Levels function returning NULL

You can only run levels on a factor vector, not on a data frame.

Example below

> df <- data.frame(a = factor(c('a','b','c'), levels = c('a','b','c','d','e')),
+                  b = factor(c('a','b','c')), 
+                  c = factor(c('a','a','c')))
> levels(df)
NULL

To see the level of every column in your data frame, you can use lapply

> lapply(df, levels)
$a
[1] "a" "b" "c" "d" "e"

$b
[1] "a" "b" "c"

$c
[1] "a" "c"

If you want the levels of a specific column, you can specify that instead:

> levels(df[, 2])
[1] "a" "b" "c"

---

EDIT: To answer question below on why apply(df, 2, levels) returns NULL.

Note the following from the documentation for apply():

In all cases the result is coerced by as.vector to one of the basic vector types before the dimensions are set, so that (for example) factor results will be coerced to a character array.

You can see this behavior when you try to take the class, and try a few other functions.

> apply(df, 2, levels)
NULL
> apply(df, 2, class)
          a           b           c 
"character" "character" "character" 
> apply(df, 2, function(i) levels(i))
NULL
> apply(df, 2, function(i) levels(factor(i)))
$`a`
[1] "a" "b" "c"

$b
[1] "a" "b" "c"

$c
[1] "a" "c"

Note that even though we can force apply() to treat the columns as factors, we lose the prior ordering/levels that were set for df when it was originally created (see column `a`). This is because it has been coerced into a character vector.

When initializing a dataframe, pass stringsAsFactors = T in the initialization

eg. dataFrame <- read.csv(file.choose(), stringsAsFactors=T)

this makes R treat the string values as factors. Hope it helped