svykappa using the survey package

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).

# calculate survey Kappa
svykappa(~xbpchek53+xcholck53, design)

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,

+                          xbpchek53 = ifelse(bpchek53 ==1, 1,
+                             ifelse(bpchek53 %in% 2:6, 2,NA)),
+                          xcholck53 = ifelse(cholck53 ==1, 1,
+                            ifelse(cholck53 %in% 2:6, 2,NA)))

# 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 )))

        xbpchek53
bpchek53     1     2   Sum
     -9      0     0     0
     -8      0     0     0
     -7      0     0     0
     -1      0     0     0
     1   19778     0 19778
     2       0  2652  2652
     3       0  1014  1014
     4       0   538   538
     5       0   737   737
     6       0   623   623
     Sum 19778  5564 25342

addmargins(with(my_hc2013_data, table(cholck53,xcholck53 )))

        xcholck53
cholck53     1     2   Sum
     -9      0     0     0
     -8      0     0     0
     -7      0     0     0
     -1      0     0     0
     1   14850     0 14850
     2       0  3153  3153
     3       0  1170  1170
     4       0   696   696
     5       0   909   909
     6       0  3764  3764
     Sum 14850  9692 24542

addmargins(with(my_hc2013_data, table(xbpchek53,xcholck53 )))

         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?

mydata[ mixedorder(mydata$prevalence_c, decreasing=TRUE),
c(mydata$indicator, mydata$prevalence_c) ]

Try instead just a vector of names for the second argument to "["

 mydata[ mixedorder(mydata$prevalence_c, decreasing=TRUE),
         c("indicator", "prevalence_c") ]

Error in `[.data.frame`(mydata, mixedorder(mydata$prevalence_c,

decreasing = TRUE),  :

 undefined columns selected

********************

str(mydata)

Classes 'tbl_df', 'tbl' and 'data.frame':     10 obs. of  10 variables:
$ indicator   : chr  "1. Health check-up" "2. Blood cholesterol checked

" "3. Recieved flu vaccine" "4. Blood pressure checked" ...

$ subgroup    : chr  "Both sexes, ages =35 yrs""| __truncated__ "Both

sexes, ages =35 yrs""| __truncated__ "Both sexes, ages =35 yrs""|
__truncated__ "Both sexes, ages =35 yrs""| __truncated__ ...

$ n           : num  2117 2127 2124 2135 1027 ...
$ prevalence_c: chr  "74.7 (1.20)" "90.3 (0.89)" "51.7 (1.35)" "93.2

(0.70)" ...

$ prevalence_p: chr  "77.2 (1.19)" "84.5 (1.14)" "50.0 (1.33)" "88.7

(0.88)" ...

$ sensitivity : chr  "87.4 (1.10)" "99.2 (0.27)" "97.0 (0.62)" "99.0

(0.27)" ...

$ specificity : chr  "68.3 (2.80)" "58.2 (3.72)" "93.5 (0.90)" "52.7

(3.90)" ...

$ ppv         : chr  "90.4 (0.94)" "92.8 (0.85)" "93.7 (0.87)" "94.3

(0.63)" ...

$ npv         : chr  "61.5 (3.00)" "92.8 (2.27)" "96.9 (0.63)" "87.5

(3.27)" ...

$ kappa       : chr  "0.536 (0.029)" "0.676 (0.032)" "0.905 (0.011)"

"0.626 (0.035)" ...

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 Daniel
Nordlund
Sent: Wednesday, June 15, 2016 6:37 PM
To: r-help at r-project.org
Subject: Re: [R] dplyr's arrange function

On 6/15/2016 2:08 PM, Muhuri, Pradip (AHRQ/CFACT) wrote:

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.

testdata$prevalence

 [1] "77.2 (1.19)" "84.5 (1.14)" "50.0 (1.33)" "88.7 (0.88)" "11.7

(1.02)"

 [6] "60.2 (1.41)" "6.1 (0.61)"  "14.6 (1.00)" "72.6 (1.82)" "73.3

(2.37)"

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

David Winsemius
Alameda, CA, USA