similar to: retaining "POSIXct" formatting when using apply(muff, FUN=MAX) on POSIXct dataframe?

Please see example code below. I have a vector ("mydata") of length 10. "mydata" can have various formats (e.g. numeric, text, POSIXct, etc) I use the matrix and data.frame functions to convert "mydata" to a dataframe ("mydf") of 2 columns and 5 rows. What is a "good" way to ensure that the format is retained when I create the

I have a dataframe ("mydf") that contains "differences of means". I wish to test whether these differences are significantly different from zero. Below, I calculate the t-statistic for each column. What is a "good" method to calculate/look-up the p-value for each column? mydf=data.frame(a=c(1,-22,3,-4),b=c(5,-6,-7,9)) mymean=mean(mydf) mysd=sd(mydf)

I have a two question regarding the "aggregate.data.frame" method of the "aggregate" function. My situation: a. My "x" variable is a data.frame ("mydf") with two columns, both columns of type/format "numeric". b. My "by" variable is a data.frame("mybys") with two columns, both columns of type/format "character". c.

Hello! I have a data frame with dates. I need to create a new "month" that starts on the 20th of each month - because I'll need to aggregate my data later by that "shifted" month. I wrote the code below and it works. However, I was wondering if there is some ready-made function in some package - that makes it easier/more elegant? Thanks a lot! # Example data:

Hi All, I've been successfully using the with function for analyses and the transform function for multiple transformations. Then I thought, why not use "with" for both? I ran into problems & couldn't figure them out from help files or books. So I created a simplified version of what I'm doing: rm( list=ls() ) x1<-c(1,3,3) x2<-c(3,2,1) x3<-c(2,5,2)

I'm writing a function and keep getting the following error message. myfunc <- function(lst) { lst <- list(roots = c("car insurance", "auto insurance"), roots2 = c("insurance"), prefix = c("cheap", "budget"), prefix2 = c("low cost"), suffix = c("quote", "quotes"), suffix2 = c("rate",

Hi all, Can anyone explain why the following use of the subset() function produces a different outcome than the use of the "[" extractor? The subset() function as used in density(subset(mydf, ht >= 150.0 & wt <= 150.0, select = c(age))) appears to me from documentation to be equivalent to density(mydf[mydf$ht >= 150.0 & mydf$wt <= 150.0, "age"])

Dear R experts, I am trying to arrange multiple plots, creating one graph for each size1 factor variable in my data frame, and each plot has the median price on the y-axis and the size2 on the x-axis grouped by clarity: library(ggplot2) df <- data.frame(price=matrix(sample(1:1000, 100, replace = TRUE), ncol = 1)) df$size1 = 1:nrow(df) df$size1 = cut(df$size1, breaks=11)

...well, I don't think this is exactly the expected result (see my post) to be noted that the columns affected should be "A" and "B" thanks for the help max ----- Messaggio originale ----- Da: "Rui Barradas" <ruipbarradas at sapo.pt> A: "Massimo Bressan" <massimo.bressan at arpa.veneto.it>, "r-help" <r-help at

Dear all, I wanted to make a two-way-table of two variables with a counting variable stored in another column of a dataframe. In version 1.9.1, the behavior is as expected as shown in the simplified example code. > sex <- rep(c("F", "M"), 5) > income <- c(rep("low", 5), rep("high", 5)) > count <- 1:10 > mydf <-

Hello, Try the following. icol <- which(grepl("flag", names(mydf))) mydf[icol] <- lapply(mydf[icol], function(x){ is.na(x) <- x == 0 x }) mydf # A A_flag B B_flag #1 8 10 5 12 #2 7 NA 6 9 #3 10 1 2 NA #4 1 NA 1 5 #5 5 2 0 NA Hope this helps, Rui Barradas On 11/22/2017 10:34 AM, Massimo Bressan

Dear all, given I have data in a data.frame which indicate the number of people in a specific year at a specific age: n <- 10 mydf <- data.frame(yr=sample(1:10, size=n, replace=FALSE), age=sample(1:12, size=n, replace=FALSE), no=sample(1:10, size=n, replace=FALSE)) Now I would like to make a matrix with (in this simple example) 10 columns (for the

I'm working with some data, and am trying to generate it in the following format. state city zipcode I like pizza 0 0 0 I live in Denver 0 1 0 All the fun stuff is in Alaska 1 0 0 he lives in 66062

## Hello there, ## I have an issue where I need to use the value of column names to multiply with the individual values in a column and I have many columns to do this over. I have data like this where the column names are numbers: mydf <- data.frame(`2.72`=runif(20, 0, 125), `3.2`=runif(20, 50, 75), `3.78`=runif(20, 0, 100), yy=

Do you mean like this: mydf <- within(mydf, { is.na(A)<- !A_flag is.na(B)<- !B_flag } ) > mydf A A_flag B B_flag 1 8 10 5 12 2 NA 0 6 9 3 10 1 NA 0 4 NA 0 1 5 5 5 2 NA 0 Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into

#I want to "merge" or "join" 2 dataframes (df1 & df2) into a 3rd (mydf). I want the 3rd dataframe to contain 1 row for each row in df1 & df2, and all the columns in both df1 & df2. The solution should "work" even if the 2 dataframes are identical, and even if the 2 dataframes do not have the same column names. The rbind.fill function seems to work. For

Hi All, I'm fiddling with an program to read a text file containing periods that SAS uses for missing values. I know that if I had the original SAS data set instead of a text file, R would handle this conversion for me. Data frames do not allow missing values in their indices but vectors do. Why is that? A search of the error message points out the problem and solution but not why they

yes, it works, even if I do not really get how and why it's working the combination of logical results (could you provide some insights for that?) moreover, and most of all, I was hoping for a compact solution because I need to deal with MANY columns (more than 40) in data frame with the same basic structure as the simplified example I posted thanks m ----- Messaggio originale ----- Da:

OPS, Sorry i did not read the post carfully. Mine will not work if you have zeros on columns A and B.. But you could modify it to work for specific columns i believe. EK On Wed, Nov 22, 2017 at 8:37 AM, Ek Esawi <esawiek at gmail.com> wrote: > Hi *Massimo,* > > *Try this.* > > *a <- mydf==0mydf[a] <- NAHTHEK* > > On Wed, Nov 22, 2017 at 5:34 AM, Massimo Bressan

Hi *Massimo,* *Try this.* *a <- mydf==0mydf[a] <- NAHTHEK* On Wed, Nov 22, 2017 at 5:34 AM, Massimo Bressan < massimo.bressan at arpa.veneto.it> wrote: > > > Given this data frame (a simplified, essential reproducible example) > > > > > A<-c(8,7,10,1,5) > > A_flag<-c(10,0,1,0,2) > > B<-c(5,6,2,1,0) > > B_flag<-c(12,9,0,5,0) >