I'm needing to subset a list which contains an array as well as a factor variable. Essentially if you imagine each component of the array is relative to a single individual which is then associated to a two factor variable (treatment).
list(array=array(rnorm(2,4,1),c(5,5,10)), treatment= rep(c(1,2),5))
Typically when sub-setting multiple components of the array from the first component of the list I would use something like
list$array[,,c(2,4,6)]
this would return the array components in location 2,4 and 6. However, for the factor component of the list this wouldn't work as subsetting is different, what you would need is this:
list$treatment[c(2,4,6)]
Need to subset a list with containing different classes (array and vector) by the same relative number.
You're treating your list of matrices as some kind of 3-dimensional object, but it's not.
Your list$matrices is of itself a list as well, which means you can index at as a list as well, it doesn't matter if it is a list of matrices, numerics, plot-objects, or whatever.
The data you provided as an example can just be indexed at one level, so list$matrices[c(2,4,6)] works fine.
And I don't really get your question about saving the indices in a numeric vector, what's to stop you from this code?
indices <- c(2,4,6)
mysubset <- list(list$matrices[indices], list$treatment[indices])
EDIT, adding new info for edited question:
I see you actually have an 3-D array now. Which is kind of weird, as there is no clear convention of what can be seen as "components". I mean, from your question I understand that list$array[,,n] refers to the n-th individual, but from a pure code-point of view there is no reason why something like list$array[n,,] couldn't refer to that.
Maybe you got the idea from other languages, but this is not really R-ish, your earlier example with a list of matrices made more sense to me. And I think the most logical would have been a data.frame with columns matrix and treatment (which is conceptually close to a list with a vector and a list of matrices, but it's clearer to others what you have).
But anyway, what is your desired output?
If it's just subsetting: with this structure, as there are no constraints on what could have been the content, you just have to tell R exactly what you want. There is no one operator that takes a subset of a vector and the 3rd index of an array at the same time. You're going to have to tell R that you want 3rd index to use for subsetting, and that you want to use the same index for subsetting a vector. Which is basically just the code you already have:
idx <- c(2,4,6)
output <- list(list$array[,,idx], list$treatment[idx])
The way that you use for subsetting multiple matrices actually gives an error since you are giving extra dimension although you already specify which sublist you are in. Hence in order to subset matrices for the given indices you can usemy_list[[1]][indices] or directly my_list$matrices[indices]. It is the same for the case treatement my_list[[2]][indices] or my_list$treatement[indices]
Related
I got this code from elsewhere and I wondering if someone can explain what the square brackets are doing.
matrix1[i,] <- df[[1]][]
I am using this to assign values to a matrix and it works but I am not sure what exactly it's doing. What does the initial set of [[]] mean followed by another []?
This might help you understand a bit. You can copy and paste this code and see the differences between different ways of indexing using [] and $. The only thing I can't answer for you is the second empty set of square brackets, from my understanding that does nothing, unless a value is within those brackets.
#Retreives the first column as a data frame
mtcars[1]
#Retrieves the first column values only (three different methods of doing the same thing)
mtcars[,1]
mtcars[[1]]
mtcars$mpg
#Retrieves the first row as a data frame
mtcars[1,]
#I can use a second set of brackets to get the 4th value within the first column
mtcars[[1]][4]
mtcars$mpg[4]
The general function of [ is that of subsetting, which is well documented both in help (as suggested in comments), and in this piece. The rest of of my answer is heavily based on that source.
In fact, there are operators for subsetting in R; [[,[, and $.
The [ and $ are useful for returning the index and named position, respectfully, for example the first three elements of vector a = 1:10 may be subsetted with a[c(1,2,3)]. You can also negatively subset to remove elements, as a[-1] will remove the first index.
The $ operator is different in that it only takes element names as input, e.g. if your df was a dataframe with a column values, df$values would subset that column. You can achieve the same [, but only with a quoted name such as df["values"].
To answer more specifically, what does df[[1]][] do?
First, the [[-operator will return the 1st element from df, and the following empty [-operator will pull everything from that output.
Sorry if this is a duplicate, I read through a number of threads but couldn't really find a good explanation.
I have a dataset (dataframe) where I calculated the mean value of each column. I now want to do some logical comparisons between these values. I used lapply to get the means
means_list <- lapply(dataset_df, mean)
which outputs a named list. But when I try to compare two elements of this list, e.g.
means_list["condition1"] > means_list["condition2"]
I get an error ("comparison of these types is not implemented").
I don't get that error if I use sapply instead so that I'm working with a named vector. I can also get around the error by converting the list to a dataframe with as.data.frame first.
So, I feel like I'm doing something wrong when subsetting a named list here but I don't quite understand how. Is there a correct way to subset the list so that I can do the logical comparison? Or is this not possible with named lists?
Thanks!
To access to the element of a list by its name, you have to use double brackets:
means_list[["condition1"]] > means_list[["condition2"]]
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 have two lists of matrices and I want to multiply the first element of the first list with the first element of the second list and so on, without writing every operatios due to may be a large number of elements on each list (both lists have the same length)
this is what I mean
'(colSums(R1*t(M1))),(colSums(R2*t(M2))),...(colSums(Rn*t(Mn)))'
Do I need to create an extra list?
Although first I must be able to transpose the matrices of one of the lists before multiplying them. The results will be used for easier operations.
I already tried to use indexes and loops and doesn't work,
first tried to transpose matrices in one list like this (M is one of the lists and the other is named R, M contains M1,M2,..Mn and the same for list R)
The complete operation looks like this:
'for (i in 1:length(M)){Mt<-list(t(M[[i]]))}'
and only applies it to the last element.
The full operation looks like this:
'(cbind((colSums(R1*t(M1))),(colSums(R2*t(M2))),...(colSums(Rn*t(Mn))))'
any step of these will be useful
you could use the rlist package.
The function
list.apply(.data, .fun, ...)
will apply a function to each list element.
You can find documentation at [https://cran.r-project.org/web/packages/rlist/rlist.pdf][1].
Ok, I'm stuck in a dumbness loop. I've read thru the helpful ideas at How to sort a dataframe by column(s)? , but need one more hint. I'd like a function that takes a matrix with an arbitrary number of columns, and sorts by all columns in sequence. E.g., for a matrix foo with N columns,
does the equivalent of foo[order(foo[,1],foo[,2],...foo[,N]),] . I am happy to use a with or by construction, and if necessary define the colnames of my matrix, but I can't figure out how to automate the collection of arguments to order (or to with) .
Or, I should say, I could build the entire bloody string with paste and then call it, but I'm sure there's a more straightforward way.
The most elegant (for certain values of "elegant") way would be to turn it into a data frame, and use do.call:
foo[do.call(order, as.data.frame(foo)), ]
This works because a data frame is just a list of variables with some associated attributes, and can be passed to functions expecting a list.