You don't have a closing parens on the 'sum'
On Mon, Jul 14, 2008 at 11:25 PM, <[EMAIL PROTECTED]> wrote:
> One more question? I am trying to iterate through this array
>
> I have:
>
> sc <- split(x, list(x$Category, x$SubCategory), drop=TRUE)
>
> I think I understand 'length(sc)' It would be the total number of non empty
> category and sub category pairs (in this case 2415).
>
> I don't seems to be able to iterate through this list. My first try is:
>
> for(i in 1:length(sc))
> {
> sum(sc[[i]]$Quantity
> }
>
> This gives an error:
>
> Error: unexpected ']' in:
> "{
> sum(sc[[i]]]"
>> }
> Error: unexpected '}' in "}"
>>
>
> sc[[1]] refers to an array of data corresponding to a specific
> Category/SubCategory pair. Since this is a vector sc[[1]]$Category and
> sc[[1]]$SubCategory are the same. Is there anyway to access just the Category
> and SubCategory? R seems to be able to access this informaiton. I would just
> like to be able to access this. Or is it just as efficient to
> sc[[1]]$Category[1]? When I do this in R I get:
>
>> sc[[4]]$Category[1]
> [1] ADDITIONAL GUEST
> 46 Levels: (Unknown) 10" Plates 7" Plates (Dessert) ... WOMEN
>>
>
> What are 'Levels'?
>
> Thank you for your assistance.
>
> Kevin
>
> ---- jim holtman <[EMAIL PROTECTED]> wrote:
>> On Sun, Jul 13, 2008 at 5:45 PM, <[EMAIL PROTECTED]> wrote:
>> > Thank you I will try drop=TRUE.
>> >
>> > In the mean time do you know how I can access the members (for lack of a
>> > better term) of the results of a split? In the sample you provided below
>> > you have:
>> >
>> > z <- split(x, list(x$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.
>
>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
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