This works perfectly. Ah, just needed a vector as output instead of a
1-column df.
Thank you!!!
On Tue, Jan 19, 2021 at 2:18 PM Bill Dunlap
wrote:
> Your translate... function seems unnecessarily complicated and reusing the
> name 'var' for both the input and the data.frame containing the input
I avoid case_when, so don't complain to me about it. Bert and I both suggested
standard evaluation approaches that are very amenable to using lookup tables.
On January 19, 2021 1:51:17 PM PST, Steven Rigatti wrote:
>I use case_when a lot - but I have a lot of dynamic tables to treat
>this
>way a
David
library(tidyverse)
char_vec <- sample(c("a", "b", "c"), 10, replace = TRUE)
recode(char_vec, a = "Apple")
works for me.
On Tue, 19 Jan 2021 at 15:13, David Winsemius
wrote:
>
> On 1/19/21 11:17 AM, Bill Dunlap wrote:
> > Your translate... function seems unnecessarily complicated and reusi
Sent from my iPhone
> On Jan 19, 2021, at 1:52 PM, Steven Rigatti wrote:
>
> I use case_when a lot - but I have a lot of dynamic tables to treat this
> way and case_when has to be hard-coded.
But, but, but my case_when-based illustration let you pass a parameter
dataframe that contain
I use case_when a lot - but I have a lot of dynamic tables to treat this
way and case_when has to be hard-coded.
On Tue, Jan 19, 2021 at 3:48 PM Jeff Newmiller
wrote:
> Second this. There is also the findInterval function, which omits the
> factor attributes and just returns integers that can be
Second this. There is also the findInterval function, which omits the factor
attributes and just returns integers that can be used in lookup tables.
On January 19, 2021 10:33:59 AM PST, Bert Gunter wrote:
>If you are willing to entertain another approach, have a look at ?cut.
>By
>defining the '
On 1/19/21 11:17 AM, Bill Dunlap wrote:
Your translate... function seems unnecessarily complicated and reusing the
name 'var' for both the input and the data.frame containing the input makes
it confusing to me. The following replacement, f, uses your algorithm but
I think gets the answer you w
Your translate... function seems unnecessarily complicated and reusing the
name 'var' for both the input and the data.frame containing the input makes
it confusing to me. The following replacement, f, uses your algorithm but
I think gets the answer you want.
f <-
function(var, upper, lookup) {
It's not that I can't get the output I want. I was able to do that.
It is just that I can't make it pipeable - I get that weird error message
that I don't understand.
On Tue, Jan 19, 2021 at 1:34 PM Bert Gunter wrote:
> If you are willing to entertain another approach, have a look at ?cut. By
>
On 1/19/21 7:50 AM, Steven Rigatti wrote:
I am having some problems with what seems like a pretty simple issue. I
have some data where I want to convert numbers. Specifically, this is
cancer data and the size of tumors is encoded using millimeter
measurements. However, if the actual measurement
If you are willing to entertain another approach, have a look at ?cut. By
defining the 'breaks' argument appropriately, you can easily create a
factor that tells you which values should be looked up and which accepted
as is. If I understand correctly, this seems to be what you want. If I have
not,
On Oct 12, 2015, at 3:11 PM, John Sorkin wrote:
> I am trying to learn how to write R functions (really to program in R). I
> want to write a function that will allow me to refer to the elements of a
> data frame by column name, much as is done in lm. I can't seem to get the
> syntax correct.
Hi John,
Here are a couple of ways to do this. Note that they produce slightly
different results.
data2<-data.frame(POSTHHMONO=c(1,2,3,4),Mo6MON=c(10,11,12,13))
doit<- function(pre,post,data) {
element <- deparse(substitute(pre))
print(element)
print(data[element])
frame<-deparse(substitut
On 25.01.2012 02:16, R. Michael Weylandt wrote:
I think you are getting stuck on the same regexp problem as before
(i.e., once again the dollar sign is being interpreted as the
beginning
You meant "end".
Uwe
of the line rather than an actual dollar sign)
If I understand your question, mi
I think you are getting stuck on the same regexp problem as before
(i.e., once again the dollar sign is being interpreted as the
beginning of the line rather than an actual dollar sign)
If I understand your question, might I suggest something much easier?
x = data.frame(a = c("$1034.23","1,230"),
On Jan 24, 2012, at 11:07 AM, Dan Abner wrote:
Hi everyone,
I am using Michael's approach (grepl()) to identify which columns
containing $ signs. I was hoping to incorporate this into a line of
code that would automatically 1) find which columns contain $ signs,
2) strip the $ and commas, and
Hi everyone,
I am using Michael's approach (grepl()) to identify which columns
containing $ signs. I was hoping to incorporate this into a line of
code that would automatically 1) find which columns contain $ signs,
2) strip the $ and commas, and 3) convert the result to a numeric
vector.
I have
On 01/24/2012 03:49 PM, Dan Abner wrote:
> Hello everyone,
>
> I am writing my own function to return the column index of all variables
> (these are currently character vectors) in a data frame that contain a
> dollar sign($). A small piece of the data look like this:
>
> can_sta can_zip ind_
Either
any(grepl("$",x, fixed = TRUE)) # You probably want grepl not grep
any(grepl("\\$",x) )
? regexpr # $ has a special value
Michael
PS -- Stop with HTML postings (seriously, it actually does mess up
what the rest of us see and I think it causes trouble for the archives
as well)
On Tue, Jan
Thanks Michael, I'll keep that in mind if I want to do anything more
complicated.
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Sent from the R help mailing list ar
Just a heads up -- I don't think your code will work with an actual
.xls(x) file, only .txt, .csv, etc (aka, plain text files). I may be
wrong about that, but if you actually need to work with Excel files
directly you will need an additional package.
Michael
On Thu, Dec 1, 2011 at 9:10 AM, AOLear
Thank you very much, that does exactly what I want it to! :)
Aodhán
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Sent from the R help mailing list archive at Nabb
Sure, change your example as follows and then you can pass the name properly:
foo <- function(x,y,NAME = "filename.csv"){
#make a matrix with x rows and y cols
M <- matrix(nrow=x,ncol=y)
#write the matrix
write.table(M, file = NAME,append=TRUE, sep = ",")
}
Bryan
Prof. Bryan Ha
mohamed nur anisah <[EMAIL PROTECTED]> [Fri, Feb 08, 2008 at 04:42:41PM CET]:
> Dear lists,
>
> I'm in my process of learning of writing a function. I tried to write a
> simple functions of a matrix and a vector. Here are the codes:
>
> mm<-function(m,n){ #matrix function
It is doing exactly what you ask. You are asking for
the last element in the matrix w[i,j] and the last
element in the vector y[i].
Try return(w) and return(y).
--- mohamed nur anisah <[EMAIL PROTECTED]>
wrote:
> Dear lists,
>
> I'm in my process of learning of writing a
> function. I t
Hi Mohamed,
Just change return(w[i,j]) by return(w), and return(y[i]) by return(y).
I hope this helps,
Jorge
On 2/8/08, mohamed nur anisah <[EMAIL PROTECTED]> wrote:
>
> Dear lists,
>
> I'm in my process of learning of writing a function. I tried to write a
> simple functions of a matrix and
Hi Mohamed,
mohamed nur anisah wrote (8.2.2008):
> Dear lists,
>
> I'm in my process of learning of writing a function. I tried to
> write a simple functions of a matrix and a vector. Here are the
> codes:
>
> mm<-function(m,n){ #matrix function
> w<-matrix(nrow=m, ncol=n)
> for
Hi Mohamed,
You want to return the matrix - you're returning an element of the
matrix. So in your formula, insert:
return(w)
instead of
return(w[i,j])
On Feb 8, 2008 8:42 AM, mohamed nur anisah <[EMAIL PROTECTED]> wrote:
> Dear lists,
>
> I'm in my process of learning of writing a function.
Anisah,
You just need to omit the indices in the return statements:
mm<-function(m,n){ #matrix function
w<-matrix(nrow=m, ncol=n)
for(i in 1:m){
for(j in 1:n){
w[i,j]=i+j
}
}
w
}
v<-function
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