I’ve tried this, looking at wikipedia. http://en.wikipedia.org/wiki/Invertible_matrix#Blockwise_inversion getInverse <- function(mat) { if(nrow(mat) == 1) { return (matrix( 1.0/ mat[1,1] )) } idx <- nrow(mat) / 2 A <- mat[1:idx, 1:idx, drop=F] B <- mat[1:idx, -1:-idx, drop=F] C <- mat[-1:-idx, 1:idx, drop=F] D <- mat[-1:-idx, -1:-idx, drop=F] invA <- getInverse(A) temp <- getInverse(D – C %*% … Read more
With ggplot, it is essential to have a single data frame with everything in it (at least for a single plotting layer, e.g., all the bars in a plot). You create a data frame of the column sums, and then try to use external vectors for the id and the grouping, which makes things difficult. … Read more
Here is the complete solution: The usage of the last command of linear interpolation solves the issue > Lines <- “D1,Value + 1,20/11/2014 16:00,0.00 + 2,20/11/2014 17:00,0.01 + 3,20/11/2014 19:00,0.05 + 4,20/11/2014 22:00,0.20 + 5,20/11/2014 23:00,0.03” > ts1 <- read.csv(text = Lines, as.is = TRUE) > library(zoo) > z <- read.zoo(ts1, tz = “”, format … Read more
I’m a bit lost on the question, but this line vec <- 1:3 embed(c(rep(0,length(vec)),vec),length(vec)+1) …produces the last part of your desired result: [,1] [,2] [,3] [,4] [1,] 1 0 0 0 [2,] 2 1 0 0 [3,] 3 2 1 0 Which you can then bind to the first number of the vector. cbind(vec[1],embed(c(rep(0,length(vec)),vec),length(vec)+1)) to … Read more
This happens because gather requires a name and value argument (supplied as strings, hence the error message. I imagine that supplying this would solve your problem. However, I strong encourage you to change from gather to the pivot arguments as gather has been retired. I think your code would work fine with pivot. Please see … Read more
The function questionr::irec opens a shiny application with the variable original values and fields to input the new values. It copies the full code to the console on exit. questionr::irec(mydf,”x.fine”) solved R / data cleaning / user-friendly construction of reference tables [closed]
We could use shift from data.table library(data.table) m1 <- na.omit(do.call(cbind, shift(df1$col1, 0:4, type=”lead”))) rowSums(m1*(1:5)[col(m1)]/5) #[1] 13.60 12.20 31.24 25.58 30.48 32.58 44.88 Or another option m1 <- embed(df1$col1,5) rowSums(m1*(5:1)[col(m1)]/5) #[1] 13.60 12.20 31.24 25.58 30.48 32.58 44.88 solved in R, How to sum by flowing row in a data frame
Assuming you mean subset columns that all of their values are greater than 1 you could do something like this (essentially you can also use the following according to any condition you might have. Just change the if condition): Example Data a <- data.frame(matrix(runif(90),ncol=3)) > a X1 X2 X3 1 0.33341130 0.09307143 0.51932506 2 0.78014395 … Read more
You could try using purrr instead of the loop as follows: require(rvest) require(purrr) require(tibble) URLs %>% map(read_html) %>% map(html_nodes, “#yw1 .spielprofil_tooltip”) %>% map_df(~tibble(Player = html_text(.), P_URL = html_attr(., “href”))) Timing: user system elapsed 2.939 2.746 5.699 The step that take the most time is the crawling via map(read_html). To paralyze that you can use e.g. … Read more
The duplicated function traverses its argument(s) sequentially and returns TRUE if there has been a prior value identical to the current value. It is a generic function, so it has a default definition (for vectors) but also a definition for other classes, such as objects of the data.frame class. The subset function treats expressions passed … Read more
If I understand what it is you’re asking, I think you could resolve your situation by using unlist(). d <- c(1:10) # creates a sample data frame to use d <- as.list(d) # converts the data frame into a list d <- unlist(d) # converts the list into a vector 0 solved Turn extraction list … Read more
A useful package for this may be dplyr. I’ll use a couple of bogus variables from the included iris dataset to give an indication of how this would work. In your case, replace iris with your dataset, and change the various calculations to what you need. require(dplyr) iris %>% mutate(# Calculate required variables at the … Read more
Assuming that the first column contain blanks (“”) and is character class, we convert the blanks to NA and use na.locf to replace the NA elements with the non-NA previous element, then paste the output with the second column and create a new column (“colN”). library(zoo) library(data.table) setDT(df1)[col1==””, col1:= NA][, colN := as.numeric(paste0(na.locf(col1), col2))] df1 … Read more
As mentioned, zoo has a last-observation-carried-forward function. We can add one to it: library(zoo) v2 <- na.locf(v) v2[is.na(v)] <- v2[is.na(v)] + 1L #[1] 2 2 3 3 3 3 8 9 2 solved Fill missing value based on previous values [duplicate]
First define a function with the elements that need to be changed as arguments. Set those arguments to defaults in the graph you really like. To change the defaults, pass them on to the function in its calls. library(ggplot2) my_theme <- function(plot_title_size = 30, colour = “black”, size = 20, angle= 0){ theme(plot.title = element_text(family … Read more