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Beginner Question: List value without Levels

4 messages · dabs, Joshua Wiley

#
Hey folks,

I'm new to the R Project so I'm facing a great problem. I've read a file
into R:
I   V   L   F   C   M   A    G    T    W    S    Y    P    H    Q    D   
N    E    K    R
1 4,5 4,2 3,8 2,8 2,5 1,9 1,8 -0,4 -0,7 -0,9 -0,8 -1,3 -1,6 -3,2 -3,5 -3,5
-3,5 -3,5 -3,9 -4,5
[1] list


Now I want to multiplicate each of this values with this one:
A  C  D  E  F  G  H  I  K  L  M  N  P  Q  R  S  T  V  W  Y 
28  8 11 14 17 34  7 26 15 26 10  9 12  8 11 21 19 33  7  7
[1] "numeric"


Because I don't know how to do that. I had in mind to write a function, in
which every single value is multiplicated with the other one. So I tried to
test, if I could multiplicate two single values. This is the result:
[1] NA
Warning:
In Ops.factor(myVal[1, 1], data[1]) :
  * nicht sinnvoll f?r Faktoren


(I'm using R with German language :))

What can I do, to multplicate each value with his adequate?

Thanks a lot :)!

dabs


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#
Hi Dabs,

If those are your two objects, could you post the output from:

dput(myVal)
dput(data)

The output of dput() can be copied and pasted into the console and
then we have the data on our systems exactly as you do on yours.  From
there it should be easy to show you how to do what you want.

Cheers!

Josh
On Tue, May 24, 2011 at 11:54 AM, dabs <ragon016 at web.de> wrote:

  
    
#
Hey Joshua, 

thanks for your help. Here are the dput outputs :)!
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
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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#
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: