It seems the R console took them out. Here is hat I tried:
> for(i in 1:length(sc))
+ {
+ sum(sc[[i]]]$Quantity)
Error: unexpected ']' in:
"{
sum(sc[[i]]]"
> }
Error: unexpected '}' in "}"
>
>
>
What I entered is in the sum that is after the '+'
Thank you.
Kevin
---- jim holtman <[EMAIL PROTECTED]> wrote:
> You don't have a closing parens on the 'sum'
>
> On Mon, Jul 14, 2008 at 11:25 PM, <[EMAIL PROTECTED]> wrote:
> > One more question? I am trying to iterate through this array
> >
> > I have:
> >
> > sc <- split(x, list(x$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
>
> 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.
