I guess you need a barplot: x <- data.frame(n_vehicles = c(10, 20, 30, 40), time_interval = c(2, 5, 4, 9)) barplot(height = x$time_interval, names.arg = x$n_vehicles) Or alternatively: plot(x = x$n_vehicles, y = x$time_interval, type = “h”) The “h” stands for “histogram”. 1 solved How to plot histogram in R?
It’s hard to say without knowing exactly what your data look like, but perhaps something like this would work: cols <- rainbow(17)[as.factor(gtex_pm$tissue)] plot(pc_gtex$x[,1], pc_gtex$x[,2], col=cols, main = “PCA”, xlab = “PC1”, ylab = “PC2”) 4 solved how to color points in 17 colors based on principal component? [closed]
I finally came to the bottom of this, I figured I could create as many new column I wanted in order to calculate what I had to so that’s what I did. I think this is the best way to do it but am open to suggestions if there are other ways or even faster. … Read more
We get the index of column names that start with ‘pd’ (nm1). Loop through those columns with lapply, match each column with the ‘nom’ column to get the row index. Using ifelse, we change the NA elements with the column values and the rest of the values from the ‘ID’ column. nm1 <- grep(‘^pd’, names(df1)) … Read more
We can use lapply to loop over the columns of dataset, then created a grouping variable with %/% and get the mean using tapply lapply(df1, function(x) tapply(x, (seq_along(x)-1)%/%128, FUN = mean, na.rm = TRUE)) data set.seed(24) df1 <- data.frame(col1 = sample(c(NA, 1:10), 140, replace=TRUE), col2 = sample(c(NA, 1:3), 140, replace=TRUE)) 8 solved Find the mean … Read more
We can try subset subset(dati, format(as.Date(Date),”%Y”)==2005) If we are trying to subset the data for each year, try split split(dati, format(as.Date(dati$Date), “%Y”)) solved R get subset from data frame filtering by year Date value
The 2.2 is the number of times that x will be repeated (not what x will be multiplied by). In your example x has length 5 and y has length 11. The 2.2 comes because 2.2 times 5 is 11, so in order to have 2 vectors of the same length to add together, the … Read more
This is a workaround, a response to your #2 Looking at your code, there is a much easier way of subsetting data. Try this. Check if this solves your issue. library(dplyr) active<- clinic %>% filter(Days.since.injury.physio>20, Days.since.injury.physio<35, Days.since.injury.F.U.1>27, Days.since.injury.F.U.1<63 ) dplyr does wonders when it comes to subsetting and manipulation of data. The %>% symbol chains … Read more
Echoing @DWin, it is not clear what Y is, given that it will try and get Y from your workspace. If you mean to set the pch by the column Y within banknote, then the following will work pairs(banknote[,-c(1,7)], panel = function(x,y,…){ points(x,y,pch = ifelse(as.logical(banknote$Y), 0,15))}) If you don’t want to have to reference the … Read more
You can drop the rownames of a dataframe by simply setting them to null. They’re null by default (see ?data.frame), but sometimes they get set by various functions: # Create a sample dataframe with row.names a_data_frame <- data.frame(letters = c(“a”, “b”, “c”), numbers = c(1, 2, 3), row.names = c(“Row1”, “Row2”, “Row3”)); # View it … Read more
ind <- df$id[df$month_key == df$month_key[1]] df[df$id %in% ind,] # month_key id #1 201504 1 #2 201504 2 #3 201505 1 #4 201505 2 7 solved Filter with a condition [closed]
Use the “data.table” package! The syntax is much easier, and the run time is faster. ### Load package require(data.table) ### Set up variables; Create data.table time <- c(0:4, 7) ColA <- c(1, 3, 0, 3, 4, 10) ColB <- c(10, 7, 8, 4, 5, 23) ColC <- c(5, 15, 9, 5, 6, 4) data <- … Read more
I created a sample data. Hope this is what you need df <- data.frame(names=c(‘A’,’A’,’A’,’A’,’B’,’B’,’B’,’C’,’C’,’C’,’C’,’C’),Length=c(1:12)) library(plyr) df2<- ddply(df, “names”, subset, Length==max(Length)) solved how to unique a data frame by the values in a specified column?
We can group by ‘A’, ‘B’ and select the first non-NA element across other columns library(dplyr) df1 %>% group_by(A, B) %>% summarise(across(everything(), ~ .[order(is.na(.))][1]), .groups=”drop”) -output # A tibble: 1 x 8 # A B C D E F G H # <chr> <chr> <int> <int> <int> <int> <int> <lgl> #1 time place 1 2 … Read more
Your desired output isn’t very clear, but I think this does what you need (you also have Email column twice) library(data.table) cols <- c(“Call”, “Callback”, “Email”) # Choose columns to modify First solution (simple version) setDT(df)[, paste0(cols, “Sum”) := lapply(.SD, function(x) c(rep(0L, .N – 1L), sum(x))), by = .(E_Add, cumsum(Action == “Event”)), .SDcols = cols][] … Read more