[R] sorting a data.frame (df) by a vector (which is not contained in the df) - unexpected behaviour of match and factor

[Berend Hasselman]

drflxms wrote
> 
> Dear R colleagues,
> 
> consider my data.frame named "df" with 3 columns - being level,
> prevalence and sensitivity - and 7 rows of data (see dump below).
> 
> df <-
> structure(list(level = structure(1:7, .Label = c("0", "1", "10",
> "100", "1010", "11", "110"), class = "factor"), prevalence =
> structure(c(4L,
> 2L, 3L, 5L, 6L, 1L, 7L), .Label = c("0.488", "0.5", "0.754",
> "0.788", "0.803", "0.887", "0.905"), class = "factor"), sensitivity =
> structure(c(6L,
> 1L, 5L, 4L, 3L, 2L, 1L), .Label = c("0", "0.05", "0.091", "0.123",
> "0.327", "0.933"), class = "factor")), .Names = c("level", "prevalence",
> "sensitivity"), class = "data.frame", row.names = c(NA, -7L))
> 
> I'd like to order df by a vector which is NOT contained in the
> data.frame. Let's call this vector desiredOrder (see dump below).
> 
> desiredOrder <- c("0", "1", "10", "100", "11", "110", "1010")
> 
> So after sorting, the order of the level column (df$level) should be in
> the order of the vector desiredOrder (as well a the associated data in
> the other columns).
> I know that this is not an easy task to achieve by order(...) as the
> order of desiredOrder isn't a natural one. But I would expect both of
> the following to work:
> 
> ## using match
> df[match(df$level,desiredOrder),]
> 
> ## using factor
> df[factor(df$level,levels=desiredOrder),]
> 
> Unfortunately the result isn't what I expected: I get a data.frame with
> the level column in the order 0,1,10,100,110,1010,11 instead of the
> order in desiredOrder (0,1,10,100,11,110,1010).
> 
> Does anybody see, what I am doing wrong?
> 

Try this:

df[match(desiredOrder,df$level),]

Berend


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