Hi Arun,
Thanks a lot for your script. Ill work on it tomorrow.
Cheers
On 24/04/2014, at 11:54 AM, "arun kirshna [via R]"
<ml-node+s789695n4689356...@n4.nabble.com> wrote:
> HI,
> I guess you got an output like this using my script:
> ##Please use ?dput() to show the example data.
>
> FA <- structure(list(Sample = c("L1 Control", "L1 Control", "L1 Control",
> "BBM Control", "BBM Control", "BBM Control", "L1 Ash", "L1 Ash",
> "L1 Ash", "BBM Ash", "BBM Ash", "BBM Ash"), C14.0 = c(0.456509192,
> 0.513989684, 0.555894496, 0.418392781, 0.405826292, 0.398633968,
> 0.504528078, 0.548667997, 0.499237645, 0.380582244, 0.395617943,
> 0.389027115), C15.0 = c(0.469562687, 0.527958026, 0.502389699,
> 0.385119329, 0.368564514, 0.391851493, 0.479125577, 0.517533922,
> 0.490619858, 0.380051535, 0.384498216, 0.370815474), C15.1 = c(0.774909216,
> 0.732083085, 0.706407924, 1.318261983, 1.114889958, 1.238411437,
> 0.793236101, 0.632962545, 0.74858627, 0.996870831, 0.963780759,
> 0.923329859)), .Names = c("Sample", "C14.0", "C15.0", "C15.1"
> ), class = "data.frame", row.names = c(NA, -12L))
>
> library(gvlma)
> y <- names(FA)[-1]
> y
> #[1] "C14.0" "C15.0" "C15.1"
>
> lst1 <- setNames(vector("list", length(y)),y)
>
> for(i in y){
> lst1[[i]] <- gvlma(lm(get(i)~Sample,data=FA))
> lst1}
>
> lst1[[1]]
> #
> #Call:
> #lm(formula = get(y[i]) ~ Sample, data = FA)
> #---------------------------------------
>
> But, you wanted to show each of the list output as in gvlmaFA.
>
> gvlmaFA <- gvlma(lm(C14.0~Sample,data=FA))
>
>
> In my previous script, I didn't name the list. Here, by setting the names as
> in "y", it could be easier. I guess you wanted to reflect that in the model
> formula as well.
>
> lst2 <- setNames(vector("list", length(y)), y)
> for(names in y){
> lst2[[names]] <- eval(bquote(gvlma(lm(.(names1)~ Sample, data=FA)),
> list(names1=as.name(names))))
> lst2}
>
> identical(gvlmaFA, lst2[[1]])
> #[1] TRUE
>
> A.K.
>
>
>
>
>
>
> Hi Arun,
> Your script works but it does not do what I was after. To be a bit more
> specific, this the table FA in which Im working on ( but the original one
> has 34 fatty acids instead of 3: C14.0, C15.0, and C15.1).
>
> Sample
>
>
> C14:0
>
>
> C15:0
>
>
> C15:1
>
> L1 Control
>
>
> 0.456509192
>
>
> 0.469562687
>
>
> 0.774909216
>
> L1 Control
>
>
> 0.513989684
>
>
> 0.527958026
>
>
> 0.732083085
>
> L1 Control
>
>
> 0.555894496
>
>
> 0.502389699
>
>
> 0.706407924
>
> BBM Control
>
>
> 0.418392781
>
>
> 0.385119329
>
>
> 1.318261983
>
> BBM Control
>
>
> 0.405826292
>
>
> 0.368564514
>
>
> 1.114889958
>
> BBM Control
>
>
> 0.398633968
>
>
> 0.391851493
>
>
> 1.238411437
>
> L1 Ash
>
>
> 0.504528078
>
>
> 0.479125577
>
>
> 0.793236101
>
> L1 Ash
>
>
> 0.548667997
>
>
> 0.517533922
>
>
> 0.632962545
>
> L1 Ash
>
>
> 0.499237645
>
>
> 0.490619858
>
>
> 0.74858627
>
> BBM Ash
>
>
> 0.380582244
>
>
> 0.380051535
>
>
> 0.996870831
>
> BBM Ash
>
>
> 0.395617943
>
>
> 0.384498216
>
>
> 0.963780759
>
> BBM Ash
>
>
> 0.389027115
>
>
> 0.370815474
>
>
> 0.923329859
>
> I just want to run the following script but with C15.0, C15.1 and the other
> 32 so I can quickly scroll up and down to see who does not meet the
> assumptions.
>
> FA.ml=lm(C14.0~Sample,data=FA)
> gvlmaFA<-gvlma(FA.ml)
> gvlmaFA
>
> This is the result when I run the script
>
> Call:
> lm(formula = C14.0 ~ Sample, data = FA)
>
> Coefficients:
> (Intercept) SampleBBM Control SampleL1 Ash SampleL1 Control
> 0.38841 0.01921 0.12907 0.12039
>
>
> ASSESSMENT OF THE LINEAR MODEL ASSUMPTIONS
> USING THE GLOBAL TEST ON 4 DEGREES-OF-FREEDOM:
> Level of Significance = 0.05
>
> Call:
> gvlma(x = FA.ml)
>
> Value p-value Decision
> Global Stat 4.757e+00 0.31312 Assumptions acceptable.
> Skewness 1.944e-02 0.88911 Assumptions acceptable.
> Kurtosis 1.462e-01 0.70219 Assumptions acceptable.
> Link Function 3.682e-16 1.00000 Assumptions acceptable.
> Heteroscedasticity 4.592e+00 0.03213 Assumptions NOT satisfied!
>
> I really appreciate if you can help me with this issue. This would be really
> useful for me since I have large tables of data.
> Cheers
>
>
> On Monday, April 21, 2014 9:19 AM, arun <[hidden email]> wrote:
> Hi,
>
> Using the example data from library(gvlma)
>
> library(gvlma)
> data(CarMileageData)
> CarMileageNew <- CarMileageData[,c(5,6,3)]
> lst1 <- list()
> y <- c("NumGallons", "NumDaysBetw")
> for(i in seq_along(y)){
> lst1[[i]] <- gvlma(lm(get(y[i])~MilesLastFill,data=CarMileageNew))
> lst1}
> pdf("gvlmaplot.pdf")
> lapply(lst1,plot)
> dev.off()
>
>
> You could also use ?lapply().
>
>
> A.K.
>
>
>
> Hi
> I have a spread sheet with a column Samples (column1) and then 34 more
> columns with different concentrations of fatty acids per sample. Im trying to
> run the same function 34 times. In this case (the first of 34), I have a
> fatty acid called C14.0 (column 2). I'm a newbie with R so I spent the last 4
> days looking for a way of doing it (without running the same function 34
> times with a different fatty acid each time). I saw that people do similar
> things with loops but I cannot get them to work.
> I have tried the script below but it does not work.
>
> y<-c("C14.0","C15.0","C16.0")
> for (i in y) {
> FA.ml=lm(i~Sample,data=FA)
> gvlmaFA<-gvlma(FA.ml)
> gvlmaFA
> }
>
>
> I really appreciate if someone can give me a hand with that. I know would
> have been finished if I had typed the 34 fatty acids but I want to learn how
> to do it with loops.
> Cheers
>
> ______________________________________________
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