I think I had an extra '[', ']'.
My mistake on this one.
Thank you.
Kevin
---- [EMAIL PROTECTED] wrote:
> With my data I get
>
> sc[["(Unknown).{Unknown)"]]
>
> returns NULL
>
> but sc[[1]] returns
>
> DayOfYear Quantity Fraction Category SubCategory
> 1 1 82 3.903927e-05 (Unknown) (Unknown)
> 2 2 78 3.713492e-05 (Unknown) (Unknown)
> 3 3 112 5.332193e-05 (Unknown) (Unknown)
> .....
>
> So there are Categories and Sub-Categories so this shouldn't be NULL. Do I
> need to escape something?
>
> Thanks again.
>
> Kevin
> ---- jim holtman <[EMAIL PROTECTED]> wrote:
> > On Sun, Jul 13, 2008 at 5:45 PM, <[EMAIL PROTECTED]> wrote:
> > > Thank you I will try drop=TRUE.
> > >
> > > In the mean time do you know how I can access the members (for lack of a
> > > better term) of the results of a split? In the sample you provided below
> > > you have:
> > >
> > > z <- split(x, list(x$cat, x$a), drop=TRUE)
> >
> > You can do 'str(z)' to see the structure of 'z'. In most cases, you
> > should be able to reference by the keys, if they exist:
> >
> > > n <- 20
> > > set.seed(1)
> > > x <- data.frame(a=sample(LETTERS[1:2], n,TRUE), b=sample(letters[1:4], n,
> > > TRUE), val=runif(n))
> > > z <- split(x, list(x$a, x$b), drop=TRUE)
> > > str(z)
> > List of 8
> > $ A.a:'data.frame': 2 obs. of 3 variables:
> > ..$ a : Factor w/ 2 levels "A","B": 1 1
> > ..$ b : Factor w/ 4 levels "a","b","c","d": 1 1
> > ..$ val: num [1:2] 0.647 0.245
> > $ B.a:'data.frame': 3 obs. of 3 variables:
> > ..$ a : Factor w/ 2 levels "A","B": 2 2 2
> > ..$ b : Factor w/ 4 levels "a","b","c","d": 1 1 1
> > ..$ val: num [1:3] 0.5530 0.0233 0.5186
> > $ A.b:'data.frame': 3 obs. of 3 variables:
> > ..$ a : Factor w/ 2 levels "A","B": 1 1 1
> > ..$ b : Factor w/ 4 levels "a","b","c","d": 2 2 2
> > ..$ val: num [1:3] 0.530 0.693 0.478
> > $ B.b:'data.frame': 4 obs. of 3 variables:
> > ..$ a : Factor w/ 2 levels "A","B": 2 2 2 2
> > ..$ b : Factor w/ 4 levels "a","b","c","d": 2 2 2 2
> > ..$ val: num [1:4] 0.789 0.477 0.438 0.407
> > $ A.c:'data.frame': 3 obs. of 3 variables:
> > ..$ a : Factor w/ 2 levels "A","B": 1 1 1
> > ..$ b : Factor w/ 4 levels "a","b","c","d": 3 3 3
> > ..$ val: num [1:3] 0.8612 0.0995 0.6620
> > $ B.c:'data.frame': 1 obs. of 3 variables:
> > ..$ a : Factor w/ 2 levels "A","B": 2
> > ..$ b : Factor w/ 4 levels "a","b","c","d": 3
> > ..$ val: num 0.783
> > $ A.d:'data.frame': 1 obs. of 3 variables:
> > ..$ a : Factor w/ 2 levels "A","B": 1
> > ..$ b : Factor w/ 4 levels "a","b","c","d": 4
> > ..$ val: num 0.821
> > $ B.d:'data.frame': 3 obs. of 3 variables:
> > ..$ a : Factor w/ 2 levels "A","B": 2 2 2
> > ..$ b : Factor w/ 4 levels "a","b","c","d": 4 4 4
> > ..$ val: num [1:3] 0.7323 0.0707 0.3163
> >
> > Here are some examples of accessing the data:
> >
> > > z$B.d
> > a b val
> > 9 B d 0.73231374
> > 15 B d 0.07067905
> > 17 B d 0.31627171
> > > # or just the value (it is a vector)
> > > z$B.d$val
> > [1] 0.73231374 0.07067905 0.31627171
> > > # or by name
> > > z[["B.d"]]$val
> > [1] 0.73231374 0.07067905 0.31627171
> > > # or by absolute number
> > > z[[8]]$val
> > [1] 0.73231374 0.07067905 0.31627171
> > > # take the mean
> > > mean(z$B.d$val)
> > [1] 0.3730882
> > > # get the length
> > > length(z$B.d$val)
> > [1] 3
> > >
> >
> >
> >
> > >
> > > Now I can print out 'z[1], z[2] etc' This is nice but what if I want the
> > > access/iterate through all of the members of a particular column in z.
> > > You have given some methods like z[[1]]$b to access the specific columns
> > > in z. I notice for your example z[[1]]$b prints out two values. Can I
> > > assume that z[[1]]$b is a vecotr? So if I want to find the mean i can
> > > 'mean(z[[1]]$b)' and it will give me the mean value of the b columns in
> > > z? (similarily sum, and range, etc.). Does nrows(z[[1]]$b) return two in
> > > your example below? I would like to find out how many elements are in
> > > z[1]. Or would it be just as fast to do 'nrows(z[1])'?
> > >
> > > Thank you for this extended session on data frames, matrices, and
> > > vectors. I feel much more comfortable with the concepts now.
> > >
> > > Kevin
> > > ---- jim holtman <[EMAIL PROTECTED]> wrote:
> > >> The reason for the empty levels was I did not put drop=TRUE on the
> > >> split to remove unused levels. Here is the revised script:
> > >>
> > >> > set.seed(1) # start with a known number
> > >> > x <-
> > >> > data.frame(cat=sample(LETTERS[1:3],20,TRUE),a=sample(letters[1:4], 20,
> > >> > TRUE), b=runif(20))
> > >> > x
> > >> cat a b
> > >> 1 A d 0.82094629
> > >> 2 B a 0.64706019
> > >> 3 B c 0.78293276
> > >> 4 C a 0.55303631
> > >> 5 A b 0.52971958
> > >> 6 C b 0.78935623
> > >> 7 C a 0.02333120
> > >> 8 B b 0.47723007
> > >> 9 B d 0.73231374
> > >> 10 A b 0.69273156
> > >> 11 A b 0.47761962
> > >> 12 A c 0.86120948
> > >> 13 C b 0.43809711
> > >> 14 B a 0.24479728
> > >> 15 C d 0.07067905
> > >> 16 B c 0.09946616
> > >> 17 C d 0.31627171
> > >> 18 C a 0.51863426
> > >> 19 B c 0.66200508
> > >> 20 C b 0.40683019
> > >> > # drop unused groups from the split
> > >> > (z <- split(x, list(x$cat, x$a), drop=TRUE))
> > >> $B.a
> > >> cat a b
> > >> 2 B a 0.6470602
> > >> 14 B a 0.2447973
> > >>
> > >> $C.a
> > >> cat a b
> > >> 4 C a 0.55303631
> > >> 7 C a 0.02333120
> > >> 18 C a 0.51863426
> > >>
> > >> $A.b
> > >> cat a b
> > >> 5 A b 0.5297196
> > >> 10 A b 0.6927316
> > >> 11 A b 0.4776196
> > >>
> > >> $B.b
> > >> cat a b
> > >> 8 B b 0.4772301
> > >>
> > >> $C.b
> > >> cat a b
> > >> 6 C b 0.7893562
> > >> 13 C b 0.4380971
> > >> 20 C b 0.4068302
> > >>
> > >> $A.c
> > >> cat a b
> > >> 12 A c 0.8612095
> > >>
> > >> $B.c
> > >> cat a b
> > >> 3 B c 0.78293276
> > >> 16 B c 0.09946616
> > >> 19 B c 0.66200508
> > >>
> > >> $A.d
> > >> cat a b
> > >> 1 A d 0.8209463
> > >>
> > >> $B.d
> > >> cat a b
> > >> 9 B d 0.7323137
> > >>
> > >> $C.d
> > >> cat a b
> > >> 15 C d 0.07067905
> > >> 17 C d 0.31627171
> > >>
> > >> > # access the value ('b' in this instance); two ways- should be the same
> > >> > z[[1]]$b
> > >> [1] 0.6470602 0.2447973
> > >> > z$B.a$b
> > >> [1] 0.6470602 0.2447973
> > >> >
> > >> >
> > >> >
> > >> >
> > >>
> > >>
> > >> On Sun, Jul 13, 2008 at 1:26 AM, <[EMAIL PROTECTED]> wrote:
> > >> > This is almost it. Maybe it is as good as can be expected. The only
> > >> > problem that I see is that this seems to form a Category/SubCategory
> > >> > pair where none existed in the original data. For example, A might
> > >> > have two sub-categories a and b, and B might have two categories c and
> > >> > d. As far as I can tell the method that you outlined forms a
> > >> > Category/SubCategory pair like B a or B b where none existed. This
> > >> > results in alot of empty lists and it seems to take a long time to
> > >> > generate. But if that is as good as it gets then I can live with it.
> > >> >
> > >> > I know that I said one more question. But I have run into a problem. c
> > >> > <- split(x, x$Category) returns a vector of the rows in each of the
> > >> > categories. Now I would like to access the "Quantity" column within
> > >> > this split vector. I can see it listed. I just can't access it. I have
> > >> > tried c[1]$Quantity and c[1,2] both which give me errors. Any ideas?
> > >> >
> > >> > Sorry this is so hard for me. I am more used to C type arrays and C
> > >> > type arrays of structures. This seems to be somewhat different.
> > >> >
> > >> > Thank you.
> > >> >
> > >> > Kevin
> > >> > ---- jim holtman <[EMAIL PROTECTED]> wrote:
> > >> >> Is this something like what you were asking for? The output of a
> > >> >> 'split' will be a list of the dataframe subsets for the categories you
> > >> >> have specified.
> > >> >>
> > >> >> > x <- data.frame(g1=sample(LETTERS[1:2],30,TRUE),
> > >> >> + g2=sample(letters[1:2], 30, TRUE),
> > >> >> + g3=1:30)
> > >> >> > y <- split(x, list(x$g1, x$g2))
> > >> >> > str(y)
> > >> >> List of 4
> > >> >> $ A.a:'data.frame': 7 obs. of 3 variables:
> > >> >> ..$ g1: Factor w/ 2 levels "A","B": 1 1 1 1 1 1 1
> > >> >> ..$ g2: Factor w/ 2 levels "a","b": 1 1 1 1 1 1 1
> > >> >> ..$ g3: int [1:7] 3 4 6 8 9 13 24
> > >> >> $ B.a:'data.frame': 7 obs. of 3 variables:
> > >> >> ..$ g1: Factor w/ 2 levels "A","B": 2 2 2 2 2 2 2
> > >> >> ..$ g2: Factor w/ 2 levels "a","b": 1 1 1 1 1 1 1
> > >> >> ..$ g3: int [1:7] 10 11 16 17 18 20 25
> > >> >> $ A.b:'data.frame': 6 obs. of 3 variables:
> > >> >> ..$ g1: Factor w/ 2 levels "A","B": 1 1 1 1 1 1
> > >> >> ..$ g2: Factor w/ 2 levels "a","b": 2 2 2 2 2 2
> > >> >> ..$ g3: int [1:6] 2 12 23 26 27 29
> > >> >> $ B.b:'data.frame': 10 obs. of 3 variables:
> > >> >> ..$ g1: Factor w/ 2 levels "A","B": 2 2 2 2 2 2 2 2 2 2
> > >> >> ..$ g2: Factor w/ 2 levels "a","b": 2 2 2 2 2 2 2 2 2 2
> > >> >> ..$ g3: int [1:10] 1 5 7 14 15 19 21 22 28 30
> > >> >> > y
> > >> >> $A.a
> > >> >> g1 g2 g3
> > >> >> 3 A a 3
> > >> >> 4 A a 4
> > >> >> 6 A a 6
> > >> >> 8 A a 8
> > >> >> 9 A a 9
> > >> >> 13 A a 13
> > >> >> 24 A a 24
> > >> >>
> > >> >> $B.a
> > >> >> g1 g2 g3
> > >> >> 10 B a 10
> > >> >> 11 B a 11
> > >> >> 16 B a 16
> > >> >> 17 B a 17
> > >> >> 18 B a 18
> > >> >> 20 B a 20
> > >> >> 25 B a 25
> > >> >>
> > >> >> $A.b
> > >> >> g1 g2 g3
> > >> >> 2 A b 2
> > >> >> 12 A b 12
> > >> >> 23 A b 23
> > >> >> 26 A b 26
> > >> >> 27 A b 27
> > >> >> 29 A b 29
> > >> >>
> > >> >> $B.b
> > >> >> g1 g2 g3
> > >> >> 1 B b 1
> > >> >> 5 B b 5
> > >> >> 7 B b 7
> > >> >> 14 B b 14
> > >> >> 15 B b 15
> > >> >> 19 B b 19
> > >> >> 21 B b 21
> > >> >> 22 B b 22
> > >> >> 28 B b 28
> > >> >> 30 B b 30
> > >> >>
> > >> >> > y[[2]]
> > >> >> g1 g2 g3
> > >> >> 10 B a 10
> > >> >> 11 B a 11
> > >> >> 16 B a 16
> > >> >> 17 B a 17
> > >> >> 18 B a 18
> > >> >> 20 B a 20
> > >> >> 25 B a 25
> > >> >> >
> > >> >> >
> > >> >> >
> > >> >>
> > >> >>
> > >> >> On Sat, Jul 12, 2008 at 8:51 PM, <[EMAIL PROTECTED]> wrote:
> > >> >> > OK. Now I know that I am dealing with a data frame. One last
> > >> >> > question on this topic. a <- read.csv() gives me a dataframe. If I
> > >> >> > have 'c <- split(x, x$Category), then what is returned by split in
> > >> >> > this case? c[1] seems to be OK but c[2] is not right in my mind. If
> > >> >> > I run ci <- split(nrow(a), a$Category). And then ci[1] seems to be
> > >> >> > the rows associated with the first category, c[2] is the
> > >> >> > indices/rows associated with the second category, etc. But this
> > >> >> > seems different than c[1], c[2], etc.
> > >> >> >
> > >> >> > Using the techniques below I can get the information on the
> > >> >> > categories. Now as an extra level of complexity there are
> > >> >> > SubCategories within each Category. Assume that the SubCategory
> > >> >> > names are not unique within the dataset so if I want the
> > >> >> > SubCategory data I need to retrive the indices (or data) for the
> > >> >> > Category and SubCategory pair. In other words if I have a Category
> > >> >> > that ranges from 'A' to 'Z', it is possible that I might have a
> > >> >> > subcategory A a, A b (where a and b are the sub category names). I
> > >> >> > also might have B a, B b. I want all of the sub categories A a. NOT
> > >> >> > the subcategories a (because that might include B a which would be
> > >> >> > different). I am guessing that this will take more than a simple
> > >> >> > 'split'.
> > >> >> >
> > >> >> > Thank you.
> > >> >> >
> > >> >> > Kevin
> > >> >> >
> > >> >> > ---- Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> > >> >> >> On 12/07/2008 3:59 PM, [EMAIL PROTECTED] wrote:
> > >> >> >> > I am sorry but if read.csv returns a dataframe and a dataframe
> > >> >> >> > is like a matrix and I have a set of input like below and a[1,]
> > >> >> >> > gives me the first row, what is the second index? From what I
> > >> >> >> > read and your input I am guessing that it is the column number.
> > >> >> >> > So a[1,1] would return the DayOfYear column for the first row,
> > >> >> >> > right? What does a$DayOfYear return?
> > >> >> >>
> > >> >> >> a$DayOfYear would be the same as a[,1] or a[,"DayOfYear"], i.e. it
> > >> >> >> would
> > >> >> >> return the entire first column.
> > >> >> >>
> > >> >> >> Duncan Murdoch
> > >> >> >>
> > >> >> >> >
> > >> >> >> > Thank you for your patience.
> > >> >> >> >
> > >> >> >> > Kevin
> > >> >> >> >
> > >> >> >> > ---- Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> > >> >> >> >> On 12/07/2008 12:31 PM, [EMAIL PROTECTED] wrote:
> > >> >> >> >>> I am using a simple R statement to read in the file:
> > >> >> >> >>>
> > >> >> >> >>> a <- read.csv("Sample.dat", header=TRUE)
> > >> >> >> >>>
> > >> >> >> >>> There is alot of data but the first few lines look like:
> > >> >> >> >>>
> > >> >> >> >>> DayOfYear,Quantity,Fraction,Category,SubCategory
> > >> >> >> >>> 1,82,0.0000390392720794458,(Unknown),(Unknown)
> > >> >> >> >>> 2,78,0.0000371349173438631,(Unknown),(Unknown)
> > >> >> >> >>> . . .
> > >> >> >> >>> 71,2,0.0000009521773677913,WOMEN,Piratesses
> > >> >> >> >>> 72,4,0.0000019043547355827,WOMEN,Piratesses
> > >> >> >> >>> 73,3,0.0000014282660516870,WOMEN,Piratesses
> > >> >> >> >>> 74,14,0.0000066652415745395,WOMEN,Piratesses
> > >> >> >> >>> 75,2,0.0000009521773677913,WOMEN,Piratesses
> > >> >> >> >>>
> > >> >> >> >>> If I read the data in as above, the command
> > >> >> >> >>>
> > >> >> >> >>> a[1]
> > >> >> >> >>>
> > >> >> >> >>> results in the output
> > >> >> >> >>>
> > >> >> >> >>> [ reached getOption("max.print") -- omitted 16193 rows ]]
> > >> >> >> >>>
> > >> >> >> >>> Shouldn't this be the first row?
> > >> >> >> >> No, the first row would be a[1,]. read.csv() returns a
> > >> >> >> >> dataframe, and
> > >> >> >> >> those are indexed with two indices to treat them like a matrix,
> > >> >> >> >> or with
> > >> >> >> >> one index to treat them like a list of their columns.
> > >> >> >> >>
> > >> >> >> >> Duncan Murdoch
> > >> >> >> >>
> > >> >> >> >>> a$Category[1]
> > >> >> >> >>>
> > >> >> >> >>> results in the output
> > >> >> >> >>>
> > >> >> >> >>> [1] (Unknown)
> > >> >> >> >>> 4464 Levels: Tags ... WOMEN
> > >> >> >> >>>
> > >> >> >> >>> But
> > >> >> >> >>>
> > >> >> >> >>> a$Category[365]
> > >> >> >> >>>
> > >> >> >> >>> gives me:
> > >> >> >> >>>
> > >> >> >> >>> [1] 7 Plates
> > >> >> >> >>> (Dessert),Western\n120,5,0.0000023804434194784,7 Plates
> > >> >> >> >>> (Dessert)
> > >> >> >> >>> 4464 Levels: Tags ... WOMEN
> > >> >> >> >>>
> > >> >> >> >>> There is something fundamental about either vectors of the
> > >> >> >> >>> read.csv command that I am missing here.
> > >> >> >> >>>
> > >> >> >> >>> Thank you.
> > >> >> >> >>>
> > >> >> >> >>> Kevin
> > >> >> >> >>>
> > >> >> >> >>> ---- jim holtman <[EMAIL PROTECTED]> wrote:
> > >> >> >> >>>> Please provide commented, minimal, self-contained,
> > >> >> >> >>>> reproducible code,
> > >> >> >> >>>> or at least a before/after of what you data would look like.
> > >> >> >> >>>> Taking a
> > >> >> >> >>>> guess at what you are asking, here is one way of doing it:
> > >> >> >> >>>>
> > >> >> >> >>>>
> > >> >> >> >>>>> x <- data.frame(cat=sample(LETTERS[1:3],20,TRUE),a=1:20,
> > >> >> >> >>>>> b=runif(20))
> > >> >> >> >>>>> x
> > >> >> >> >>>> cat a b
> > >> >> >> >>>> 1 B 1 0.65472393
> > >> >> >> >>>> 2 C 2 0.35319727
> > >> >> >> >>>> 3 B 3 0.27026015
> > >> >> >> >>>> 4 A 4 0.99268406
> > >> >> >> >>>> 5 C 5 0.63349326
> > >> >> >> >>>> 6 A 6 0.21320814
> > >> >> >> >>>> 7 C 7 0.12937235
> > >> >> >> >>>> 8 A 8 0.47811803
> > >> >> >> >>>> 9 A 9 0.92407447
> > >> >> >> >>>> 10 A 10 0.59876097
> > >> >> >> >>>> 11 A 11 0.97617069
> > >> >> >> >>>> 12 A 12 0.73179251
> > >> >> >> >>>> 13 B 13 0.35672691
> > >> >> >> >>>> 14 C 14 0.43147369
> > >> >> >> >>>> 15 C 15 0.14821156
> > >> >> >> >>>> 16 C 16 0.01307758
> > >> >> >> >>>> 17 B 17 0.71556607
> > >> >> >> >>>> 18 B 18 0.10318424
> > >> >> >> >>>> 19 C 19 0.44628435
> > >> >> >> >>>> 20 B 20 0.64010105
> > >> >> >> >>>>> # create a list of the indices of the data grouped by 'cat'
> > >> >> >> >>>>> split(seq(nrow(x)), x$cat)
> > >> >> >> >>>> $A
> > >> >> >> >>>> [1] 4 6 8 9 10 11 12
> > >> >> >> >>>>
> > >> >> >> >>>> $B
> > >> >> >> >>>> [1] 1 3 13 17 18 20
> > >> >> >> >>>>
> > >> >> >> >>>> $C
> > >> >> >> >>>> [1] 2 5 7 14 15 16 19
> > >> >> >> >>>>
> > >> >> >> >>>>> # or do you want the data
> > >> >> >> >>>>> split(x, x$cat)
> > >> >> >> >>>> $A
> > >> >> >> >>>> cat a b
> > >> >> >> >>>> 4 A 4 0.9926841
> > >> >> >> >>>> 6 A 6 0.2132081
> > >> >> >> >>>> 8 A 8 0.4781180
> > >> >> >> >>>> 9 A 9 0.9240745
> > >> >> >> >>>> 10 A 10 0.5987610
> > >> >> >> >>>> 11 A 11 0.9761707
> > >> >> >> >>>> 12 A 12 0.7317925
> > >> >> >> >>>>
> > >> >> >> >>>> $B
> > >> >> >> >>>> cat a b
> > >> >> >> >>>> 1 B 1 0.6547239
> > >> >> >> >>>> 3 B 3 0.2702601
> > >> >> >> >>>> 13 B 13 0.3567269
> > >> >> >> >>>> 17 B 17 0.7155661
> > >> >> >> >>>> 18 B 18 0.1031842
> > >> >> >> >>>> 20 B 20 0.6401010
> > >> >> >> >>>>
> > >> >> >> >>>> $C
> > >> >> >> >>>> cat a b
> > >> >> >> >>>> 2 C 2 0.35319727
> > >> >> >> >>>> 5 C 5 0.63349326
> > >> >> >> >>>> 7 C 7 0.12937235
> > >> >> >> >>>> 14 C 14 0.43147369
> > >> >> >> >>>> 15 C 15 0.14821156
> > >> >> >> >>>> 16 C 16 0.01307758
> > >> >> >> >>>> 19 C 19 0.44628435
> > >> >> >> >>>>
> > >> >> >> >>>>
> > >> >> >> >>>> On Sat, Jul 12, 2008 at 3:32 AM, <[EMAIL PROTECTED]> wrote:
> > >> >> >> >>>>> I have search the archive and I could not find what I need
> > >> >> >> >>>>> so I will try to ask the question here.
> > >> >> >> >>>>>
> > >> >> >> >>>>> I read a table in (read.table)
> > >> >> >> >>>>>
> > >> >> >> >>>>> a <- read.table(.....)
> > >> >> >> >>>>>
> > >> >> >> >>>>> The table has column names like DayOfYear, Quantity, and
> > >> >> >> >>>>> Category.
> > >> >> >> >>>>>
> > >> >> >> >>>>> The values in the row for Category are strings (characters).
> > >> >> >> >>>>>
> > >> >> >> >>>>> I want to get all of the rows grouped by Category. The
> > >> >> >> >>>>> number of unique category names could be around 50. Say for
> > >> >> >> >>>>> argument sake the number of categories is exactly 50. Can I
> > >> >> >> >>>>> somehow get a vector of length 50 containing the rows
> > >> >> >> >>>>> corresponding to the category (another vector)? I realize I
> > >> >> >> >>>>> can access any row a[i]$Category (right?). But I wanta
> > >> >> >> >>>>> vector containing the rows corresponding to each distinct
> > >> >> >> >>>>> Category name.
> > >> >> >> >>>>>
> > >> >> >> >>>>> Thank you.
> > >> >> >> >>>>>
> > >> >> >> >>>>> Kevin
> > >> >> >> >>>>>
> > >> >> >> >>>>> ______________________________________________
> > >> >> >> >>>>> [email protected] 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
> > >> >> >> >>>> Cincinnati, OH
> > >> >> >> >>>> +1 513 646 9390
> > >> >> >> >>>>
> > >> >> >> >>>> What is the problem you are trying to solve?
> > >> >> >> >>> ______________________________________________
> > >> >> >> >>> [email protected] 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
> > >> >> Cincinnati, OH
> > >> >> +1 513 646 9390
> > >> >>
> > >> >> What is the problem you are trying to solve?
> > >> >
> > >> >
> > >>
> > >>
> > >>
> > >> --
> > >> Jim Holtman
> > >> Cincinnati, OH
> > >> +1 513 646 9390
> > >>
> > >> What is the problem you are trying to solve?
> > >
> > >
> >
> >
> >
> > --
> > Jim Holtman
> > Cincinnati, OH
> > +1 513 646 9390
> >
> > What is the problem you are trying to solve?
> >
> > ______________________________________________
> > [email protected] 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.
>
> ______________________________________________
> [email protected] 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.
______________________________________________
[email protected] 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.
