Às 00:22 de 25/10/2023, Sorkin, John escreveu:
Colleagues,
I have written an R function (see fully annotated code below), with which I
want to process a dataframe within levels of the variable StepType. My program
works, it processes the data within levels of StepType, but the usual headers
t
Dear John,
Printing inside the function is problematic. Your function itself does
NOT print the labels.
Just as a clarification:
F = factor(rep(1:2, 2))
by(data.frame(V = 1:4, F = F), F, function(x) { print(x); return(NULL); } )
# V F
# 1 1 1
# 3 3 1
# V F
# 2 2 2
# 4 4 2
# F: 1 <- this i
Hello,
These two gives the same results:
aggregate(values ~ sex + status, mydata, sum)
# sex status values
#1 0 0 224
#2 1 0 5227
#3 0 1 11
#4 1 1 552
by(mydata$values, list(mydata$sex, mydata$status), sum)
#: 0
#: 0
#[1] 224
#-
On 2020-07-24 01:48 +0200, Rasmus Liland wrote:
> aggregate(x=list("values"=mydata$values),
> by=list("sex"=mydata$sex,
> "status"=mydata$status),
> FUN=sum)
>
> yields
>
> sex status values
> 1 0 0224
>
On 2020-07-23 18:54 -0400, Duncan Murdoch wrote:
> On 23/07/2020 6:15 p.m., Sorkin, John wrote:
> > Colleagues,
> > The by function in the R program below is not giving me the sums
> > I expect to see, viz.,
> > 382+170=552
> > 4730+170=4900
> > 5+6=11
> > 199+25=224
> > ###
On 23/07/2020 6:15 p.m., Sorkin, John wrote:
Colleagues,
The by function in the R program below is not giving me the sums
I expect to see, viz.,
382+170=552
4730+170=4900
5+6=11
199+25=224
###
#full R program:
mydata <- data.frame(covid=c(0,0,0,
by() chooses **data frame** subsets -- sum() is acting on these frames,
adding up everything in them.
Try this instead:
> by(mydata,list(mydata$sex,mydata$status),function(x)sum(x$values))
: 0
: 0
[1] 224
---
: 1
: 0
[1] 5227
Hi
The problem is in your function, not in by.
> dstats(mtcars[,c('mpg', 'hp')])
Error in is.data.frame(x) :
(list) object cannot be coerced to type 'double'
In addition: Warning message:
In mean.default(x) : argument is not numeric or logical: returning NA
>
Your function expects vector but y
10 24 43
6B H 13 17 28
---------
David L Carlson
Department of Anthropology
Texas A&M University
College Station, TX 77840-4352
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf O
Good day,
Yes, exactly. I found that aggregate is another alternative which doesn't
require a package dependency, although the column formatting is less suitable,
always prepending x.
aggregate(warpbreaks[, 1], warpbreaks[, 2:3], function(breaks) c(Min =
min(breaks), Med = median(breaks), Max
Do you want something like the following?
> library(dplyr, quietly=TRUE, warn.conflicts=FALSE)
> warpbreaks %>% group_by(wool, tension) %>% summarize(Min=min(breaks),
> Median=median(breaks), Max=max(breaks))
Source: local data frame [6 x 5]
Groups: wool
wool tension Min Median Max
1A
Hi,
You could try this:
dat1<-read.table(text="
id,age,weight,height,gender
1,22,180,72,m
2,13,100,67,f
3,5,40,40,f
4,6,42,,f
5,12,98,66,
6,50,255,60,m
",sep=",",header=TRUE,stringsAsFactors=FALSE,na.strings="")
list1<-by(dat1[c("weight","height")],dat1[c("age","gender")],colMeans,na.rm=TR
On Jan 3, 2013, at 9:00 PM, Ray DiGiacomo, Jr. wrote:
Hello,
I have the following dataset. Please note that there are missing
values on
records 4 and 5:
id,age,weight,height,gender
1,22,180,72,m
2,13,100,67,f
3,5,40,40,f
4,6,42,,f
5,12,98,66,
6,50,255,60,m
I'm using the "By" function lik
Liviu,
Thanks for your attention and help!
Sorry for attached csv file, I didn't read R-Help Posting guide
careful. Another R-helper thaught me use de dput() function, and this
output are in email's end.
Your correction in my code works pretty well!! Now I can understand a
little more how the R
Hello
On Fri, Apr 8, 2011 at 12:38 AM, Raoni Rodrigues
wrote:
> Hello all!
>
> I have a data frame with nine variables and 293 cases. (attached goes
> the csv file).
>
The CSV didn't get through so it's difficult to replicate your
example. Please post the output of:
str(cpue)
> I need to calcul
sage
> news:77eb52c6dd32ba4d87471dcd70c8d7000243c...@na-pa-vbe03.na.tibco.com...
>> -Original Message-
>> From: r-help-boun...@r-project.org
>> [mailto:r-help-boun...@r-project.org] On Behalf Of L.A.
>> Sent: Saturday, December 12, 2009 12:39 PM
>> To: r-help@
You asked how to 'create confidence intervals around the median'. Since
intervals was plural and you asked within the context of 'by' LEAID then
that made sense and I guessed you meant 'confidence interval around the
median ratio of each group'. I already assumed you had posted a small subset
o
Thanks, but that produces what I think is an estimated interval.
I really want to use the above formula. I just can't figure out how to
get it to run by the LEAID.
It does require 9 observations to produce an interval, but I was showing a
sample.
Thanks again.
L.A.
Matthew Dowle-3 wrote:
>
>
Maybe this (with enough data for a CI) ? :
> Dataset = data.table(Dataset)
> Dataset[,as.list(wilcox.test(ratio,conf.int=TRUE)$conf.int),by="LEAID"]
LEAID V1 V2
[1,] 6307 0.720 0.92
[2,] 8300 0.5678462 0.83
Warning messages:
1: In switch(alternative, two.sided
Well, I'm back again.
Thanks for all the help. Besides working, it's helping me begin to
understand how these functions work.
I still have trouble reading or following the process of a function, which
brings my next question:
Dataset:
LEAID ratio
3 6307 0.720
1 6307 0.7623810
-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of L.A.
> Sent: Saturday, December 12, 2009 12:39 PM
> To: r-help@r-project.org
> Subject: Re: [R] by function ??
>
>
>
> Thanks for all the help, They all worked, But I'm stuck again.
> I've
On Dec 12, 2009, at 3:38 PM, L.A. wrote:
Thanks for all the help, They all worked, But I'm stuck again.
I've tried searching, but I not sure how to word my search as
nothing came
up.
Here is my new hurdle, my data has 7 abservations and my results
have 2
answers:
Here is my data
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of L.A.
> Sent: Saturday, December 12, 2009 12:39 PM
> To: r-help@r-project.org
> Subject: Re: [R] by function ??
>
>
>
> Thanks for all the help,
Thanks for all the help, They all worked, But I'm stuck again.
I've tried searching, but I not sure how to word my search as nothing came
up.
Here is my new hurdle, my data has 7 abservations and my results have 2
answers:
Here is my data
LEAID ratio
3 6307 0.720
1 6307
other
situations.
Michael
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Ista Zahn
> Sent: Mittwoch, 9. Dezember 2009 02:54
> To: L.A.
> Cc: r-help@r-project.org
> Subject: Re: [R] by function ??
&
Hi,
I think you want
by(TestData[ , "RATIO"], LEAID, median)
-Ista
On Tue, Dec 8, 2009 at 8:36 PM, L.A. wrote:
>
> I'm just learning and this is probably very simple, but I'm stuck.
> I'm trying to understand the by().
> This works.
> by(TestData, LEAID, summary)
>
> But, This doesn't.
>
> by
Please give minimal reproducible examples.
The data and a test driver were missing.
See last line to every message to r-help.
The problem is discussed here:
http://tolstoy.newcastle.edu.au/R/help/06/08/32017.html
On Mon, Feb 25, 2008 at 10:05 AM, vincent chouraki <[EMAIL PROTECTED]> wrote:
> Dear
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