similar to: Correlation Network Diagram?

Hi, I want to do a visualisation of a matrix plot made up of several plots of correlation matrices (using corrplot()). My data is in csv format. Here's an example: id,category,attribute1,attribute2,attribute3,attribute4 661,SCHS,43.2,0,56.5,1 12202,SCHS,161.7,5.7,155,16 1182,SCHS,21.4,0,29,0 1356,SSS, 8.8182,0.1818,10.6667,0.6667 1864,SCHS,443.7273,9.9091,537,46 12360,SOA,6.6364,0,10,0

Hi all If I have a string, say "ABCDA", and I want to convert this to the sum of the letter values, e.g. A -> 1 B -> 2 etc, so "ABCDA" = 1+2+3+4+1 = 11 Is there an elegant way to do this? Trying something like which(LETTERS %in% unlist(strsplit("ABCDA", ""))) is not quite correct, as it does not count repeated characters. I guess what I need is

Hi all Does anyone know if it is possible when plotting a line or scatter plot, to selectively color the data points based on the data value? i.e. if plotting say the percentage change in stock price movements, to color +ve points in green and -ve points in red? And extending this to a user-defined range of colors based on the quartile of the data points? Thanks Rory Rory Winston RBS Global

Hi (If you're wondering, this is a Project Euler question :)) If I wanted to calculate the sum of the digits in the decimal representation of 2^1000, what would be a good way to go about that? I've tried the following methods: # Calculate the sum of digits in the decimal representation of 2^n # Only works for smaller values of n bsum <- function(n) { s <- 0 e <-

Created a new data set with 3 numeric variable to find correlation CR1<- mar%>% as_data_frame%>% select(AGE, OLD_CAR_PURCHASE_YRS, Total.Spend.With.AA) had to convert it to a data frame, code: as.matrix(as.data.frame(CR1)) Now i need to run a correlation plot for these 3 variables: corrplot(CR1, method = "circle") But i am getting this error: Error in

Hi Sarah, Thank you for your help. I tried using CR1<-as.matrix(CR1) but gives error Error in corrplot(CR1, method = "circle") : The matrix is not in [-1, 1]!. I am using a corrplot library. Please find the reproducible example: dput(head(CR1,10)) structure(c(26L, 46L, 39L, 38L, 47L, 59L, 56L, 61L, 43L, 60L, 78L, 63L, 2L, 58L, 8L, 1L, 1L, 9L, 11L, 2L, 1037500L, 46747L, 346300L,

Hi Consider the following graph: x <- rnorm(1000) x <- x + exp(-x/2) layout(matrix(rep(c(1,1,2), 2), 2, 3, byrow=TRUE)) boxplot(x) rug(jitter(x), side=2) plot(density(x)) What I would really like to do is to have the density plot rotated by 90 degrees so that I can see it line up with the rug plot on the side axis. I cant seem to do this. I have seen references to the grid

That does clarify for me that you're missing a step: I didn't clearly follow your description at first. corrplot expects a correlation matrix, not your original data. You need to use cor() first. That's pretty clear in the documentation. See for instance the examples: data(mtcars) M <- cor(mtcars) corrplot(M) Sarah On Sat, Mar 17, 2018 at 12:00 PM Shivi Bhatia <shivipmp82 at

I'm assuming you are using the corrplot package. If so, your data object does need to be a matrix, not a data frame. Since it's already a data frame, your line of code: as.matrix(as.data.frame(CR1)) doesn't need the as.data.frame function, but more importantly, you didn't assign the result to anything: as.matrix() does not work in place. CR1 <- as.matrix(CR1) Now try. If

Hi all (I'm sure this question has been asked before, but I cant find it). If I have two character vectors: > x <- c("aaa","bbb","ccc") > y <- c("1","2","3") How can I get the cartesian product of the string values? > expand.grid(x,y) Gives me a data frame with separate columns...however, I cant seem to get *apply

Hi Firstly , happy Christmas to R-Help! Secondly, I wonder if anyone can help me with the following query: I am trying to reproduce some explicit probability calculations performed in APPL (a Maple extension for computational probability). For instance, in APPL, to compute the probability that the sum of 10 iid uniform variables [0,1] will be between 4 and 6, (i..e Pr( 4 < \sum_{i=1}^{10}X_i

Hi Does anyone know how I might pick out diagonal elements of a matrix using a vector? If I create a matrix a: a <- matrix(c(1:16), 4, byrow=TRUE) and I want to pick out the elements (1,1),(2,2),(3,3), or another arbitrary diagonal (upper or lower), is there any way I can use a vector to do this? So if I want a diagonal of size 3, I could create a vector like x <- c(0:2) and then pick

Hi I am having a bit of difficulty with changing the canvas size on a trellis/lattice plot. I am plotting two "cubes" of 3-dimensional random numbers, as follows: library(gsl) library(lattice) q <- qrng_alloc(type="sobol", 3) npoints <- 200 rs <- qrng_get(q,npoints) # Plot the normal variates in a 3-dim cube p1 <- cloud(rnorm(npoints) ~ rnorm(npoints) +

Hi R, Can we download Reuters (3000 Xtra) data from R? Does ODBC package help me in this? Or otherwise, is there a way to extract daily closing prices data of Reuters from R? Thank you very much, Shubha This e-mail may contain confidential and/or privileged i...{{dropped:13}}

Hi all I have a question about correct usage of persp(). I have a simple neural net-based XOR example, as follows: library(nnet) xor.data <- data.frame(cbind(expand.grid(c(0,1),c(0,1)), c(0,1,1,0))) names(xor.data) <- c("x","y","o") xor.nn <- nnet(o ~ x + y, data=xor.data, linout=FALSE, size=1) # Create an (x.y) surface and predict over all points d <-

Hi I am currently (for pedagogical purposes) writing a simple numerical analysis library in R. I have come unstuck when writing a simple Newton-Raphson implementation, that looks like this: f <- function(x) { 2*cos(x)^2 + 3*sin(x) + 0.5 } root <- newton(f, tol=0.0001, N=20, a=1) My issue is calculating the symbolic derivative of f() inside the newton() function. I cant seem to get R to

Hi I have a dataset that looks like this (dput'd below): > head(x, 20) time status 1 2009-07-02 10:32:37 1 2 2009-07-02 10:32:43 0 3 2009-07-02 10:32:43 1 4 2009-07-02 10:32:44 0 5 2009-07-02 10:32:44 1 6 2009-07-02 10:32:48 0 7 2009-07-02 10:32:48 1 8 2009-07-02 10:32:54 0 9 2009-07-02 10:33:04 1 10 2009-07-02 10:33:04 0 11 2009-07-02 10:33:05 1 12 2009-07-02 10:33:07 0 13 2009-07-02

Hi all Here is a rather naive implementation of the SEPA algorithm for generating lexical permutations: lexperm3 <- function(x, n=length(x)) { perms <- list() k <- 1 perms[[k]] <- x k <- k + 1 for (y in 1:(factorial(n)-1)) { i <- n-1 while (x[i] > x[i+1] && i > 0) { i <- i - 1 } # i is largest index st x[i] > x[i+1] j <- n #

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Hi I have two irregular time series, which are of different lengths and being and end at different times. For the common subset of time that they both span, they should have the same values, but the values may occur at slightly different time intervals. I am trying to "line up" the identical values and reconcile them. I have merged the two series into a zoo object which looks