Your desiredOrder vector is a vector of strings. Convert it to numeric and it
should work.
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drflxms <drfl...@googlemail.com> 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?
>I'd appreciate any kind of help very much!
>Best regards, Felix
>
>______________________________________________
>R-help@r-project.org mailing list
>https://stat.ethz.ch/mailman/listinfo/r-help
>PLEASE do read the posting guide
>http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.
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