Apply factor levels to multiple columns with missing factor levels

问题

I have a dataframe with many factors and want to create statistical tables that show the distribution for each factor, including factor levels with zero observations. For instance, these data:

structure(list(engag11 = structure(c(5L, 4L, 4L), .Label = c("Strongly Disagree", "Disagree", "Neither A or D", "Agree", "Strongly Agree"), class = "factor"), encor11 = structure(c(1L, 1L, 1L), .Label = c("Agree", "Neither Agree or Disagree", "Strongly Agree"), class = "factor"), know11 = structure(c(3L, 
1L, 1L), .Label = c("Agree", "Neither Agree or Disagree", "Strongly Agree"), class = "factor")), .Names = c("engag11", "encor11", "know11"), row.names = c(NA, 3L), class = "data.frame")

show 6 rows, but only some of the factor levels are observed for each column. When I produce a table, I'd like to display not only counts for the levels observed, but also levels NOT observed (such as "Strongly Disagree"). Like this:

# define the factor and levels
library(dplyr);library(pander);library(forcats)
eLevels<-factor(c(1,2,3,4,5), levels=1:5, labels=c("Strongly    Disagree","Disagree","Neither A or D","Agree","Strongly Agree"),ordered =TRUE )

# apply the factor to one variable
csc2$engag11<-factor(csc2$engag11,eLevels)

t1<-table(csc2$engag11)
pander(t1)

Which results in a frequency table that shows counts for each level, including zeroes for levels not reported / observed.

But I have dozens of variables to convert. A simple lapply function recommended on Stackoverflow doesn't seem to work, such as this one:

csc2[1:3]<-lapply(csc[1:3],eLevels)

I also tried a simple function (n=list of columns) for this, but failed:

facConv<-function(df,n)
{   df$n<-factor(c(1,2,3,4,5), levels=1:5, labels=c("Strongly 
Disagree","Disagree","Neither A or D","Agree","Strongly Agree") )
return(result)   }

Can someone offer a solution?

回答1:

An lapply should work fine, you just need to specify the factor() function:

csc2[1:3] <- lapply(csc2[1:3], function(x) factor(x, eLevels))

Then you can call table like:

table(csc2[1])

#Strongly    Disagree             Disagree       Neither A or D                Agree       Strongly Agree 
#                   0                    0                    0                    2                    1 
table(csc2[2])

#Strongly    Disagree             Disagree       Neither A or D                Agree       Strongly Agree 
#                   0                    0                    0                    3                    0

回答2:

The inelegant quick and dirty way is to use for loop:

df <- data.frame(A = c("A", "A", "B"),
                 B = c("A", "C", "A"),
                 C = c("A", "A", "D"))
lvl <- c("A", "B", "C", "D", "E")

for (i in 1:ncol(df)) {
  df[,i] <- factor(df[,i], levels=lvl)
}

table(df$A)

And if your original data is numbers then:

df <- data.frame(A = c(1,1,2),
                 B = c(1,3,1),
                 C = c(1,1,4))
lvl <- c("A", "B", "C", "D", "E")
for (i in 1:ncol(df)) {
  df[,i] <- factor(df[,i], levels=1:5, labels=lvl)
}
df
table(df$A)

来源：https://stackoverflow.com/questions/48196217/apply-factor-levels-to-multiple-columns-with-missing-factor-levels

标签

multiple-columns

lapply