Your SQL CTE syntax is incorrect: you have a comma after the last CTE’s close-paren. The only time you need a comma there is if you have another CTE following it. Try changing to with unit as ( select t.RefDate,…
There is nothing wrong that I can see. Your code works. Some times such errors arise due to not flowing the ggplot expression properly. Try it like this: ggplot(train, aes(x= Item_Visibility, y = Item_Outlet_Sales)) + geom_point(size = 2.5, color=”navy”) +…
apriori returns rules object, not a vector: data(“Adult”) ## Mine association rules. rules <- apriori(Adult, parameter = list(supp = 0.5, conf = 0.9, target = “rules”)) class(rules) # [1] “rules” if you want to compare lists of rules you will…
Actually figured it out – was very simple. NewDataFrame <- data.frame(colnames(Original)) solved R: create new dataframe rows are columns from another dataframe
dataset <- dataset[!(dataset %in% c(‘A’,’B’))] solved Removing contents from a dataset in R
Plyr solution: library(plyr) ddply(df,.(ID),transform,AGE_HEAD=head(AGE,1)) OR ddply(df,.(ID),transform,AGE_HEAD=AGE[PERNO==1]) ID AGE PERNO AGE_HEAD 1 1 30 1 30 2 1 25 2 30 3 2 25 1 25 4 2 24 2 25 5 2 3 3 25 6 3 65 1 65…
A possible solution: library(lubridate) # example time data time = c( “2015-10-05 15:44:41.986797”, “2015-10-05 15:59:23.389583”, “2015-10-05 16:59:44.99402”, “2015-10-05 16:59:44.99402”, “2015-10-05 16:59:44.99402”, “2015-10-05 16:59:44.99402”, “2015-10-05 17:59:59.594524”, “2015-10-05 17:59:59.594524”, “2015-10-05 18:00:05.555564” ) # transform time strings to POSIXct objects for count time…
Based on your description: data2[] <- lapply(data2, function(x) vapply(x, paste, collapse = “, “, character(1L))) Or data2[] <- lapply(data2, function(x) vapply(x, toString, character(1L))) 2 solved how do you apply a function to all of the columns in data frame? [closed]
You could try using the GenomicRanges package. library(dplyr) library(GenomicRanges) Here we load in the the example input data. (This is an inelegant way to do this — I know… but I was lazy and the sublime multiline edit made it…
The problem appears to be here: sieve <- c(sieve[(sieve %% i) ! = 0], i). I’m assuming that you want “is not equal to”, so should be sieve <- c(sieve[(sieve %% i) != 0], i). Here’s the formatted code that…
Just write df1-df2[colnames(df1)] # a b c # 1 1 -3 1 # 2 2 -5 8 # 3 3 -6 -1 2 solved Subtract two data sets where column names are same [closed]
One approach is to use the tidy function from the broom package. It is versatile and can organize most R statistical results into a data frame. library(broom) set.seed(1) x <- rnorm(1000) x[sample(1:length(x), 100)] <- NA df <- tidy(summary(x)) df minimum…
This is what I was looking for avgDelay <- function(carrier, dest) { TempTotal = 0 TempCount = 0 for(i in 1:dim(delays)[1]) { if(delays$CARRIER[i] == carrier & delays$ARR_DELAY[i] >0 & is.na(delays$ARR_DELAY[i]) == FALSE & delays$DEST[i] == dest) { TempTotal <-TempTotal +…
You defined one and only one object when you executed: tino=read.delim(“clipboard”) The column names of that object are not handled as other objects. If you wnated to create a new object from that dataframe you could do this: x <-…
As Gregor points out, you need a more fine-grained representation of your time variable if you want to look at how temperature changes by time. There are two solutions. The best option is if your data are ordered in the…