The following code finds the indexes in the 50th column of p where the value is equal to 1.
findall(p[:,50].== 1)
But suppose I was interested in screening for multiple criteria. For example, if I was also interested in the indexes where the value is 0.5. I have tried the following in that case, but something goes wrong:
findall(p[:,50].== 1 | p[:,50].== 0.5)
You're forgetting to dot the | operator. But you also need to use parens:
findall((p[:,50].== 1) .| (p[:,50].== 0.5))
But still, this is a bit wasteful, since you are making two copies of the same column, and are allocating five intermediate vectors that you don't need. You should try to use a predicate function to avoid this, like e.g. here:
findall(x->x in (0.5, 1.0), p[:,50])
or
findall(x->x==0.5||x==1, p[:,50])
On top of this, you can use view to avoid allocations due to p[:,50]:
findall(x->x==0.5||x==1, view(p, :,50))
Related
I have the following function: problema_firma_emprestimo(r,w,r_emprestimo,posicao,posicao_banco), where all input are scalars.
This function return three different matrix, using
return demanda_k_emprestimo,demanda_l_emprestimo,lucro_emprestimo
I need to run this function for a series of values of posicao_banco that are stored in a vector.
I'm doing this using a for loop, because I need three separate matrix with each of them storing one of the three outputs of the function, and the first dimension of each matrix corresponds to the index of posicao_banco. My code for this part is:
demanda_k_emprestimo = zeros(num_bancos,na,ny);
demanda_l_emprestimo = similar(demanda_k_emprestimo);
lucro_emprestimo = similar(demanda_k_emprestimo);
for i in eachindex(posicao_bancos)
demanda_k_emprestimo[i,:,:] , demanda_l_emprestimo[i,:,:] , lucro_emprestimo[i,:,:] = problema_firma_emprestimo(r,w,r_emprestimo[i],posicao,posicao_bancos[i]);
end
Is there a fast and clean way of doing this using vectorized functions? Something like problema_firma_emprestimo.(r,w,r_emprestimo[i],posicao,posicao_bancos) ? When I do this, I got a tuple with the result, but I can't find a good way of unpacking the answer.
Thanks!
Unfortunately, it's not easy to use broadcasting here, since then you will end up with output that is an array of tuples, instead of a tuple of arrays. I think a loop is a very good approach, and has no performance penalty compared to broadcasting.
I would suggest, however, that you organize your output array dimensions differently, so that i indexes into the last dimension instead of the first:
for i in eachindex(posicao_bancos)
demanda_k_emprestimo[:, :, i] , ...
end
This is because Julia arrays are column major, and this way the output values are filled into the output arrays in the most efficient way. You could also consider making the output arrays into vectors of matrices, instead of 3D arrays.
On a side note: since you are (or should be) creating an MWE for the sake of the people answering, it would be better if you used shorter and less confusing variable names. In particular for people who don't understand Portuguese (I'm guessing), your variable names are super long, confusing and make the code visually dense. Telling the difference between demanda_k_emprestimo and demanda_l_emprestimo at a glance is hard. The meaning of the variables are not important either, so it's better to just call them A and B or X and Y, and the functions foo or something.
I'm creating a Monte Carlo model using R. My model creates matrices that are filled with either zeros or values that fall within the constraints. I'm running a couple hundred thousand n values thru my model, and I want to find the average of the non zero matrices that I've created. I'm guessing I can do something in the last section.
Thanks for the help!
Code:
n<-252500
PaidLoss_1<-numeric(n)
PaidLoss_2<-numeric(n)
PaidLoss_3<-numeric(n)
PaidLoss_4<-numeric(n)
PaidLoss_5<-numeric(n)
PaidLoss_6<-numeric(n)
PaidLoss_7<-numeric(n)
PaidLoss_8<-numeric(n)
PaidLoss_9<-numeric(n)
for(i in 1:n){
claim_type<-rmultinom(1,1,c(0.00166439057698873, 0.000810856947763742, 0.00183509730283373, 0.000725503584841243, 0.00405428473881871, 0.00725503584841243, 0.0100290201433936, 0.00529190850119495, 0.0103277569136224, 0.0096449300102424, 0.00375554796858996, 0.00806589279617617, 0.00776715602594742, 0.000768180266302492, 0.00405428473881871, 0.00226186411744623, 0.00354216456128371, 0.00277398429498122, 0.000682826903379993))
claim_type<-which(claim_type==1)
claim_Amanda<-runif(1, min=34115, max=2158707.51)
claim_Bob<-runif(1, min=16443, max=413150.50)
claim_Claire<-runif(1, min=30607.50, max=1341330.97)
claim_Doug<-runif(1, min=17554.20, max=969871)
if(claim_type==1){PaidLoss_1[i]<-1*claim_Amanda}
if(claim_type==2){PaidLoss_2[i]<-0*claim_Amanda}
if(claim_type==3){PaidLoss_3[i]<-1* claim_Bob}
if(claim_type==4){PaidLoss_4[i]<-0* claim_Bob}
if(claim_type==5){PaidLoss_5[i]<-1* claim_Claire}
if(claim_type==6){PaidLoss_6[i]<-0* claim_Claire}
}
PaidLoss1<-sum(PaidLoss_1)/2525
PaidLoss3<-sum(PaidLoss_3)/2525
PaidLoss5<-sum(PaidLoss_5)/2525
PaidLoss7<-sum(PaidLoss_7)/2525
partial output of my numeric matrix
First, let me make sure I've wrapped my head around what you want to do: you have several columns -- in your example, PaidLoss_1, ..., PaidLoss_9, which have many entries. Some of these entries are 0, and you'd like to take the average (within each column) of the entries that are not zero. Did I get that right?
If so:
Comment 1: At the very end of your code, you might want to avoid using sum and dividing by a number to get the mean you want. It obviously works, but it opens you up to a risk: if you ever change the value of n at the top, then in the best case scenario you have to edit several lines down below, and in the worst case scenario you forget to do that. So, I'd suggest something more like mean(PaidLoss_1) to get your mean.
Right now, you have n as 252500, and your denominator at the end is 2525, which has the effect of inflating your mean by a factor of 100. Maybe that's what you wanted; if so, I'd recommend mean(PaidLoss_1) * 100 for the same reasons as above.
Comment 2: You can do what you want via subsetting. Take a smaller example as a demonstration:
test <- c(10, 0, 10, 0, 10, 0)
mean(test) # gives 5
test!=0 # a vector of TRUE/FALSE for which are nonzero
test[test!=0] # the subset of test which we found to be nonzero
mean(test[test!=0]) # gives 10, the average of the nonzero entries
The middle three lines are just for demonstration; the only necessary lines to do what you want are the first (to declare the vector) and the last (to get the mean). So your code should be something like PaidLoss1 <- mean(PaidLoss_1[PaidLoss_1 != 0]), or perhaps that times 100.
Comment 3: You might consider organizing your stuff into a dataframe. Instead of typing PaidLoss_1, PaidLoss_2, etc., it might make sense to organize all this PaidLoss stuff into a matrix. You could then access elements of the matrix with [ , ] indexing. This would be useful because it would clean up some of the code and prevent you from having to type lots of things; you could also then make use of things like the apply() family of functions to save you from having to type the same commands over and over for different columns (such as the mean). You could also use a dataframe or something else to organize it, but having some structure would make your life easier.
(And to be super clear, your code is exactly what my code looked like when I first started writing in R. You can decide if it's worth pursuing some of that optimization; it probably just depends how much time you plan to eventually spend in R.)
I have a basic question in regards to the R programming language.
I'm at a beginners level and I wish to understand the meaning behind two lines of code I found online in order to gain a better understanding. Here is the code:
as.data.frame(y[1:(n-k)])
as.data.frame(y[(k+1):n])
... where y and n are given. I do understand that the results are transformed into a data frame by the function as.data.frame() but what about the rest? I'm still at a beginners level so pardon me if this question is off-topic or irrelevant in this forum. Thank you in advance, I appreciate every answer :)
Looks like you understand the as.data.frame() function so let's look at what is happening inside of it. We're looking at y[1:(n-k)]. Here, y is a vector which is a collection of data points of the same type. For example:
> y <- c(1,2,3,4,5,6)
Try running that and then calling back y. What you get are those numbers listed out. Now, consider the case you want to just call out the number 1 in that vector. How would you do that? Well, this is where the brackets come into play. If you wanted to just call the number 1 in y:
> y[1]
[1] 1
Therefore, the brackets are a way of calling out or indexing specific items in the vector. Note that the indexing starts at the value 1 and goes up to the number of items in the vector, or length. One last thing before we go back to the example you gave. What if we want to index the numbers 1, 2, and 3 from the vector but not the rest?
> y[1:3]
[1] 1 2 3
This is where the colon comes into play. It allows us to reference a subset of the numbers. However, it will reference all the numbers between the index left of the colon and right of it. Try this out for yourself in R! Play around and see what happens.
Finally going back to your example:
y[1:(n-k)]
How would this work based on what we discussed? Well, the colon means that we are indexing all values in the vector y from two index values. What are those values? Well, they are the numbers to the left and right of the colon. Therefore, we are asking R to give us the values from the first position (index of 1) to the (n-k) position. Therefore, it's important to know what n and k are. If n is 4 and k is 1 then the command becomes:
y[1:3]
The same logic can apply to the second as.data.frame() command in your question. Essentially, R is picking out different numbers from a vector y and multiplying them together.
Hope this helps. The best way to learn R is to play around with a command, throw different numbers at it, guess what will happen, and then see what happens!
Suppose I have an 2 dimensional array and I want to apply several functions to each of its columns. Ideally I would like to get the results back in the form of a matrix (with one row per function, and one column per input column).
The following code generates the values I want, but as an Array of Arrays.
A = rand(10,10)
[mapslices(f, A, 1) for f in [mean median iqr]]
Another similar example is here [Julia: use of pmap with matrices
Is there a better syntax for getting the results back in the form of a 2 dimensional array, instead of an array of arrays?
What I'd really like is something with a functionality similar to sapply from R. [https://stat.ethz.ch/R-manual/R-devel/library/base/html/lapply.html]
You can use an anonymous function as in
mapslices(t -> [mean(t), median(t), iqr(t)], A, 1)
but using comprehensions and splatting, as in your last example, is also fine. For very large arrays, you might want to avoid the temporary allocations introduced by transpose and splatting, but in most cases you don't have to pay attention to that.
After playing around a bit I found one option, but I am still interested in hearing if there are any better ways of doing it.
[[mapslices(f, A, 1)' for f in [mean median iqr]]...]
I'm trying to make a function that determines what bucket a certain value goes into based off of a given vector. So my function has two inputs: a vector determining the break points for the bucket
(ex: if the vector is (1,4,5,10) the buckets would be <=1, 110)
and a certain number. I want the function to output a certain value determining the bucket.
For example if I input .9 the output could be 1, 1.6, the output could be 4, 5.8 the output could be 10, and 13, the output could be "10+".
The way I'm doing it right now is I first check if the input number is bigger than the vector's largest element or smaller than the vector's smallest element. If not, I then run a for loop (can't figure out how to use apply) to check if the number is in each specific interval. The problem is this is way too inefficient because I'm dealing with a large data set. Does anyone know an efficient way to do this?
The cut() function is convenient for bucketing: cut(splitme,breaks=vectorwithsplits) .
However, it looks like you're actually trying to figure out an insertion point. You need something like binary search.