# Re: [R] Beginner Question: List value without Levels

From: Joshua Wiley <jwiley.psych_at_gmail.com>
Date: Tue, 24 May 2011 14:26:00 -0700

Hi dabs,

Thanks for that. Part of your problem is that the data in myVals is not being stored as numbers, it is stored as factors. If "," is your decimal separator, then something like:

myVal2 <- unlist(myVal)
myVal2 <- as.numeric(levels(myVal2))[myVal2]

should convert it to numeric class data, at which point, all you need to do is:

myVal2 * data

and because R is vectorized, each element of "myVal2" will be multiplied by each element of "data". If "," is not a decimal separator and indicates coordinates or something, then you will need to handle it a little differently.

Cheers,

Josh

On Tue, May 24, 2011 at 12:31 PM, dabs <ragon016_at_web.de> wrote:
> Hey Joshua,
>
> thanks for your help. Here are the dput outputs :)!
>
>>dput(myVal)
> structure(list(I = structure(1L, .Label = "4,5", class = "factor"),
>    V = structure(1L, .Label = "4,2", class = "factor"), L = structure(1L,
> .Label = "3,8", class = "factor"),
>    F = structure(1L, .Label = "2,8", class = "factor"), C = structure(1L,
> .Label = "2,5", class = "factor"),
>    M = structure(1L, .Label = "1,9", class = "factor"), A = structure(1L,
> .Label = "1,8", class = "factor"),
>    G = structure(1L, .Label = "-0,4", class = "factor"), T = structure(1L,
> .Label = "-0,7", class = "factor"),
>    W = structure(1L, .Label = "-0,9", class = "factor"), S = structure(1L,
> .Label = "-0,8", class = "factor"),
>    Y = structure(1L, .Label = "-1,3", class = "factor"), P = structure(1L,
> .Label = "-1,6", class = "factor"),
>    H = structure(1L, .Label = "-3,2", class = "factor"), Q = structure(1L,
> .Label = "-3,5", class = "factor"),
>    D = structure(1L, .Label = "-3,5", class = "factor"), N = structure(1L,
> .Label = "-3,5", class = "factor"),
>    E = structure(1L, .Label = "-3,5", class = "factor"), K = structure(1L,
> .Label = "-3,9", class = "factor"),
>    R = structure(1L, .Label = "-4,5", class = "factor")), .Names = c("I",
> "V", "L", "F", "C", "M", "A", "G", "T", "W", "S", "Y", "P", "H",
> "Q", "D", "N", "E", "K", "R"), class = "data.frame", row.names = c(NA,
> -1L))
>
>
> And
>
>
>>dput(data)
> structure(c(28L, 8L, 11L, 14L, 17L, 34L, 7L, 26L, 15L, 26L, 10L,
> 9L, 12L, 8L, 11L, 21L, 19L, 33L, 7L, 7L), .Dim = 20L, .Dimnames =
> structure(list(
>    data= c("A", "C", "D", "E", "F", "G", "H", "I", "K",
>    "L", "M", "N", "P", "Q", "R", "S", "T", "V", "W", "Y")), .Names =
> "data"), class = "table")
>
>
> Cheers!
>
> dabs
>
>
> --
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> Sent from the R help mailing list archive at Nabble.com.
>
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> R-help_at_r-project.org mailing list
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> and provide commented, minimal, self-contained, reproducible code.
>

```--
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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