[R] help please: put output into dataframe

jim holtman jholtman at gmail.com
Fri Mar 18 17:20:23 CET 2011


The easiest thing is to use 'save' so that you write the object out as
binary.  If you don't need 'text', then save/load is the way to
operate with the data.

On Fri, Mar 18, 2011 at 10:53 AM, Ram H. Sharma <sharma.ram.h at gmail.com> wrote:
> Thanks, Jim for the idea.
>
> I tried with save as list. I can not write to a table with "write.table", I
> could not find a function that is write.list or equivalent. Even if it is
> list I think it would be difficult to post-processing than as table.
>
> outx<- as.list(apply(datafr1, 2, fout))
> write.table (outx, "outlier.csv", sep=",")
>
> Ram
>
>
> On Fri, Mar 18, 2011 at 10:04 AM, jim holtman <jholtman at gmail.com> wrote:
>>
>> I think it was suggested that you save your output to a 'list' and
>> then you will have it in a format that can accept variable numbers of
>> items in each element and it is also in a form that you can easily
>> process it to create whatever other output you might need.
>>
>> On Fri, Mar 18, 2011 at 7:24 AM, Ram H. Sharma <sharma.ram.h at gmail.com>
>> wrote:
>> > Hi Dennis and R-users
>> >
>> > Thank you for more help. I am pretty close, but challenge still remain
>> > is
>> > forcing the output with different length to output dataframe.
>> >
>> >> x <- data.frame(apply(datafr1, 2, fout))
>> > Error in data.frame(var1 = c(-0.70777998321315, 0.418602152926712,
>> > 2.08356737154810,  :
>> >  arguments imply differing number of rows: 28, 12, 20, 19
>> >
>> > As I need to work with >2000 variables, my intension here is to save
>> > this
>> > output to such way that it would be further manipulated. Topline is to
>> > save
>> > in dataframe that have extreme values for the variable concerned and
>> > bottomline is automate to save the output printed in the screen to a
>> > textfile.
>> >
>> > Thank you for help once again.
>> >
>> > Ram
>> >
>> >
>> > On Fri, Mar 18, 2011 at 3:16 AM, Dennis Murphy <djmuser at gmail.com>
>> > wrote:
>> >
>> >> Hi:
>> >>
>> >> Is this what you're after?
>> >>
>> >> fout <- function(x) {
>> >>      lim <- median(x) + c(-2, 2) * mad(x)
>> >>      x[x < lim[1] | x > lim[2]]
>> >>    }
>> >> > apply(datafr1, 2, fout)
>> >> $var1
>> >>  [1] 17.5462078 18.4548214  0.7083442  1.9207578 -1.2296787 17.4948240
>> >>  [7] 19.5702558  1.6181150 20.9791652 -1.3542099  1.8215087 -1.0296303
>> >> [13] 20.5237930 17.5366497 18.5657566  0.9335419 19.7519983 17.8607968
>> >> [19] 19.1307524 19.6145711 21.8037136 19.1532175 -2.6688409 19.6949309
>> >> [25] 1.9712347
>> >>
>> >> $var2
>> >>  [1]  37.3822087  35.6490641  35.6000785  38.5981086  -1.6504275
>> >> 37.1419290
>> >>  [7]  37.7605230  40.3508689   0.6639900   2.4695841  38.8209491
>> >> 39.9087921
>> >> [13]  38.9907585  35.8279437   2.7870799  37.0941113   0.6308583
>> >> 36.4556638
>> >> [19] -10.2384849   2.8480199  -7.7680457  35.7076539  -0.5467739
>> >> 3.4702765
>> >> [25]  40.4818580   3.2864273   1.4917174
>> >>
>> >> $var3
>> >>  [1]  74.252563  68.396391  68.845461  -5.006545  66.083402  76.036577
>> >>  [7]  75.112586  -6.374241  63.883549  64.041216 -19.764360 -15.051017
>> >> [13]  -9.782767  64.696013  70.970648  -4.562031 -22.135003  70.549310
>> >> [19]  69.495915  -4.095587  86.612375  87.029526  70.072126  -6.421695
>> >> [25] 65.737536
>> >>
>> >> $var4
>> >>  [1]  81.476483  87.098767 -10.451616  91.927329  86.588952  85.080950
>> >>  [7]  84.958645  -9.456368  86.270876 -22.936779  83.314032
>> >>
>> >> Double checks:
>> >> > apply(datafr1, 2, function(x) median(x) + c(-2, 2) * mad(x))
>> >>          var1      var2      var3      var4
>> >> [1,]  2.12167  3.779415 -3.736066 -3.471752
>> >> [2,] 17.37176 34.929800 62.969733 80.224799
>> >> > apply(datafr1, 2, range)
>> >>           var1      var2      var3      var4
>> >> [1,] -2.668841 -10.23848 -22.13500 -22.93678
>> >> [2,] 21.803714  40.48186  87.02953  91.92733
>> >>
>> >> Assuming you wanted to do this columnwise (by variable), it appears to
>> >> be
>> >> doing the right thing.
>> >>
>> >> HTH,
>> >> Dennis
>> >>
>> >>
>> >> On Thu, Mar 17, 2011 at 7:04 PM, Ram H. Sharma
>> >> <sharma.ram.h at gmail.com>wrote:
>> >>
>> >>> Dear R community members
>> >>>
>> >>> I have been struggling on this simple question, but never get
>> >>> appropriate
>> >>> solution. So please help.
>> >>>
>> >>>  # my data, though I have a large number of variables
>> >>> var1 <- rnorm(500, 10,4)
>> >>> var2 <- rnorm(500, 20, 8)
>> >>> var3 <- rnorm(500, 30, 18)
>> >>> var4 <- rnorm(500, 40, 20)
>> >>> datafr1 <- data.frame(var1, var2, var3, var4)
>> >>>
>> >>> # my unsuccessful codes
>> >>>  nvar <- ncol(datafr1)
>> >>> for (i in 1:nvar) {
>> >>>              out1 <- NULL
>> >>>              out2 <- NULL
>> >>>              medianx <- median(getdata[,i], na.rm = TRUE)
>> >>>              show(madx <- mad(getdata[,i], na.rm = TRUE))
>> >>>              MD1 <- c(medianx + 2*madx)
>> >>>              MD2 <- c(medianx - 2*madx)
>> >>>              out1[i] <- which(getdata[,i] > MD1) # store data that are
>> >>> greater than median + 2 mad
>> >>>              out2[i] <- which (getdata[,1] < MD2) # store data that
>> >>> are
>> >>> greater than median - 2 mad
>> >>>             resultdf <- data.frame(out1, out2)
>> >>>             write.table (resultdf, "out.csv", sep=",")
>> >>>              }
>> >>>
>> >>>
>> >>> My idea here is to store those value which are either greater than
>> >>> median
>> >>> +
>> >>> 2 *MAD or less than median - 2*MAD. Each variable have different
>> >>> length of
>> >>> output.
>> >>>
>> >>> The following last error message:
>> >>> Error in data.frame(out1, out2) :
>> >>>  arguments imply differing number of rows: 2, 0
>> >>> In addition: Warning messages:
>> >>> 1: In out1[i] <- which(getdata[, i] > MD1) :
>> >>>  number of items to replace is not a multiple of replacement length
>> >>> 2: In out2[i] <- which(getdata[, 1] < MD2) :
>> >>>  number of items to replace is not a multiple of replacement length
>> >>> 3: In out1[i] <- which(getdata[, i] > MD1) :
>> >>>  number of items to replace is not a multiple of replacement length
>> >>>
>> >>> Thank you in advance for helping me.
>> >>>
>> >>> Best regards;
>> >>> RHS
>> >>>
>> >>>        [[alternative HTML version deleted]]
>> >>>
>> >>> ______________________________________________
>> >>> R-help at 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.
>> >>>
>> >>
>> >>
>> >
>> >        [[alternative HTML version deleted]]
>> >
>> > ______________________________________________
>> > R-help at 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.
>> >
>>
>>
>>
>> --
>> Jim Holtman
>> Data Munger Guru
>>
>> What is the problem that you are trying to solve?
>
>



-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?



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