similar to: if statement

Displaying 20 results from an estimated 50000 matches similar to: "if statement"

2010 Dec 13
3
curve
Hi All, I generated 5000 samples using the following script test<- rnorm(5000,1000,100) test1 <- subset(test, subset=(test > 1100)) d <- density(test) plot(d, main="Density of production") abline(v=mean(test1) I wanted to do the following but faced difficulties 1. to shade or color (blue) the curve using the criterion that any
2012 Nov 21
4
help with if statement
Hi all, I had a dataset A like: TIME DV 0 0 1 10 5 20 24 30 36 80 48 60 72 15 I would like to add 24 to those values higher than 24 in the TIME column. I did the following: If (A$TIME>=24) { A$TIME <- A$TIME+24} It did not work. How should I do it? Thanks, -- View this message in context: http://r.789695.n4.nabble.com/help-with-if-statement-tp4650315.html Sent from the
2011 Jul 11
1
Ifelse statement
Hello everyone, I have a (small) issue. I already googled a lot, so I decided to use ifelse instead of if (){} else{} All the elements seem to work seperately, but combined in the ifelse statement, it doesn't seem to work. #The price function is a function which is normally distributed with only positive answers price<-function() {abs(rnorm(1,10,25))} #Before I use pieceprice in the
2012 May 30
5
problem with ifelse
Dear all, ?The code below is used to generate interval censored data but unfortunately there is an error with the ifelse which i am not able to rectify. ?Can somebody help correct it for me. Thank you t<-rexp(20,0.2)? v<-c(0,m,999)? y<-function(t,v){ ? z<-numeric(length(t (( ? ? s<-numeric(length(t (( ? ? ? for(i in 1:length(t)){ ? ? ? ? for(j in 1:length(v-1))? ? ? ? ? { ifelse
2012 May 08
1
grouping function
Hello, I would like to write a function that makes a grouping variable for some panel data . The grouping variable is made conditional on the begin year and the end year. Here is the code I have written so far. name <- c(rep('Frank',5), rep('Tony',5), rep('Edward',5)); begin <- c(seq(1990,1994), seq(1991,1995), seq(1992,1996)); end <- c(seq(1995,1999),
2012 Jun 01
1
trouble with append() in a for loop
Hello all, * * I'm having some difficulty, and I think the problem is with how I'm using append() nested inside a for loop. The data are: y,x 237537.61,873 5007.148438,227 17705.77306,400 12396.64369,427 228703.4021,1173 350181.9752,1538 59967.79376,630 140322.7774,710 42650.07251,630 5382.858702,264 34405.82429,637 92261.34614,980 144927.1713,1094 362998.7355,1420 203313.6442,1070
2011 Apr 26
2
matrix
Hi all, Assume I have a matrix xv= [1 0 0 0 0 12, 0 1 0 0 0 10, * 0 0 1 0 0 -9,* 0 0 0 1 0 20, * 0 0 0 0 1 -5]* if the last column of "xv" less than 0 then I want to set zero the entire row. The desired output looks like the following. In this case row 3 and row 5 are set zero. [ 1 0 0 0 0 12, 0 1 0 0 0 10, * 0 0 0 0 0 0,* 0 0 0 1 0 20,
2011 Dec 03
4
Data alignment
Hello! I have a data.frame which looks like: Name - Value A - 400 A - 300 B - 200 B - 350 C - 500 C - 350 D - 450 D - 600 E - 700 E - 750 F - 630 F - 650 I want to add another column where all A,B should get an index 1, all C,D an index of 2 and all E,F an index of 3 so that the data.frame looks like: ID - Name - Value 1 - A - 400 1 - A - 300 1 - B - 200 1 - B - 350 2 - C - 500 2 - C - 350 2
2012 Mar 21
2
Check results between two data.frame
Dear R-user, I'm trying to compare two sets of results and wanted to find out which element in the two data frame/matrix are different. I wrote the following function and it works ok, and gives me a long list of "good" as outcomes. CHECK<- function (x = "file1", y = "file2") { for (i in 1:nrow(x)) { for (j in 1:ncol(x)) { if (x[i, j]
2013 Jul 23
1
Heat Map for species - code from Numerical Ecology with R
Hello, I am relatively new to R and I am working through the code that is provided in the book Numerical Ecology with R and I have run across an error message that I can't seem to figure out. I am using the vegan, ade4, gclus and cluster packages. The code is as follows: # Ordered community table # Species are ordered by their weighted averages on site scores or <- vegemite(spe,
2017 Dec 14
1
match and new columns
Hi Bill, I put stringsAsFactors = FALSE still did not work. tdat <- read.table(textConnection("A B C Y A12 B03 C04 0.70 A23 B05 C06 0.05 A14 B06 C07 1.20 A25 A23 A12 3.51 A16 A25 A14 2,16"),header = TRUE ,stringsAsFactors = FALSE) tdat$D <- 0 tdat$E <- 0 tdat$D <- (ifelse(tdat$B %in% tdat$A, tdat$A[tdat$B], 0)) tdat$E <- (ifelse(tdat$B %in% tdat$A, tdat$A[tdat$C], 0))
2017 Dec 13
2
match and new columns
Thank you Rui, I did not get the desired result. Here is the output from your script A B C Y D E 1 A12 B03 C04 0.70 0 0 2 A23 B05 C06 0.05 0 0 3 A14 B06 C07 1.20 0 0 4 A25 A23 A12 3.51 1 1 5 A16 A25 A14 2,16 4 4 On Wed, Dec 13, 2017 at 4:36 PM, Rui Barradas <ruipbarradas at sapo.pt> wrote: > Hello, > > Here is one way. > > tdat$D <- ifelse(tdat$B %in% tdat$A,
2011 Mar 10
1
How to use conditional statement
Dear R helpers Suppose val1 = c(10, 20, 35, 80, 12) val2 = c(3, 8, 11, 7) I want to select either val1 or val2 depending on value of third quantity val3. val3 assumes either of the values "Monthly" or "Yearly". If val3 = "Monthly", then val = val1 and if val3 = "Yearly", then val = val2. I tried the ifelse statement as ifelse(val3 =
2012 Jan 13
2
function to replace values doesn't work on vectors
I've got a numeric vector with values ranging from 1 to 5, I would like to catagorize these values like this: 1 becomes catagory 1 3 becomes catagory 3 And everything else in catagory 2. The simple function I wrote beneath works for single numeric data, but for some reason I am unable to feed it vectors. Any help would be appreciated, as I'm fairly new to R. -- View this message in
2012 Oct 22
4
creating a function using for if
Hi all, I'm trying to create a function where it can process a vector and also give a vector output accordingly #input: a,b anc c are constants, data is the vector #set the function fun<-function(a,b,c,data) { N=as.vector() for (i in min(data):max(data)){ if(i>c){ N<-(a*(i-c)^0.5)+(b*(i-c))} else {N<-0}} return(N) } #try dummy data=c(100,210,320,130,170,120,220,90,55,45)
2017 Dec 14
0
match and new columns
Use the stringsAsFactors=FALSE argument to read.table when making your data.frame - factors are getting in your way here. Bill Dunlap TIBCO Software wdunlap tibco.com On Wed, Dec 13, 2017 at 3:02 PM, Val <valkremk at gmail.com> wrote: > Thank you Rui, > I did not get the desired result. Here is the output from your script > > A B C Y D E > 1 A12 B03 C04 0.70 0 0
2017 Jun 03
2
New var
Thank you all for the useful suggestion. I did some of my homework. library(data.table) DFM <- read.table(header=TRUE, text='obs start end 1 2/1/2015 1/1/2017 2 4/11/2010 1/1/2011 3 1/4/2006 5/3/2007 4 10/1/2007 1/1/2008 5 6/1/2011 1/1/2012 6 10/5/2004 12/1/2004',stringsAsFactors = FALSE) DFM DFM$D =as.numeric(difftime(as.Date(DFM$end,format="%m/%d/%Y"),
2011 Nov 10
3
Creating dummys in R
Dear R-project! How do i create 1 dummy from 2 already existing dummys. To be more precise, I want to create a dummy from a dummy called "sex" and another called "sex1" when both thoose dummys are 1 I want my created dummy "samesex" to take 1. Thanks for the help! Paulie [[alternative HTML version deleted]]
2011 Jul 29
3
help with plot.rpart
? data=read.table("http://statcourse.com/research/boston.csv", , sep=",", header = TRUE) ? library(rpart) ? fit=rpart (MV~ CRIM+ZN+INDUS+CHAS+NOX+RM+AGE+DIS+RAD+TAX+ PT+B+LSTAT) Please: Show me the tree. Mark -------- Original Message -------- Subject: Re: [R] help with rpart From: "Stephen Milborrow" <[1]milbo at sonic.net>
2012 Oct 04
1
R help - Adding a column in a data frame with multiple conditions
Hi, I am trying to add a column of numbers to a data frame in R with multiple conditions. Here is a simplified example df: [A] [B] [C] [D] [E] [1] 1 X 90 88 [2] 1 Y 72 70 [3] 1 Z 67 41 [4] 2 X 74 49 [5] 2 Y 42 50 [6] 2 Z 81 56 [7] 3 X 92 59 [8] 3 Y 94 80 [9] 3 Z 80 82 I would like column [E] to have a certain value (found either in [C] or [D]) based on conditions in columns [A] *and* [B].