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,
--
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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.
--
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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
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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].