Convert from matrix to list matrix - r
Sorry for the noob question but I can't seem to get this to work!
X=cbind(rep(1,m), h2(x), h3(x)) #obs
So I have a 17*3 matrix X I have to create a matrix(list(),17,3) version of this matrix. I did manually below so you can see the desired result, but there must be an easier way to do this?
Z=matrix(list(X[1,1],X[2,1],X[3,1],X[4,1],X[5,1],X[6,1],X[7,1],X[8,1],X[9,1],X[10,1],X[11,1],X[12,1],X[13,1],X[14,1],X[15,1],X[16,1],X[17,1],X[1,2],X[2,2],X[3,2],X[4,2],X[5,2],X[6,2],X[7,2],X[8,2],X[9,2],X[10,2],X[11,2],X[12,2],X[13,2],X[14,2],X[15,2],X[16,2],X[17,2],X[1,3],X[2,3],X[3,3],X[4,3],X[5,3],X[6,3],X[7,3],X[8,3],X[9,3],X[10,3],X[11,3],X[12,3],X[13,3],X[14,3],X[15,3],X[16,3],X[17,3]),17,3)
I tried this (amongst others)
Z2=list(X[1:17,1],X[1:17,2],X[1:17,3])
Z3=matrix(Z2[1:3],17,3)
But it doesn't give the correct results! It just repeats the three column vectors over and over.
Can someone please explain how to do this correctly.
Apparently you want Z <- matrix(as.list(X), ncol = 3). However, I don't see how this structure could be useful.
Related
r: A column disappears and other columns' names turn to Xs
I've got a piece of code and I don't quite understand how it works. I'm sorry if I may sound stupid... I'm making a network, so I created an adjacency matrix that looks like this: Part of my adjacency matrix So, when I try to calculate cosine similarity via lsa::cosine function the column "authors$book_id" disappears and all the other columns' names that contain authors' IDs are replaced with Xs. My code is this: a_cos <- lsa::cosine(t(as.matrix(adj_matrix[, -1]))) %>% data.frame() %>% +0.01 %>% round() This is what this code returns: After lsa::cosine Since I know this is isn't an error I want to know how this happened. Can someone please explain to me what this piece of code does and why it removes the first column (book IDs) completely and turns authors' IDs to Xs? How can I apply lsa::cosine so that this doesn't happen? I'm only a beginner, but I really need help and I will appreciate any comments! This is my first question on stackoverflow, I hope I've described the problem extensively enough...
closed/fixed:Interpertation of basic R code
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!
Accessing and changing list element attributes
The background to this is that I'm mostly a Python programmer who has some passing familiarity with R. I've been tasked to look at an R script that was written by a Perl programmer who used for and while loops a lot, to see if I can make it more R-like and get it to run faster. For example purposes, I have the following list: > lnums <- list(1:5, 6:7, 8:12) For the elements that have a length less than 5 (lnums[[2]]), I want to change the length to be 5. The original code uses a for loop to tack NA values to the end of any shorter vectors, and I know that there's got to be a better way than that. I was playing around with ways to get to it and came up with > sapply(lnums, FUN=function(x) length(x) < 5) which gets the right element, but I'm unable to figure out how to incorporate this into the subscript of a length(lnums[]) <- 5 statement. I know this is probably a really novice question, but I'd appreciate any help I can get. Additionally, the reason that I want to increase the length of the shorter list elements is so that I can put the list into a data frame. It would be great if there was a way to do that without messing around with lengths, although I still wouldn't mind an answer to my first question to satisfy my curiosity if nothing else. Thanks all. I've been digging through some topics in here and you've already helped me out quite a bit!
Here's one way: lapply(lnums, 'length<-', 5)
R or MATLAB: permute a large sparse matrix into a block diagonal matrix
I have a large sparse matrix, and I want to permute its rows or columns to turn the original matrix into a block diagonal matrix. Anyone knows which functions in R or MATLAB can do this? Thanks a lot.
I'm not really set up to test this, but for a matrix m I would try: p = symrcm(m); block_m = m(p,p); If that doesn't work, look through the other functions listed in help sparfun to see if any will help you out.
The seriation package in R has a number of tools for problems related to this one.
Not exactly sure if this is what you want, but in MATLAB this is what I have used in the past. Probably not the most elegant solution. I go from sparse to full and then chop the thing into square blocks. A=full(A); Then: blockedmatrix = mat2cell(A, (n*ones(1,size(A,1)/n)), ... (n*ones(1,size(A,1)/n))); %found somewhere on internetz This returns a cell, where each entry is of size nxn. It's easy to extract the blocks of interest, manipulate them, and then restore them to a matrix with cell2mat.
Maybe a bit late to the game, but since there are available commands, here is a simple one. If you have a matrix H and the block diagonal form is needed, you can obtain it through the following lines (MATLAB): [p,q] = dmperm(H); H(p,q) which is equivalent to Dulmage - Mendelsohn permutation.
Cumulative sum for n rows
I have been trying to produce a command in R that allows me to produce a new vector where each row is the sum of 25 rows from a previous vector.
I've tried making a function to do this, this allows me to produce a result for one data point.
I shall put where I haver got to; I realise this is probably a fairly basic question but it is one I have been struggling with... any help would be greatly appreciated;
example<-c(1;200)
fun.1<-function(x)
{sum(x[1:25])}
checklist<-sapply(check,FUN=fun.1)
This then supplies me with a vector of length 200 where all values are NA.
Can anybody help at all?
Your example is a bit noisy (e.g., c(1;200) has no meaning, probably you want 1:200 there, or, if you would like to have a list of lists then something like rep, there is no check variable, it should have been example, etc.).
Here's the code what I think you need probably (as far as I was able to understand it):
x <- rep(list(1:200), 5)
f <- function(y) {y[1:20]}
sapply(x, f)
Next time please be more specific, try out the code you post as an example before submitting a question.