Thanks.
That works great!
> df <- data.frame(x=c(1,1,NA,NA,2), y=c('a','a','a','b','b'),
> z=c(TRUE,FALSE,TRUE,FALSE,TRUE))
> cond1 <- 'x==1'
> cond2 <- 'x==1 & z'
> df
x y z
1 1 a TRUE
2 1 a FALSE
3 NA a TRUE
4 NA b FALSE
5 2 b TRUE
> subset(df, subset =
You are missing a comma between "MARITAL" and "JOBSTATUS".
On Tue, Apr 10, 2018 at 10:27 AM, Saif Tauheed
wrote:
> I run this command for converting the numerical variable into factor.
> However, I get the following error message.
>
> > cols<- c(“GrMM", "RELG", "CASTE1",
Thanks.
S. Elison provided a similar but apparently more general solution (see other
post in thread).
- Original Message -
From: "David Winsemius"
To: "Sebastien Bihorel"
Cc: r-help@r-project.org
Sent: Monday, April 9, 2018
Thank you very much.
After that I have the following error:
cols<- c("GrMM", "RELG", "CASTE1", "SECTOR", "SECTOR4","AGE", "MARITAL",
"JOBSTATUS", "ENG", "EDU", "PARENT_EDU", "MASSMEDIA_F", "MASSMEDIA_M",
"HomeComputer", "HomeInternet")
> for (I in cols) {data.frame[,i]=
I run this command for converting the numerical variable into factor. However,
I get the following error message.
> cols<- c(“GrMM", "RELG", "CASTE1", "SECTOR", "SECTOR4","AGE", "MARITAL"
> "JOBSTATUS", "ENG", "EDU", "PARENT_EDU", "MASSMEDIA_F", "MASSMEDIA_M",
> "HomeComputer", "HomeInternet")
Not sure whether this is the problem but calling your data frame
data.frame is not a good idea.
On 10/04/2018 11:48, Saif Tauheed wrote:
Thank you very much.
After that I have the following error:
cols<- c("GrMM", "RELG", "CASTE1", "SECTOR", "SECTOR4","AGE", "MARITAL", "JOBSTATUS", "ENG",
Please provide a reproducible example of the problem, with sample data.
Notes:
1) The Posting Guide points out that this is a plain text mailing list, but
does not emphasize how damaged your sent email may be if you fail to set your
email program to plain text mode.
2) Technically, 3.4.3 is not
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Saif Tauheed
> After that I have the following error:
>
> cols<- c("GrMM", "RELG", "CASTE1", "SECTOR", "SECTOR4","AGE", "MARITAL",
> "JOBSTATUS", "ENG", "EDU", "PARENT_EDU", "MASSMEDIA_F",
> "MASSMEDIA_M", "HomeComputer",
stats::spectrum for starters.
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Apr 10, 2018 at 5:49 AM, Danniel Lafeta Machado <
danniel.laf...@bcb.gov.br> wrote:
> Dear all,
> Is there any spectral analisys functionality available for R version 3.4.3?
> Series() functionality doesn't work
Dear all,
Is there any spectral analisys functionality available for R version 3.4.3?
Series() functionality doesn't work in this version.
Regards
Danniel
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R-help@r-project.org mailing list -- To
Hi
You could instead of selecting columns by name select them by number
Something like
your.data[,2:3] <- do.call(data.frame, lapply(your.data[,2:3], factor))
or you could construct a vector of column numbers
x <- c(2,3)
Cheers
Petr
-Original Message-
From: R-help
Dear Mr. Savicky,
I am currently working on a project where I want to test a random number
generator, which is supposed to create 10.000 continuously uniformly
distributed random numbers between 0 and 1. I am now wondering if I can use the
Chi-Squared-Test to solve this problem or if the
Have you ever noticed that when you run
x <- 1:5
y <- 2:6
plot( x, y+1 )
you get the expressions you used in your call to plot on the axis labels? `x`
is an expression consisting of a single symbol and y+1 is an expression
consisting of the addition operator and two arguments: the symbol x and
Disculpad el error...
La reunión es el miércoles de la semana que viene. *Miércoles 18 de abril.*
Gracias y siento la confusión...
Carlos.
El 8 de abril de 2018, 17:34, Carlos Ortega
escribió:
> Buenas a todos,
>
> Para los que puedan y quieran asistir.
>
> El
Hola, ¿qué tal?
Pues casi seguro que tienes un caso de "separación perfecta". Aunque no
exista una variable única (p.e., un nivel en una variable categórica) que
tenga asociados solo valores 0, es posible que exista una combinación
lineal de variables que separe regiones donde solo hay ceros del
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