Hi
> Dear expeRts,
>
> I have two questions concerning data frames:
> (1) How can I apply the class function to each component in a
data.frame?
> As you can see below, applying class to each column is not the right
> approach; applying it to each component seems bulky.
> (2) After transforming the data frame a bit, the classes of certain
> components change to factor. How can I remove the factor structure?
>
> Cheers,
>
> Marius
>
> x <- c(2004:2010, 2002:2011, 2000:2011)
> df <- data.frame(x=x, group=c(rep("low",7), rep("middle",10),
rep("high",12)),
> y=x+100*runif(length(x)))
>
> ## Question (1): why do the following lines do not give the same
"class"?
from help page
?apply
Arguments
X
an array, including a matrix.
array is not a data frame
> apply(df, 2, class)
> class(df$x)
> class(df$group)
> class(df$y)
sapply(df, class)
x group y
"integer" "factor" "numeric"
>
> df. <- as.data.frame(xtabs(y ~ x + group, data=df))
>
> class(df.$x)
> class(df.$group)
> class(df.$Freq)
>
> ## Question (2): how can I remove the factor structure from x?
> df.$x <- as.numeric(as.character(df.$x)) # seems bulky; note that
> as.numeric(df.$x) is not correct
Actually it is correct in a sense it behaves as documented
?factor
Warning
The interpretation of a factor depends on both the codes and the "levels"
attribute. Be careful only to compare factors with the same set of levels
(in the same order). In particular, as.numeric applied to a factor is
meaningless, and may happen by implicit coercion. To transform a factor f
to approximately its original numeric values, as.numeric(levels(f))[f] is
recommended and slightly more efficient than as.numeric(as.character(f)).
Regards
Petr
> class(df.$x)
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