[ad_1] file$Efficiency <- cut(file$Duration, breaks=c(-1,5,15,30,200), labels=c(“Above Par”, “Par”, “Below Par”, “Unacceptable”)) Here is a little example of data from me: x <- c(0,1, 4:6, 14:16, 29:31) y <- cut(x, breaks=c(-1,5,15,30,200), labels=c(“Above Par”, “Par”, “Below Par”, “Unacceptable”)) data.frame(x=x, y=y) 1 [ad_2] solved Explanation for If statement in R
[ad_1] You are using ==, which tests for pairwise equality (e.g., is the first row of df1$subject equal to the first row of df2$subject, are the second rows equal, etc.). Consider c(1, 1, 2, 3) == c(1, 2, 3, 4) # [1] TRUE FALSE FALSE FALSE Instead, you want to be testing if each row … Read more
[ad_1] We can use duplicated dat[!(duplicated(dat$ID)|duplicated(dat$ID, fromLast = TRUE)),] 0 [ad_2] solved Keep only non-duplicate rows based on a Column Value [duplicate]
[ad_1] x <- c(.5, 0.2, 2, 3.4, 0.4, 1.2, .2, .8, 2.3, 7.4) result.final <- NULL cumul.total <- 0 for (i in x) { cumul.total <- cumul.total + i if(i > 1){ result <- cumul.total } else { next } result.final <- c(result.final, result) } result.final 8 [ad_2] solved R programming – if/else command [closed]
[ad_1] More elementary solution could be cbind(df[,1:8], pmax(df[,9], df[,10], df[,11])) where df is your data frame. 3 [ad_2] solved print first 8 columns and one of the last three non-zero columns linux/awk/sed/R [closed]
[ad_1] Although I agree with joran in this case, sometimes it can be useful to use the assign() function, for instance as follows: for (i in 1:3) { assign(paste0(“data.”, i), i) } This results in the following: > ls() [1] “data.1” “data.2” “data.3” “i” > data.1 [1] 1 > data.2 [1] 2 > data.3 [1] … Read more
[ad_1] The easiest way would just be to shuffle the column… df$col3 <- sample(df$col3) 3 [ad_2] solved How to sample a column based on frequency in R?
[ad_1] Using a toy example… df <- data.frame(a=LETTERS[1:10],b=LETTERS[3:12],stringsAsFactors = FALSE) limits <- c(“E”,”H”) sapply(df,function(x){ del.min <- grep(limits[1],x) del.max <- grep(limits[2],x) x[del.min:del.max] <- “” return(x)}) a b [1,] “A” “C” [2,] “B” “D” [3,] “C” “” [4,] “D” “” [5,] “” “” [6,] “” “” [7,] “” “I” [8,] “” “J” [9,] “I” “K” [10,] “J” … Read more
[ad_1] 1) The issue is that sales_clean$Year is a factor. 2) ggplot interprit your x-value as categorical, y-value as continous and aggregated value into the bar plot (instead bar there are lines). Please see the simulation: library(ggplot2) set.seed(123) sales_clean <- data.frame(Year = rep(factor(2014:2018), 1000), Net_Rev = abs(rnorm(5000))) plotLine <- ggplot(sales_clean, aes(Year, Net_Rev, na.rm = FALSE)) … Read more
[ad_1] We loop through the columns, remove the NA elements, then select the minimum number of observations after comparing all the elements in the list. lst <- lapply(df1, function(x) x[complete.cases(x)]) res <- data.frame(lapply(lst, `length<-`,min(lengths(lst)))) res # cost customer.satisfaction safety time #1 40 57 32 24 #2 38 72 30 40 #3 36 73 58 22 … Read more
[ad_1] We can use data.table to get the fastest extraction using either unique with by option unique(df2, by = “Person”) Or extracting with row index setDT(df2)[df2[, .I[1L],Person]$V1] Update If we need the minimum ‘Age’ row per ‘Person setDT(df2)[, .SD[which.min(Age)], Person] Or if we prefer dplyr, then library(dplyr) df2 %>% group_by(Person) %>% slice(1L) Update df2 %>% … Read more
[ad_1] Try this: patients[patients$time > ‘2017-01-02 11:41:53’ & patients$time < ‘2017-08-07 09:06:07’,] 3 [ad_2] solved Fetch the time between two timestamps in R [duplicate]
[ad_1] The MClust package contains the function densityMclust which produces an object that contains parameter estimates for the fitted Gaussian mixture model along with the density itself. From the MClust manual: > densWaiting <- densityMclust(faithful$waiting) > summary(densWaiting, parameters = TRUE) ——————————————————- Density estimation via Gaussian finite mixture modeling ——————————————————- Mclust E (univariate, equal variance) model … Read more
[ad_1] you should not use all variables as id.vars try this code dfm <- melt(df, id.vars=c(“iteration”)) What do you really need to plot?? What is your x and y axis? line plot or point? As far as i understood this should work. ggplot(dfm,aes(iterartion,value,colour=variable))+geom_point() [ad_2] solved how to plot with ggplot?
[ad_1] Do you just want to code the table up? Would something like this suffice?: PCP <- c(0, 4.9, 5, 9.9, 10, 14.9, 15) seq2max <- seq(0,max(PCP)+5,5) result <- data.frame(min=seq2max,max=seq2max+4.9,DRI=seq_along(seq2max)-1) min max DRI 1 0 4.9 0 2 5 9.9 1 3 10 14.9 2 4 15 19.9 3 5 20 24.9 4 result$DRI # … Read more