[R] dplyr's arrange function - 3 solutions received - 1 New Question
Muhuri, Pradip (AHRQ/CFACT)
Pradip.Muhuri at ahrq.hhs.gov
Thu Jun 16 15:12:00 CEST 2016
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$prevalence_c) ]
Error in `[.data.frame`(mydata, mixedorder(mydata$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
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