21:47
Sabah LalaSabah Lala
31
asked Dec 5 '18 at 21:47
Sabah LalaSabah Lala
31
asked Dec 5 '18 at 21:47
Sabah LalaSabah Lala
31
31
$begingroup$
multiply with $(1+score)$ to give the bump. It's like score% increase. So you'll multiple with 1.1 instead of 0.1
$endgroup$
– karakfa
Dec 5 '18 at 21:52
$begingroup$
The number $0.1$ means "only 10% as good." If your boss says "we'll pay you 10% of your old salary next year" then you are getting a very large pay cut. As karakfa wrote, if you want "10% better", use $1.1$.
$endgroup$
– David K
Dec 5 '18 at 21:55
$begingroup$
@karakfa You mean add 1 to both scores or just the multiplier score of cold foods?
$endgroup$
– Sabah Lala
Dec 6 '18 at 15:40
$begingroup$
you use the same term for both, it gets confusing. I'll use different words with larger numbers. Assume I got recommendation for A: 4, B: 5, C: 10. Based background you know that I value A more, so you want to raise it by 30%. Now my updated values will be A4:*1.3=5.2, B:5, C:10.
$endgroup$
– karakfa
Dec 6 '18 at 15:55
add a comment |
$begingroup$
multiply with $(1+score)$ to give the bump. It's like score% increase. So you'll multiple with 1.1 instead of 0.1
$endgroup$
– karakfa
Dec 5 '18 at 21:52
$begingroup$
The number $0.1$ means "only 10% as good." If your boss says "we'll pay you 10% of your old salary next year" then you are getting a very large pay cut. As karakfa wrote, if you want "10% better", use $1.1$.
$endgroup$
– David K
Dec 5 '18 at 21:55
$begingroup$
@karakfa You mean add 1 to both scores or just the multiplier score of cold foods?
$endgroup$
– Sabah Lala
Dec 6 '18 at 15:40
$begingroup$
you use the same term for both, it gets confusing. I'll use different words with larger numbers. Assume I got recommendation for A: 4, B: 5, C: 10. Based background you know that I value A more, so you want to raise it by 30%. Now my updated values will be A4:*1.3=5.2, B:5, C:10.
$endgroup$
– karakfa
Dec 6 '18 at 15:55
$begingroup$
multiply with $(1+score)$ to give the bump. It's like score% increase. So you'll multiple with 1.1 instead of 0.1
$endgroup$
– karakfa
Dec 5 '18 at 21:52
$begingroup$
multiply with $(1+score)$ to give the bump. It's like score% increase. So you'll multiple with 1.1 instead of 0.1
$endgroup$
– karakfa
Dec 5 '18 at 21:52
$begingroup$
The number $0.1$ means "only 10% as good." If your boss says "we'll pay you 10% of your old salary next year" then you are getting a very large pay cut. As karakfa wrote, if you want "10% better", use $1.1$.
$endgroup$
– David K
Dec 5 '18 at 21:55
$begingroup$
The number $0.1$ means "only 10% as good." If your boss says "we'll pay you 10% of your old salary next year" then you are getting a very large pay cut. As karakfa wrote, if you want "10% better", use $1.1$.
$endgroup$
– David K
Dec 5 '18 at 21:55
$begingroup$
@karakfa You mean add 1 to both scores or just the multiplier score of cold foods?
$endgroup$
– Sabah Lala
Dec 6 '18 at 15:40
$begingroup$
@karakfa You mean add 1 to both scores or just the multiplier score of cold foods?
$endgroup$
– Sabah Lala
Dec 6 '18 at 15:40
$begingroup$
you use the same term for both, it gets confusing. I'll use different words with larger numbers. Assume I got recommendation for A: 4, B: 5, C: 10. Based background you know that I value A more, so you want to raise it by 30%. Now my updated values will be A4:*1.3=5.2, B:5, C:10.
$endgroup$
– karakfa
Dec 6 '18 at 15:55
$begingroup$
you use the same term for both, it gets confusing. I'll use different words with larger numbers. Assume I got recommendation for A: 4, B: 5, C: 10. Based background you know that I value A more, so you want to raise it by 30%. Now my updated values will be A4:*1.3=5.2, B:5, C:10.
$endgroup$
– karakfa
Dec 6 '18 at 15:55
add a comment |
1 Answer
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$begingroup$
Make Ice Cream 1.00004, Hot dogs 1.00005, etc.
Multiplying them will now result in a larger number
If you want, you can subtract 1 at the end, but I think making it all in terms of 1.x is more helpful because they are positive-influence multipliers. A multiplier <1 means that it has a negative influence
answered Dec 5 '18 at 23:17
zach274zach274
161
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add a comment |
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$begingroup$
Make Ice Cream 1.00004, Hot dogs 1.00005, etc.
Multiplying them will now result in a larger number
If you want, you can subtract 1 at the end, but I think making it all in terms of 1.x is more helpful because they are positive-influence multipliers. A multiplier <1 means that it has a negative influence
answered Dec 5 '18 at 23:17
zach274zach274
161
$endgroup$
add a comment |
$begingroup$
Make Ice Cream 1.00004, Hot dogs 1.00005, etc.
Multiplying them will now result in a larger number
If you want, you can subtract 1 at the end, but I think making it all in terms of 1.x is more helpful because they are positive-influence multipliers. A multiplier <1 means that it has a negative influence
answered Dec 5 '18 at 23:17
zach274zach274
161
$endgroup$
add a comment |
$begingroup$
Make Ice Cream 1.00004, Hot dogs 1.00005, etc.
Multiplying them will now result in a larger number
If you want, you can subtract 1 at the end, but I think making it all in terms of 1.x is more helpful because they are positive-influence multipliers. A multiplier <1 means that it has a negative influence
answered Dec 5 '18 at 23:17
zach274zach274
161
$endgroup$
Make Ice Cream 1.00004, Hot dogs 1.00005, etc.
Multiplying them will now result in a larger number
If you want, you can subtract 1 at the end, but I think making it all in terms of 1.x is more helpful because they are positive-influence multipliers. A multiplier <1 means that it has a negative influence
answered Dec 5 '18 at 23:17
zach274zach274
161
answered Dec 5 '18 at 23:17
zach274zach274
161
answered Dec 5 '18 at 23:17
zach274zach274
161
answered Dec 5 '18 at 23:17
zach274zach274
161
161
add a comment |
add a comment |
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$begingroup$
multiply with $(1+score)$ to give the bump. It's like score% increase. So you'll multiple with 1.1 instead of 0.1
$endgroup$
– karakfa
Dec 5 '18 at 21:52
$begingroup$
The number $0.1$ means "only 10% as good." If your boss says "we'll pay you 10% of your old salary next year" then you are getting a very large pay cut. As karakfa wrote, if you want "10% better", use $1.1$.
$endgroup$
– David K
Dec 5 '18 at 21:55
$begingroup$
@karakfa You mean add 1 to both scores or just the multiplier score of cold foods?
$endgroup$
– Sabah Lala
Dec 6 '18 at 15:40
$begingroup$
you use the same term for both, it gets confusing. I'll use different words with larger numbers. Assume I got recommendation for A: 4, B: 5, C: 10. Based background you know that I value A more, so you want to raise it by 30%. Now my updated values will be A4:*1.3=5.2, B:5, C:10.
$endgroup$
– karakfa
Dec 6 '18 at 15:55