I have a database of 817 items, each given a "rank" of 1 to 817 (the smaller the number, the "better" the item). This rank is based off of many factors that indicate quality.
Now, I need to assign a "value" to these items, with the item at rank 1 being valued the most, and the value decreasing with rank (non-linear).
The easiest first attempt was to simply choose an arbitrary base (100,000) and divide by the rank:
$value = 100000 / $rank;
/**
* Rank : Value
* 1 : 100,000
* 2 : 50,000
* 3 : 33,333
* etc.
*/
This produces exponential decay, as shown in the red line in this image:
However, I wish to value these items in a manner that looks more like the blue line above. How can I change my formula to achieve this?
Try 1/sqrt(x) (i.e, pow(x, -1/2)) for starters. If that's still not slow enough, try a smaller fractional power.
Why don't you go with linear?
value = n - rank
where n is the count of your items, i.e. 817.
I haven't tried but use exponent instead of dividing by 1000 of a base 2.
UPDATES
value = 2 pow (n-rank)
Related
So in my text book there is this example of a recursive function using f#
let rec gcd = function
| (0,n) -> n
| (m,n) -> gcd(n % m,m);;
with this function my text book gives the example by executing:
gcd(36,116);;
and since the m = 36 and not 0 then it ofcourse goes for the second clause like this:
gcd(116 % 36,36)
gcd(8,36)
gcd(36 % 8,8)
gcd(4,8)
gcd(8 % 4,4)
gcd(0,4)
and now hits the first clause stating this entire thing is = 4.
What i don't get is this (%)percentage sign/operator or whatever it is called in this connection. for an instance i don't get how
116 % 36 = 8
I have turned this so many times in my head now and I can't figure how this can turn into 8?
I know this is probably a silly question for those of you who knows this but I would very much appreciate your help the same.
% is a questionable version of modulo, which is the remainder of an integer division.
In the positive, you can think of % as the remainder of the division. See for example Wikipedia on Euclidean Divison. Consider 9 % 4: 4 fits into 9 twice. But two times four is only eight. Thus, there is a remainder of one.
If there are negative operands, % effectively ignores the signs to calculate the remainder and then uses the sign of the dividend as the sign of the result. This corresponds to the remainder of an integer division that rounds to zero, i.e. -2 / 3 = 0.
This is a mathematically unusual definition of division and remainder that has some bad properties. Normally, when calculating modulo n, adding or subtracting n on the input has no effect. Not so for this operator: 2 % 3 is not equal to (2 - 3) % 3.
I usually have the following defined to get useful remainders when there are negative operands:
/// Euclidean remainder, the proper modulo operation
let inline (%!) a b = (a % b + b) % b
So far, this operator was valid for all cases I have encountered where a modulo was needed, while the raw % repeatedly wasn't. For example:
When filling rows and columns from a single index, you could calculate rowNumber = index / nCols and colNumber = index % nCols. But if index and colNumber can be negative, this mapping becomes invalid, while Euclidean division and remainder remain valid.
If you want to normalize an angle to (0, 2pi), angle %! (2. * System.Math.PI) does the job, while the "normal" % might give you a headache.
Because
116 / 36 = 3
116 - (3*36) = 8
Basically, the % operator, known as the modulo operator will divide a number by other and give the rest if it can't divide any longer. Usually, the first time you would use it to understand it would be if you want to see if a number is even or odd by doing something like this in f#
let firstUsageModulo = 55 %2 =0 // false because leaves 1 not 0
When it leaves 8 the first time means that it divided you 116 with 36 and the closest integer was 8 to give.
Just to help you in future with similar problems: in IDEs such as Xamarin Studio and Visual Studio, if you hover the mouse cursor over an operator such as % you should get a tooltip, thus:
Module operator tool tip
Even if you don't understand the tool tip directly, it'll give you something to google.
i would like to know if you can help me with this problem for my game. I'm currently using lots of switch, if-else, etc on my code and i'm not liking it at all.
I would like to generate 2 random arithmethic expressions that have one of the forms like the ones bellow:
1) number
e.g.: 19
2) number operation number
e.g.: 22 * 4
3) (number operation number) operation number
e.g.: (10 * 4) / 5
4) ((number operation number) operation number) operation number
e.g.: ((25 * 2) / 10) - 2
After i have the 2 arithmetic expresions, the game consist in matching them and determine which is larger.
I would like to know how can i randomly choose the numbers and operations for each arithmetic expression in order to have an integer result (not float) and also that both expression have results that are as close as possible. The individual numbers shouldn't be higher than 30.
I mean, i wouldn't like a result to be 1000 and the other 14 because they would be probably too easy to spot which side is larger, so they should be like:
expresion 1: ((25 + 15) / 10) * 4 (which is 16)
expression 2: (( 7 * 2) + 10) / 8 (which is 3)
The results (16 and 3) are integers and close enough to each other.
the posible operations are +, -, * and /
It would be possible to match between two epxressions with different forms, like
(( 7 * 2) + 10) / 8
and
(18 / 3) * 2
I really appreciate all the help that you can give me.
Thanks in advance!!
Best regards.
I think a reasonable way to approach this is to start with a value for the total and recursively construct a random expression tree to reach that total. You can choose how many operators you want in each equation and ensure that all values are integers. Plus, you can choose how close you want the values of two equations, even making them equal if you wish. I'll use your expression 1 above as an example.
((25 + 15) / 10) * 4 = 16
We start with the total 16 and make that the root of our tree:
16
To expand a node (leaf), we select an operator and set that as the value of the node, and create two children containing the operands. In this case, we choose multiplication as our operator.
Multiplication is the only operator that will really give us trouble in trying to keep all of the operands integers. We can satisfy this constraint by constructing a table of divisors of integers in our range [1..30] (or maybe a bit more, as we'll see below). In this case our table would have told us that the divisors of 16 are {2,4,8}. (If the list of divisors for our current value is empty, we can choose a different operator, or a different leaf altogether.)
We choose a random divisor, say 4 and set that as the right child of our node. The left child is obviously value/right, also an integer.
*
/ \
4 4
Now we need to select another leaf to expand. We can randomly choose a leaf, randomly walk the tree until we reach a leaf, randomly walk up and right from our current child node (left) until we reach a leaf, or whatever.
In this case our selection algorithm chooses to expand the left child and the division operator. In the case of division, we generate a random number for the right child (in this case 10), and set left to value*right. (Order is important here! Not so for multiplication.)
*
/ \
÷ 4
/ \
40 10
This demonstrates why I said that the divisor table might need to go beyond our stated range as some of the intermediate values may be a bit larger than 30. You can tweak your code to avoid this, or make sure that large values are further expanded before reaching the final equation.
In the example we do this by selecting the leftmost child to expand with the addition operator. In this case, we can simply select a random integer in the range [1..value-1] for the right child and value-right for the left.
*
/ \
÷ 4
/ \
+ 10
/ \
25 15
You can repeat for as many operations as you want. To reconstruct the final equation, you simply need to perform an in-order traversal of the tree. To parenthesize as in your examples, you would place parentheses around the entire equation when leaving any interior (operator) node during the traversal, except for the root.
Ok I have the following problem:
I have several ranks in a matrix in r. (I've got this by ranking asset returns. Ranks>=3 get an NA, Ranks <3 get the rank number. If some assets share a rank, less NAs are in a row). Here are two example rows and two example rows of a matrix with returns.
ranks<-matrix(c(1,1,2,NA,NA, 1,2,NA,NA,NA),nrow=2,ncol=5)
returns<-matrix(c(0.3,0.1,-0.5,-0.7,0.2,0.1,0.4,0.05,-0.7,-0.3),nrow=2,ncol=5)
Now if all assets are equally bought for our portfolio, I can calculate the average return with:
Mat.Ret<-returns*ranks
Mean.Ret<-rowMeans(Mat.Ret,na.rm=TRUE)
However I want to have the option of giving a vector of weights for the two ranks and these weights say how big of a percentage this particular rank should have in my portfolio. As an example we have a vector of
weights<-c(0.7,0.3)
Now how would I use this in my code? I want to calculate basically ranks*returns*weights. If only ONE rank 1 and ONE rank 2 are in the table, the code works. But how would I do this variable? I mean a solution would be to calculate for each rank how many times it exists in a particular row and then divide the weight by this count. And then I would multiply this "net weight" * rank * returns.
But I have no clue how to do this in code..any help?
UPDATE AFTER FIRST COMMENTS
Ok I want to Keep it flexible to adjust the weights depending on how many times a certain rank is given. A user can choose the top 5 ranked assets, so none or several assets may share ranks. So the distribution of weights must be very flexible. I've programmed a formula which doesn't work yet since I'm obviously not yet experienced enough with the whole matrix and vector selection syntax I guess. This is what I got so far:
ranks<-apply(ranks,1,function(x)distributeWeightsPerMatrixRow(x,weights))
distributeWeightsPerMatrixRow<-function(MatrixRow,Weights){
if(length(Weights)==length(MatrixRow[!is.na(MatrixRow)])){
MatrixRow <- Weights[MatrixRow]
} else {
for(i in 1:length(MatrixRow)){
if(!is.na(MatrixRow[i])){
EqWeights<-length(MatrixRow[MatrixRow==MatrixRow[i]])
MatrixRow[i]<-sum(Weights[MatrixRow[i]:(MatrixRow[i]+EqWeights-1)])/EqWeights
}
}
}
return(MatrixRow)
}
EDIT2:
Function seems to work, however now the resulting ranks object is the transposed version of the original matrix without the column names..
Since your ranks are integers above zero, you can use this matrix for indexing the vector ranks:
mat.weights <- weights[ranks]
mat.weighted.ret <- returns * ranks * mat.weights
Update based on comment.
I suppose you're looking for something like this:
if (length(unique(na.omit(as.vector(ranks)))) == 1)
mat.weights <- (!is.na(ranks)) * 0.5
else
mat.weights <- weights[ranks]
mat.weighted.ret <- returns * ranks * mat.weights
If there is only one rank. All weights become 0.5.
I have written an algorithm that calculates the number of zero-crossings within a signal. By this, I mean the number of times a value changes from + to - and vice-versa.
The algorithm is explained like this:
If there are the following elements:
v1 = {90, -4, -3, 1, 3}
Then you multiply the value by the value next to it. (i * i+1)
Then taking the sign value sign(val) determine if this is positive or negative. Example:
e1 = {90 * -4} = -360 -> sigum(e1) = -1
e2 = {-4 * -3} = 12 -> signum(e2) = 1
e3 = {-3 * 1} = -3 -> signum(e3) = -1
e4 = {1 * 3} = 3 -> signum(e4) = 1
Therefore the total number of values changed from negative to positive is = 2 ..
Now I want to put this forumular, algorithm into an equation so that I can present it.
I have asked a simular question, but got really confused so went away and thought about it and came up with (what I think the equation should look like).. It's probably wrong, well, laughably wrong. But here it is:
Now the logic behind it:
I pass in a V (val)
I get the absolute value of the summation of the signum from calculating (Vi * Vi+1) .. The signum(Vi * Vi+1) should produce -1, 1, ..., values
If and only if the value is -1 (Because I'm only interested in the number of times zero is crossed, therefore, the zero values.
Does this look correct, if not, can anyone suggest improvements?
Thank you :)!
EDIT:
Is this correct now?
You are doing the right thing here but your equation is wrong simply because you only want to count the sign of the product of adjacent elements when it is negative. Dont sum the sign of products since positive sign products should be neglected. For this reason, an explicit mathematical formula is tricky as positive products between adjacent elements should be ignored. What you want is a function that takes 2 arguments and evaluates to 1 when their product is negative and zero when non-negative
f(x,y) = 1 if xy < 0
= 0 otherwise
then your number of crossing points is simply given by
sum(f(v1[i],v1[i+1])) for i = 0 to i = n-1
where n is the length of your vector/array v1 (using C style array access notation based on zero indexing). You also have to consider edge conditions such as 4 consecutive points {-1,0,0,1} - do you want to consider this as simply one zero crossing or 2??? Only you can answer this based on the specifics of your problem, but whatever your answer adjust your algorithm accordingly.
I am a newbie to R. I would like to create a barplot which is visually divided into different parts.
My data looks like the following:
"1.0";"0.0";"1.0";"2.0";"710";"12500"
first four numbers give the number of parts that need to be ordered from the two parts list below. The fifth number gives the result sum of the first part, the sixth then the result sum of the second part.
part 1: 10;50;100;300
part 2: 500;1000;2000;5000;
this is how it is calculated.
1 * 10 + 0 * 50 + 1 * 100 + 2 * 300 = 710 ;
1 * 500 + 0 * 1000 + 1 * 2000 + 2 * 5000 = 12500
So what I now want to plot is for example the value 12500, but I want to visually divide this value into the different part (stacked bars) like two five thousands, then one two thousand then one five hundred -> the bar should consist of these parts which can visually be seen or marked with the value (would be nice to have different colors for each part in the part)
How can I do it? Folks, I did my homework, I searched a lot and did try it on my own, but couldn't achieve what I want.
daten <- matrix(c(10,50,100,300,500,1000,2000,5000),ncol=2)
multiplier <- c(1,0,1,2)
barplot(daten*multiplier)
To display bar segments in reverse order, you need to rearrange the rows in the daten*multiplier array:
barplot((daten*multiplier)[nrow(daten):1,])