Hello,
My goal is to calculate the weighted kappa measure of agreement between two factors using the R survey package. I am getting the following error message (the console is appended below; sorry no data provided).
Error in names(probs) <- nms :
'names' attribute [15] must be the same length as the vector [8]
I have followed the following major steps:
1) Used the "haven" package to read the sas data set into R.
2) Used the dplyr mutate() to create 2 new variables and converted to factors [required for the svykappa()?].
3) Created an object (named design) using the survey design variables and the data file.
4) Used the svykappa() to compute the kappa measure of agreement.
I will appreciate if someone could give me hints on how to resolve the issue.
Thanks,
Pradip Muhuri
############### The detailed console is appended below ####################
setwd ("U:/A_PSAQ")
library(haven)
library(dplyr)
library(survey)
library(srvyr)
library(Hmisc)
my_hc2013_data <- read_sas("pc2013.sas7bdat")
# Function to convert var names in upper cases to var names in lower cases
lower <- function (df) {
+ names(df) <- tolower(names(df))
+ df
+ }
my_hc2013_data <- lower(my_hc2013_data)
# Check the contents - Hmisc package (as above) required
# contents(my_hc2013_data)
# create two new variables
my_hc2013_data <- mutate(my_hc2013_data,
# convert the numeric variables to factors for the kappa measure
my_hc2013_data$xbpchek53 <- as.factor(my_hc2013_data$xbpchek53)
my_hc2013_data$xcholck53 <- as.factor(my_hc2013_data$xcholck53)
# check whether the variables are factors
is.factor(my_hc2013_data$xbpchek53)
[1] TRUE
is.factor(my_hc2013_data$xcholck53)
[1] TRUE
# check the data from the cross table
addmargins(with(my_hc2013_data, table(bpchek53,xbpchek53 )))
xcholck53
xbpchek53 1 2 Sum
1 14667 4379 19046
2 163 5225 5388
Sum 14830 9604 24434
# create an object with design variables and data
design<-svydesign(id=~varpsu,strat=~varstr, weights=~perwt13f, data=my_hc2013_data, nest=TRUE)
# calculate survey Kappa
svykappa(~xbpchek53+xcholck53, design)
Error in names(probs) <- nms :
'names' attribute [15] must be the same length as the vector [8]
#################################################################
Pradip K. Muhuri, AHRQ/CFACT
5600 Fishers Lane # 7N142A, Rockville, MD 20857
Tel: 301-427-1564
-----Original Message-----
From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Muhuri, Pradip (AHRQ/CFACT)
Sent: Thursday, June 16, 2016 2:06 PM
To: David Winsemius
Cc: r-help at r-project.org
Subject: Re: [R] dplyr's arrange function - 3 solutions received - 1 New Question
Hello David,
Your revisions to the earlier code have given me desired results.
library("gtools")
mydata[ mixedorder(mydata$prevalence_c, decreasing=TRUE), c("indicator", "prevalence_c") ]
Thanks,
Pradip
Pradip K. Muhuri, AHRQ/CFACT
5600 Fishers Lane # 7N142A, Rockville, MD 20857
Tel: 301-427-1564
-----Original Message-----
From: David Winsemius [mailto:dwinsemius at comcast.net]
Sent: Thursday, June 16, 2016 12:54 PM
To: Muhuri, Pradip (AHRQ/CFACT)
Cc: r-help at r-project.org
Subject: Re: [R] dplyr's arrange function - 3 solutions received - 1 New Question
On Jun 16, 2016, at 6:12 AM, Muhuri, Pradip (AHRQ/CFACT) <Pradip.Muhuri at ahrq.hhs.gov> wrote:
Hello,
I got 3 solutions to my earlier code. Thanks to the contributors. May I bring your attention to a new question below (with respect to David's solution)?
1) Thanks to Daniel Nordlund for the tips - replacing leading space with a 0 in the data.
2) Thanks to David Winsemius for his solution with the gtools::mixedorder function. I have added an argument to his.
mydata[ mixedorder(mydata$prevalence_c, decreasing=TRUE), ]
3) Thanks to Jim Lemon's for his solution. I have prepended a minus sign to reverse the order.
numprev<-as.numeric(sapply(strsplit(trimws(mydata$prevalence_c),"
"),"[",1)) mydata[order(-numprev), ]
(New)Question for solution 2:
I want to keep only 2 variables (say, indicator and prevalence_c) in the output. Where to insert the additional code? Why does the following code fail?
Try instead just a vector of names for the second argument to "["
mydata[ mixedorder(mydata$prevalence_c, decreasing=TRUE),
c("indicator", "prevalence_c") ]
Hello,
I am using the dplyr's arrange() function to sort one of the many data frames on a character variable (named "prevalence").
Issue: I am not getting the desired output (line 7 is the problem, which should be the very last line in the sorted data frame) because the sorted field is character, not numeric.
The reproducible example and the output are appended below.
Is there any work-around to convert/treat this character variable (named "prevalence" in the data frame below) as numeric before using the arrange() function within the dplyr package?
Any hints will be appreciated.
Thanks,
Pradip Muhuri
# Reproducible Example
library("readr")
testdata <- read_csv(
"indicator, prevalence
1. Health check-up, 77.2 (1.19)
2. Blood cholesterol checked, 84.5 (1.14) 3. Recieved flu vaccine,
50.0 (1.33) 4. Blood pressure checked, 88.7 (0.88) 5. Aspirin
use-problems, 11.7 (1.02) 6.Colonoscopy, 60.2 (1.41) 7.
Sigmoidoscopy,
6.1 (0.61) 8. Blood stool test, 14.6 (1.00) 9.Mammogram, 72.6 (1.82)
10. Pap Smear test, 73.3 (2.37)")
# Sort on the character variable in descending order
arrange(testdata,
desc(prevalence))
# Results from Console
indicator prevalence
(chr) (chr)
1 4. Blood pressure checked 88.7 (0.88)
2 2. Blood cholesterol checked 84.5 (1.14)
3 1. Health check-up 77.2 (1.19)
4 10. Pap Smear test 73.3 (2.37)
5 9.Mammogram 72.6 (1.82)
6 6.Colonoscopy 60.2 (1.41)
7 7. Sigmoidoscopy 6.1 (0.61)
8 3. Recieved flu vaccine 50.0 (1.33)
9 8. Blood stool test 14.6 (1.00)
10 5. Aspirin use-problems 11.7 (1.02)
Pradip K. Muhuri, AHRQ/CFACT
5600 Fishers Lane # 7N142A, Rockville, MD 20857
Tel: 301-427-1564
The problem is that you are sorting a character variable.
Notice that the 7th element is "6.1 (0.61)". The first CHARACTER is a "6", so it is going to sort BEFORE the "50.0 (1.33)" (in descending order). If you want the character value of line 7 to sort last, it would need to be "06.1 (0.61)" or " 6.1 (0.61)" (notice the leading space).
Hope this is helpful,
Dan
Daniel Nordlund
Port Townsend, WA USA