When you define the new variable, convert results$name to character, not results. Also, use ” ” (a blank space) to split the words, not “” (no space). If you fix that one line, the code seems to work. Change this: new=strsplit(as.character(results),””) to this: new=strsplit(as.character(results$name),” “) 0 solved How to split the name for student mark … Read more
We need to use names instead of name in the OP’s function, use paste0 instead of paste if we don’t need a whitespace between the names and the new string, and return ‘theNamedFile’, then apply the function directly on the ‘mylist’ ReportOp<-function(x){ theNamedFile <- paste0(names(x),”~\\Myfile.pdf”) theNamedFile } ReportOp(mylist) If we apply it using lapply lapply(mylist, … Read more
If you are interested in the optimisation result from NbClust, you will find it in nc$Best.partition where each number is the cluster number for respective row in the data matrix. for example > nc$Best.partition [1] 1 2 3 4 5 1 3 5 1 1 4 1 4 1 5 1 5 1 4 2 … Read more
I think it would be less confusing to revert to a for loop. Try: plotTimeSeries <- list(temp1,temp2,temp3,temp4) for (i in plotTimeSeries) { i$dt <- strptime(i$dt, “%Y-%m-%d %H:%M:%S”) ggplot(i, aes(dt, ambtemp)) + geom_line() + scale_x_datetime(breaks = date_breaks(“2 hour”), labels=date_format(“%H:%M”)) + labs(x=”Time 00.00 ~ 24:00 (2007-09-29)”,y=”Ambient Temperature”, title = (paste(“Node”,i))) } 7 solved Syntax error in ggplot … Read more
You could use cbind function: C <- cbind(A, B[,2]) 1 solved how to combine desired columns from different matrix in R
Something like this (assuming your data is named df) # get the column names that follow this pattern cols = grepl(“[0-9]+_parameter”, names(df)) # see if any of those columns have a 97 any(df[cols] == 97) A base R way to filter for rows that have a 97 in any of those columns is this: df[rowSums(df[cols] … Read more
This piece makes it: $ awk ‘{a[$1,$2]=a[$1,$2]$3} END{for (i in a) {print i, a[i]}}’ file B118791136 x A118791136 XxXX B23456433 XXx Just stores the result in an array, having 1st and 2nd fields as indexes. At the end, it prints the result. The result gives B23456433 instead of B 23456433, trying to split it… sed … Read more
The problem is that R is trying to be “helpful”, and simplifying your data for you. The solution is to do the following (note two commas, not one): df[-1,, drop = FALSE] This will remove the specified row, and leave your data.frame otherwise untouched. solved How do I delete rows from a data frame when … Read more
According to your new rules, may be this helps: data1 <- data #changing some elements to test the code data1[2,8] <-2 data1[4,1] <- 1 ##The code is for the row indx <- which(t(sapply(split(data1 == pattern, row(data1)), { function(x) colSums(sapply(1:(length(x) – 3), function(i) x[seq(i, i + 3)]), na.rm = TRUE) == 4 })), arr.ind = TRUE) … Read more
If I understand well, your requirement is to compare sold quantity in month t with the sum of quantity sold in months t-1 and t-2. If so, I can suggest using dplyr package that offer the nice feature of grouping rows and mutating columns in your data frame. resultData <- group_by(data, KId) %>% arrange(sales_month) %>% … Read more
Try: gsub(“^[^\\-]+\\-“, “”, “0266-VN22.5P-AC”) 10 solved Remove anything before first hypen [closed]
Here are a couple of ideas. This is what I think your data look like? before <- data.frame(val=c(11330,2721,52438,6124), lab=c(“STAT1″,”STAT2″,”STAT3″,”SUZY”)) after <- data.frame(val=c(17401,3462,0,72), lab=c(“STAT1″,”STAT2″,”STAT3″,”SUZY”)) Combine them into a single data frame with a period variable: combined <- rbind(data.frame(before,period=”before”), data.frame(after,period=”after”)) Reformat to a matrix and plot with (base R) dotchart: library(reshape2) m <- acast(combined,lab~period,value.var=”val”) dotchart(m) Plot with … Read more
The idiomatic r way is to use [<-, in the form x[index] <- result If you are dealing with integers / factors or character variables, then == will work reliably for the indexing, x <- rep(1:5,3) x[x==3] <- 1 x[x==4] <- 2 x ## [1] 1 2 1 2 5 1 2 1 2 5 … Read more
What are other patterns your data has? If it’s always “(B)” you can do sub(“\\(B\\)”, “”, df$code) #[1] “1234” “5678” “1234” “5678” “0123” Or if it could be any character do sub(“\\([A-Z]\\)”, “”, df$code) You could also extract only the numbers from Code sub(“.*?(\\d+).*”, “\\1”, df$code) You might want to wrap output of sub in … Read more
Ok I finally found it. On the x axis I wanted to plot dates either as POSIXct or as number of day of recording (integer). On the y axis I wanted the time of day so that the graph would present a dark bar on each day (x-axis) and between the time (y-axis) that the … Read more