Mixing
Normalize numerical values
$begingroup$
In a AI course, we have to normalize a set of numerical inputs such that the highest input becomes $1$ and the lowest becomes $0$. The set is as follows:
$ 1\3\5\2 $
So the first entry should become $0$ and the third one becomes $1$.
Now how do I calculate the second and last one? I can't just divide by 5 because dividing one by five would give $0.2$ and not $0$.
machine-learning
asked Jan 26 at 12:45
Wouter VandenputteWouter Vandenputte
1295
$endgroup$
add a comment |
$begingroup$
In a AI course, we have to normalize a set of numerical inputs such that the highest input becomes $1$ and the lowest becomes $0$. The set is as follows:
$ 1\3\5\2 $
So the first entry should become $0$ and the third one becomes $1$.
Now how do I calculate the second and last one? I can't just divide by 5 because dividing one by five would give $0.2$ and not $0$.
machine-learning
asked Jan 26 at 12:45
Wouter VandenputteWouter Vandenputte
1295
$endgroup$
$begingroup$
what's the problem with0.2?
$endgroup$
– OmG
Jan 26 at 12:46
$begingroup$
the lowest value (here 1) should become $ 0 $
$endgroup$
– Wouter Vandenputte
Jan 26 at 12:48
add a comment |
$begingroup$
In a AI course, we have to normalize a set of numerical inputs such that the highest input becomes $1$ and the lowest becomes $0$. The set is as follows:
$ 1\3\5\2 $
So the first entry should become $0$ and the third one becomes $1$.
Now how do I calculate the second and last one? I can't just divide by 5 because dividing one by five would give $0.2$ and not $0$.
machine-learning
asked Jan 26 at 12:45
Wouter VandenputteWouter Vandenputte
1295
$endgroup$
In a AI course, we have to normalize a set of numerical inputs such that the highest input becomes $1$ and the lowest becomes $0$. The set is as follows:
$ 1\3\5\2 $
So the first entry should become $0$ and the third one becomes $1$.
Now how do I calculate the second and last one? I can't just divide by 5 because dividing one by five would give $0.2$ and not $0$.
machine-learning
machine-learning
asked Jan 26 at 12:45
Wouter VandenputteWouter Vandenputte
1295
asked Jan 26 at 12:45
Wouter VandenputteWouter Vandenputte
1295
edited Jan 26 at 12:49
Wouter Vandenputte
asked Jan 26 at 12:45
Wouter VandenputteWouter Vandenputte
1295
asked Jan 26 at 12:45
Wouter VandenputteWouter Vandenputte
1295
asked Jan 26 at 12:45
Wouter VandenputteWouter Vandenputte
1295
1295
$begingroup$
what's the problem with0.2?
$endgroup$
– OmG
Jan 26 at 12:46
$begingroup$
the lowest value (here 1) should become $ 0 $
$endgroup$
– Wouter Vandenputte
Jan 26 at 12:48
add a comment |
$begingroup$
what's the problem with0.2?
$endgroup$
– OmG
Jan 26 at 12:46
$begingroup$
the lowest value (here 1) should become $ 0 $
$endgroup$
– Wouter Vandenputte
Jan 26 at 12:48
$begingroup$
what's the problem with
0.2?$endgroup$
– OmG
Jan 26 at 12:46
$begingroup$
what's the problem with
0.2?$endgroup$
– OmG
Jan 26 at 12:46
$begingroup$
the lowest value (here 1) should become $ 0 $
$endgroup$
– Wouter Vandenputte
Jan 26 at 12:48
$begingroup$
the lowest value (here 1) should become $ 0 $
$endgroup$
– Wouter Vandenputte
Jan 26 at 12:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You just need the following formula:
$$frac{x-min}{max-min}$$
Which $min$ and $max$ are the minimum and maximum value in data which is $1$ and $5$ here, respectively.
answered Jan 26 at 12:51
OmGOmG
2,512824
$endgroup$
add a comment |
$begingroup$
You want to map all the values you have to the interval $[0,1]$ -- that's what's meant by normalise. So, if $A = {1,3,5,2}$ is your set of values, you need the width of $A$, i.e.
$$ w(a) := max_{x in A} x - min{xin A} x $$
Then you need to rebase $A$ to have zero in it. This can be done by subtracting the minimum element from the set. So for each element in the set $A$ you translate and then divide by the width, i.e.
$$ x mapsto frac{x-min_{yin A} y}{w(A)} $$
answered Jan 26 at 12:54
postmortespostmortes
2,21031322
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You just need the following formula:
$$frac{x-min}{max-min}$$
Which $min$ and $max$ are the minimum and maximum value in data which is $1$ and $5$ here, respectively.
answered Jan 26 at 12:51
OmGOmG
2,512824
$endgroup$
add a comment |
$begingroup$
You just need the following formula:
$$frac{x-min}{max-min}$$
Which $min$ and $max$ are the minimum and maximum value in data which is $1$ and $5$ here, respectively.
answered Jan 26 at 12:51
OmGOmG
2,512824
$endgroup$
add a comment |
$begingroup$
You just need the following formula:
$$frac{x-min}{max-min}$$
Which $min$ and $max$ are the minimum and maximum value in data which is $1$ and $5$ here, respectively.
answered Jan 26 at 12:51
OmGOmG
2,512824
$endgroup$
You just need the following formula:
$$frac{x-min}{max-min}$$
Which $min$ and $max$ are the minimum and maximum value in data which is $1$ and $5$ here, respectively.
answered Jan 26 at 12:51
OmGOmG
2,512824
answered Jan 26 at 12:51
OmGOmG
2,512824
answered Jan 26 at 12:51
OmGOmG
2,512824
answered Jan 26 at 12:51
OmGOmG
2,512824
2,512824
add a comment |
add a comment |
$begingroup$
You want to map all the values you have to the interval $[0,1]$ -- that's what's meant by normalise. So, if $A = {1,3,5,2}$ is your set of values, you need the width of $A$, i.e.
$$ w(a) := max_{x in A} x - min{xin A} x $$
Then you need to rebase $A$ to have zero in it. This can be done by subtracting the minimum element from the set. So for each element in the set $A$ you translate and then divide by the width, i.e.
$$ x mapsto frac{x-min_{yin A} y}{w(A)} $$
answered Jan 26 at 12:54
postmortespostmortes
2,21031322
$endgroup$
add a comment |
$begingroup$
You want to map all the values you have to the interval $[0,1]$ -- that's what's meant by normalise. So, if $A = {1,3,5,2}$ is your set of values, you need the width of $A$, i.e.
$$ w(a) := max_{x in A} x - min{xin A} x $$
Then you need to rebase $A$ to have zero in it. This can be done by subtracting the minimum element from the set. So for each element in the set $A$ you translate and then divide by the width, i.e.
$$ x mapsto frac{x-min_{yin A} y}{w(A)} $$
answered Jan 26 at 12:54
postmortespostmortes
2,21031322
$endgroup$
add a comment |
$begingroup$
You want to map all the values you have to the interval $[0,1]$ -- that's what's meant by normalise. So, if $A = {1,3,5,2}$ is your set of values, you need the width of $A$, i.e.
$$ w(a) := max_{x in A} x - min{xin A} x $$
Then you need to rebase $A$ to have zero in it. This can be done by subtracting the minimum element from the set. So for each element in the set $A$ you translate and then divide by the width, i.e.
$$ x mapsto frac{x-min_{yin A} y}{w(A)} $$
answered Jan 26 at 12:54
postmortespostmortes
2,21031322
$endgroup$
You want to map all the values you have to the interval $[0,1]$ -- that's what's meant by normalise. So, if $A = {1,3,5,2}$ is your set of values, you need the width of $A$, i.e.
$$ w(a) := max_{x in A} x - min{xin A} x $$
Then you need to rebase $A$ to have zero in it. This can be done by subtracting the minimum element from the set. So for each element in the set $A$ you translate and then divide by the width, i.e.
$$ x mapsto frac{x-min_{yin A} y}{w(A)} $$
answered Jan 26 at 12:54
postmortespostmortes
2,21031322
answered Jan 26 at 12:54
postmortespostmortes
2,21031322
answered Jan 26 at 12:54
postmortespostmortes
2,21031322
answered Jan 26 at 12:54
postmortespostmortes
2,21031322
2,21031322
add a comment |
add a comment |
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$begingroup$
what's the problem with
0.2?$endgroup$
– OmG
Jan 26 at 12:46
$begingroup$
the lowest value (here 1) should become $ 0 $
$endgroup$
– Wouter Vandenputte
Jan 26 at 12:48