Here is a hint as to how to replace the loops:
> hours <- 0:5
> result <- lapply(hours, function(.hr){
+ cbind(.hr, replicate(5, mean(sample(1:100, 1000, replace=TRUE))))
+ })
> do.call(rbind, result)
.hr
[1,] 0 50.232
[2,] 0 51.135
[3,] 0 49.169
[4,] 0 50.264
[5,] 0 52.621
[6,] 1 51.852
[7,] 1 50.557
[8,] 1 50.501
[9,] 1 49.710
[10,] 1 49.867
[11,] 2 50.536
[12,] 2 51.562
[13,] 2 51.269
[14,] 2 50.638
[15,] 2 51.083
[16,] 3 49.785
[17,] 3 48.843
[18,] 3 51.156
[19,] 3 50.877
[20,] 3 51.012
[21,] 4 52.214
[22,] 4 50.470
[23,] 4 50.388
[24,] 4 49.600
[25,] 4 51.057
[26,] 5 51.202
[27,] 5 51.022
[28,] 5 52.084
[29,] 5 49.944
[30,] 5 49.946
On Wed, Dec 30, 2009 at 12:16 PM, James Rome <jamesr...@gmail.com> wrote:
> Oops. Thank you. I guess I needed mn[i*1000 + j]. I should have known
> better.
>
> But I had wanted to make this into a data frame with
> df=data.frame(mn, il)
> so that I could plot it using factors. Can I do that with a matrix?
> And can I eliminate the loops?
>
>
> On 12/30/09 12:11 PM, jim holtman wrote:
>
> First of all your expression 'i*j' is storing on top of one another. Look
> at what happens in the case of i=1,j=8 i=2,j=4 i=4,j=2, .... These are
> all storing to the same location. Instead make 'mn' & 'li' matrices:
>
> mn <- matrix(length(alist), 1000)
>
> and then within the loop:
>
> mn[i,j] <- value
>
> On Wed, Dec 30, 2009 at 11:52 AM, James Rome <jamesr...@gmail.com> wrote:
>
>> Dear kind list people:
>>
>> I have the following code:
>> >hours
>> [1] "0" "1" "2" "4" "5" "6" "7" "8" "9" "10" "11" "12" "13"
>> "14" "15"
>> [16] "16" "17" "18" "19" "20" "21" "22" "23"
>> > alist
>> $`0`
>> [1] 3 10 10 6 5 6 4 8 9 3 7 5 8 3 6 7 2 6 6 1 4 8
>> 10 4 10
>> [26] 13 6 2 8 4 7 3 4 7 9 6 4 7 4 4 4 3
>>
>> $`1`
>> [1] 1 1 3 2 3 4 2 1
>>
>> $`2`
>> [1] 1 1 3 3
>> . . .
>> mn=c(length(alist)*1000)
>> il=c(length(alist)*1000)
>> # Now calculate the means
>> for(i in 1:length(alist)) {
>> for(j in 1:1000) {mn[i*j]=mean(sample(alist[[i]], 1000, replace=TRUE));
>> il[i*j]= hours[i]}}
>>
>> But not even the il vector is correct:
>> >il
>> [1] "0" "1" "2" "4" "5" "6" "7" "8" "9" "10" "11" "12" "13"
>> "14"
>> [15] "15" "16" "17" "18" "19" "20" "21" "22" "23" "12" "5" "13" "9"
>> "14"
>> [29] "0" "15" "0" "16" "11" "17" "7" "18" "0" "19" "13" "20" "0"
>> "21"
>> [43] "0" "22" "15" "23" "0" "16" "7" "10" "17" "13" "0" "18" "11"
>> "14"
>> [57] "19" "1" "0" "20" "0" "1" "21" "16" "13" "22" "0" "17" "23"
>> "14"
>> [71] "0" "18" "0" "1" "15" "19" "11" "13" "0" "20" "9" "1" "0"
>> "21"
>> . . .
>> and after a while, both mn and il get lots of NAs. The first 1000
>> entries should be "0".
>>
>> What am I doing wrong?
>>
>> And is there a way to do this without the two for loops?
>>
>> Thanks,
>> Jim Rome
>>
>> ______________________________________________
>> 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<http://www.r-project.org/posting-guide.html>
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
>
> --
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem that you are trying to solve?
>
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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