[R] Applying a user-defined function

Wed Jan 9 04:55:12 CET 2013

Hi Pradip,
I didn't check the mode at that time.   It generated a "matrix" "test1$newcols<- sapply(....)" 
You can do this:
test2<-data.frame(test1[,-7],test1$newcols)

str(test2)
#'data.frame':    51 obs. of  9 variables:
# $ ObtMj_P       : num  49.6 55 52.5 50.5 51.1 55.1 56.3 53.6 53.5 52.7 ...
# $ ObtMj_SE      : num  1.37 1.41 1.56 1.22 0.65 1.26 1.28 1.3 1.22 0.67 ...
# $ ExpPrevMed_P  : num  80 81.8 79.6 78 80.5 81.7 85 79.5 76.2 78.9 ...
# $ ExpPrevMed_SE : num  0.91 1.08 1.2 0.78 0.53 1.03 0.93 1.04 1.03 0.52 ...
# $ ParMon_P      : num  12.1 12.4 15.8 12.8 13 12.1 14.6 14.7 14.3 14.1 ...
# $ ParMon_SE     : num  0.68 0.9 1.08 0.72 0.41 0.72 0.77 0.97 1.13 0.45 ...
# $ ObtMj_P.1     : Factor w/ 5 levels "[42,48.7]","(48.7,50.9]",..: 2 5 3 2 3 5 5 4 4 3 ...
# $ ExpPrevMed_P.1: Factor w/ 5 levels "[76.2,79.2]",..: 2 3 2 1 2 3 5 2 1 1 ...
# $ ParMon_P.1    : Factor w/ 5 levels "[11.9,12.6]",..: 1 1 5 2 2 1 4 4 3 3 ..

levels(test2[,7])
#[1] "[42,48.7]"   "(48.7,50.9]" "(50.9,52.8]" "(52.8,54.2]" "(54.2,58.7]"
Do you want to replace this with 1:5?
levels(test2[,8])
#[1] "[76.2,79.2]" "(79.2,80.5]" "(80.5,81.9]" "(81.9,83.5]" "(83.5,85]"  
 as.numeric(test2[,7])
 #[1] 2 5 3 2 3 5 5 4 4 3 3 2 1 3 2 1 1 4 2 5 4 5 3 3 1 1 5 1 4 5 4 5 4 3 1 2 1 4
#[39] 4 5 2 1 2 1 1 5 3 4 3 2 2

A.K.

----- Original Message -----
From: "Muhuri, Pradip (SAMHSA/CBHSQ)" <Pradip.Muhuri at samhsa.hhs.gov>
To: R help <r-help at r-project.org>
Cc: 
Sent: Tuesday, January 8, 2013 10:06 PM
Subject: Re: [R] Applying a user-defined function

Hello List,

Last time, Arun's following solution worked to create 3 new columns (1,3,5).  Now how would I tweak this function to create corresponding (additional) columns (7,8,9) of mode factor (levels = 1,2,3,4,5)?

Thanks for your continued support.

Pradip

####### cut and paste from the reproducible example
CutQuintiles <- function( x) {
  cut (x,quantile (x, (0:5/5)),include.lowest=TRUE)
}

#apply the CutQuintile () on every odd-numbered columns of the "test1" data frame
test1$newcols <- sapply(test1 [, seq (1,6,2)], CutQuintiles)

# name 3 new columns based on the odd-numbered columns
names(test1$newcols) <- paste (names(test1 [, seq (1,6,2)]), "_cat")

###### Reproducible Example

test1 <- read.table (text=
"State,ObtMj_P,ObtMj_SE,ExpPrevMed_P,ExpPrevMed_SE,ParMon_P,ParMon_SE
Alabama,49.60,1.37,80.00,0.91,12.10,0.68
Alaska,55.00,1.41,81.80,1.08,12.40,0.90
Arizona,52.50,1.56,79.60,1.20,15.80,1.08
Arkansas,50.50,1.22,78.00,0.78,12.80,0.72
California,51.10,0.65,80.50,0.53,13.00,0.41
Colorado,55.10,1.26,81.70,1.03,12.10,0.72
Connecticut,56.30,1.28,85.00,0.93,14.60,0.77
Delaware,53.60,1.30,79.50,1.04,14.70,0.97
District of Columbia,53.50,1.22,76.20,1.03,14.30,1.13
Florida,52.70,0.67,78.90,0.52,14.10,0.45
Georgia,52.50,1.15,79.30,1.02,15.90,0.98
Hawaii,49.40,1.33,83.80,1.12,16.00,1.06
Idaho,48.30,1.23,82.40,0.99,11.90,0.74
Illinois,52.70,0.63,81.00,0.46,13.60,0.40
Indiana,49.60,1.16,80.90,0.91,12.60,0.82
Iowa,46.30,1.37,82.10,1.01,13.60,0.87
Kansas,44.30,1.43,79.20,0.98,12.90,0.79
Kentucky,52.90,1.37,78.70,1.05,14.60,0.98
Louisiana,49.70,1.23,76.80,1.06,14.50,0.76
Maine,55.60,1.44,82.90,0.93,16.70,0.83
Maryland,53.90,1.46,83.60,0.95,14.00,0.80
Massachusetts,55.40,1.41,81.00,1.15,14.70,0.80
Michigan,52.40,0.62,80.50,0.47,15.00,0.43
Minnesota,51.50,1.20,84.40,0.87,14.40,0.86
Mississippi,43.20,1.14,76.60,0.91,12.30,0.78
Missouri,48.70,1.20,80.30,0.90,13.70,0.12
Montana,56.40,1.16,83.70,0.95,12.10,0.68
Nebraska,45.70,1.51,83.40,0.95,12.40,0.90
Nevada,54.20,1.17,80.60,1.07,15.80,1.08
New Hampshire,56.10,1.30,83.30,0.93,12.80,0.72
New Jersey,53.20,1.45,83.70,0.95,13.00,0.41
New Mexico,57.60,1.34,78.90,1.03,12.10,0.72
New York,53.70,0.67,82.60,0.48,14.60,0.77
North Carolina,52.20,1.26,81.90,0.84,14.70,0.97
North Dakota,48.60,1.34,84.20,0.88,14.30,1.13
Ohio,50.90,0.61,82.70,0.49,14.10,0.45
Oklahoma,47.20,1.42,78.80,1.33,15.90,0.98
Oregon,54.00,1.35,80.60,1.14,16.00,1.06
Pennsylvania,53.00,0.63,79.90,0.47,11.90,0.74
Rhode Island,57.20,1.20,79.50,1.02,13.60,0.40
South Carolina,50.50,1.21,79.50,0.95,12.60,0.82
South Dakota,43.40,1.30,81.70,1.05,13.60,0.87
Tennessee,48.90,1.35,78.40,1.35,12.90,0.79
Texas,48.70,0.62,79.00,0.48,14.60,0.98
Utah,42.00,1.49,85.00,0.93,14.50,0.76
Vermont,58.70,1.24,83.70,0.84,16.70,0.83
Virginia,51.80,1.18,82.00,1.04,14.00,0.80
Washington,53.50,1.39,84.10,0.96,14.70,0.80
West Virginia,52.80,1.07,79.80,0.93,15.00,0.43
Wisconsin,49.90,1.50,83.50,1.02,14.40,0.86
Wyoming,49.20,1.29,82.00,0.85,12.30,0.78
", sep=",", row.names='State',  header=TRUE, as.is=TRUE)

# change names () to lower case

names (test1) <- tolower (names (test1))

#Write a cut/quantile function to apply on different columns of the data frame

CutQuintiles <- function( x) {
  cut (x,quantile (x, (0:5/5)),include.lowest=TRUE)
}

#apply the CutQuintile () on every odd-numbered columns of the "test1" data frame
test1$newcols <- sapply(test1 [, seq (1,6,2)], CutQuintiles)

# name 3 new columns based on the odd-numbered columns
names(test1$newcols) <- paste (names(test1 [, seq (1,6,2)]), "_cat")

dim (test1)
options (width=100)
test1

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