R: Replacing a factor with an integer value in numerous cells across numerous columns

Question

So, my challenge has been to convert a raw scale csv to a scored csv. Within numerous columns, the file has cells filled with "Strongly Agree" to "Strongly Disagree", 6 levels. These factors need to be converted in integers 5 to 0 respectively.

I have tried unsuccessfully to use sapply and convert the table to a string. Sapply works on the vector, but it destroys the table structure.

Method 1:

dat$Col<-sapply(dat$Col,switch,'Strongly Disagree'=0,'Disagree'=1,'Slightly Disagree'=2,'Slightly Agree'=3,'Agree'=4, 'Strongly Agree'=5)

My second approach is to convert the csv into a string. When I examined the dput output, I saw the area I wanted to target that started with a .Label="","Strongly Agree"... Mistake. My changes did not result in a useful outcome.

My third approach came from the internet gods of destruction who seemed to express that gsub() might handle the string approach as well. Nope, again the underlying table structure was destroyed.

Method #3: Convert into a string and pattern match

dat <- textConnection("control/Surveys/StudyDat_1.csv")
#Score Scales
##"Strongly Agree"= 5
##"Agree"= 4
##"Strongly Disagree" = 0
#levels(dat$Col) <- gsub("Strongly Agree", "5", levels(dat$Col))
    df<- gsub("Strongly Agree", "5",dat)
    dat<-read.csv(textConnection(df),header=TRUE)

In the end, I am wanting to replace ALL "Strongly Agree" to 5 across numerous columns without the consequence of destroying the retrievability of the data.

Maybe I used the wrong search string and you know the resource I need to address this problem. I would rather avoid ALL character vector approaches as that this would require labeling each column if you provide a code response. It will need to go across ALL COLUMNS.

Thanks

Data Sample Problem

    structure(list(last_updated = structure(c(3L, 1L, 7L, 2L, 10L, 6L, 8L, 9L, 7L, 5L, 4L), .Label = c("2016-05-13T12:53:56.704184Z", 
"2016-05-13T12:54:09.273359Z", "2016-05-13T12:54:22.757251Z", 
"2016-05-14T12:44:13.474992Z", "2016-05-14T12:44:31.736469Z", 
"2016-05-16T16:45:10.623410Z", "2016-05-16T16:46:17.881402Z", 
"2016-05-16T16:46:55.122257Z", "2016-05-16T16:47:14.160793Z", 
"2016-05-24T02:26:04.770799Z"), class = "factor"), feedback = c(NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), A = structure(c(NA, 
NA, 2L, NA, 1L, NA, NA, NA, 2L, NA, NA), .Label = c("", "Slightly Disagree"
), class = "factor"), B = structure(c(NA, NA, 2L, NA, 1L, NA, 
NA, NA, 3L, NA, NA), .Label = c("", "Disagree", "Strongly Agree"
), class = "factor"), C = structure(c(NA, NA, 2L, NA, 1L, NA, 
NA, NA, 3L, NA, NA), .Label = c("", "Agree", "Disagree"), class = "factor"), 
    D = structure(c(NA, NA, 2L, NA, 1L, NA, NA, NA, 2L, NA, NA
    ), .Label = c("", "Agree"), class = "factor"), E = structure(c(NA, 
    NA, 2L, NA, 1L, NA, NA, NA, 3L, NA, NA), .Label = c("", "Agree", 
    "Strongly Disagree"), class = "factor")), .Names = c("last_updated", 
"feedback", "A", "B", "C", "D", "E"), class = "data.frame", row.names = c(NA, 
-11L))

Data Sample Solution

df<-dget(structure(list(last_updated = structure(c(3L, 1L, 7L, 2L, 10L,   6L,8L, 9L, 7L, 5L, 4L), .Label = c("2016-05-13T12:53:56.704184Z", 
"2016-05-13T12:54:09.273359Z", "2016-05-13T12:54:22.757251Z", 
"2016-05-14T12:44:13.474992Z", "2016-05-14T12:44:31.736469Z", 
"2016-05-16T16:45:10.623410Z", "2016-05-16T16:46:17.881402Z", 
"2016-05-16T16:46:55.122257Z", "2016-05-16T16:47:14.160793Z", 
"2016-05-24T02:26:04.770799Z"), class = "factor"), feedback = c(NA, 
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), A = c(NA, NA, 2L, NA, 
NA, NA, NA, NA, 2L, NA, NA), B = c(NA, NA, 1L, NA, NA, NA, NA, 
NA, 5L, NA, NA), C = c(NA, NA, 4L, NA, NA, NA, NA, NA, 1L, NA, 
NA), D = c(NA, NA, 4L, NA, NA, NA, NA, NA, 4L, NA, NA), E = c(NA, 
NA, 4L, NA, NA, NA, NA, NA, 0L, NA, NA)), .Names = c("last_updated", 
"feedback", "A", "B", "C", "D", "E"), class = "data.frame", row.names =    c(NA,-11L)))

Answer 1

we can use factor with levels specified

 nm1 <- c('Strongly Disagree', 'Disagree',
     'Slightly Disagree','Slightly Agree','Agree', 'Strongly Agree')

 factor(dat$col, levels = nm1,
         labels = 0:5))

If there are multiple factor columns with the same levels, identify the factor columns ('i1'), loop through it with lapply and specify the levels and labels.

i1 <- sapply(dat, is.factor)
dat[i1] <- lapply(dat[i1], factor, levels = nm1, labels= 0:5)

Update

Using the OP's dput output

dat[-(1:2)] <- lapply(dat[-(1:2)], factor, levels = nm1, labels = 0:5)
dat
#                last_updated feedback    A    B    C    D    E
#1  2016-05-13T12:54:22.757251Z       NA <NA> <NA> <NA> <NA> <NA>
#2  2016-05-13T12:53:56.704184Z       NA <NA> <NA> <NA> <NA> <NA>
#3  2016-05-16T16:46:17.881402Z       NA    2    1    4    4    4
#4  2016-05-13T12:54:09.273359Z       NA <NA> <NA> <NA> <NA> <NA>
#5  2016-05-24T02:26:04.770799Z       NA <NA> <NA> <NA> <NA> <NA>
#6  2016-05-16T16:45:10.623410Z       NA <NA> <NA> <NA> <NA> <NA>
#7  2016-05-16T16:46:55.122257Z       NA <NA> <NA> <NA> <NA> <NA>
#8  2016-05-16T16:47:14.160793Z       NA <NA> <NA> <NA> <NA> <NA>
#9  2016-05-16T16:46:17.881402Z       NA    2    5    1    4    0
#10 2016-05-14T12:44:31.736469Z       NA <NA> <NA> <NA> <NA> <NA>
#11 2016-05-14T12:44:13.474992Z       NA <NA> <NA> <NA> <NA> <NA>

Another option is set from data.table

library(data.table)
for(j in names(dat)[-(1:2)]){
  set(dat, i = NULL, j= j, value = factor(dat[[j]], levels = nm1, labels = 0:5))
 }

Answer 2

Previous answers might meet your needs, but note that changing the labels of a factor isn't the same as changing a factor to an integer variable. One possibility would be to use ifelse (I've made a new data frame as the one you posted didn't actually have variables with these levels in it):

lev <- c('Strongly disagree', 'Disagree', 'Slightly disagree', 'Slightly agree', 'Agree', 'Strongly agree')

dta <- sample(lev, 55, replace = TRUE)
dta <- data.frame(matrix(dta, nrow = 11))
names(dta) <- LETTERS[1:5]

f_to_int <- function(f) {
  if (is.factor(f)){
   ifelse(f == 'Strongly disagree', 0, 
     ifelse(f == 'Disagree', 1, 
       ifelse(f == 'Slightly disagree', 2,`` 
         ifelse(f == 'Slightly agree', 3,
           ifelse(f == 'Agree', 4,
             ifelse(f == 'Strongly agree', 5, f))))))
  } else f
}

dta2 <- sapply(dta, f_to_int)

Note that this returns a matrix, but it is easily converted to a data frame if necessary.

Answer 3

I would just match each target column vector into a precomputed character vector to get an integer index. You can subtract 1 afterward to change the range from 1:6 to 0:5.

## define desired value order, ascending
o <- c(
    'Strongly Disagree',
    'Disagree',
    'Slightly Disagree',
    'Slightly Agree',
    'Agree',
    'Strongly Agree'
);

## convert target columns
for (cn in names(df)[-(1:2)]) df[[cn]] <- match(as.character(df[[cn]]),o)-1L;
df;
##                   last_updated feedback  A  B  C  D  E
## 1  2016-05-13T12:54:22.757251Z       NA NA NA NA NA NA
## 2  2016-05-13T12:53:56.704184Z       NA NA NA NA NA NA
## 3  2016-05-16T16:46:17.881402Z       NA  2  1  4  4  4
## 4  2016-05-13T12:54:09.273359Z       NA NA NA NA NA NA
## 5  2016-05-24T02:26:04.770799Z       NA NA NA NA NA NA
## 6  2016-05-16T16:45:10.623410Z       NA NA NA NA NA NA
## 7  2016-05-16T16:46:55.122257Z       NA NA NA NA NA NA
## 8  2016-05-16T16:47:14.160793Z       NA NA NA NA NA NA
## 9  2016-05-16T16:46:17.881402Z       NA  2  5  1  4  0
## 10 2016-05-14T12:44:31.736469Z       NA NA NA NA NA NA
## 11 2016-05-14T12:44:13.474992Z       NA NA NA NA NA NA