Here a brute force way based on the format of you input data. Basically it
reads a line in and then 'splits' it apart based on blanks and then
processes based on the 'tag'. Information is stored in some global data and
the '.result' is converted into a dataframe that you can work with.
================================
> xIN <- scan('/treedata.txt', what='', sep='\n') # read in entire line
Read 59 items
> xIN <- strsplit(xIN, ' ') # split out fields separated by blanks
> # initialize 'global' variables to collect the information
> Out <- list() # individual results
> .result <- list(); r.n <- 0
> # process the data into a list '.result'
> # make use of the '<<-' to assign to a 'global' value
> invisible(lapply(xIN, function(x){
+ if (x[1] == "A") Out$inv <<- x[2]
+ else if (x[1] == "X") {
+ Out$strat <<- x[2]
+ Out$total <<- x[3]
+ Out$year <<- x[4]
+ } else if (x[1] == "P"){
+ Out$plot <<- x[2]
+ Out$age <<- x[3]
+ Out$slope <<- x[4]
+ Out$species <<- x[5]
+ } else if (x[1] == "T"){
+ Out$tree <<- x[2]
+ Out$freq <<- x[3]
+ } else if (x[1] == "L"){
+ Out$leader <<- x[2]
+ Out$diam <<- x[3]
+ Out$height <<- x[4]
+ } else if (x[1] == "F") {
+ Out$start <<- x[2]
+ Out$finish <<- x[3]
+ Out$feature <<- x[4]
+ .result[[r.n <<- r.n + 1]] <<- Out # store the result
+ }
+ }))
> # convert the list to a dataframe for processing
> myData <- lapply(.result, function(x) do.call('cbind', x))
> myData <- as.data.frame(do.call('rbind', myData))
> myData[order(myData$inv, myData$strat, myData$plot, myData$tree,
myData$leader),]
inv strat total year plot age slope species tree freq leader diam height
start finish feature
1 BENALLA_1 1 10 YE=1985 1 20.25 slope=14 SPP:P.RAD 1 25 0 28.5 21.3528 0
21.3528 SFNSW_DIC:P
2 BENALLA_1 1 10 YE=1985 1 20.25 slope=14 SPP:P.RAD 1 25 0 28.5 21.3528
21.3528 100 SFNSW_DIC:P
3 BENALLA_1 1 10 YE=1985 1 20.25 slope=14 SPP:P.RAD 2 25 0 32 23.1 0
6.5SFNSW_DIC:A
4 BENALLA_1 1 10 YE=1985 1 20.25 slope=14 SPP:P.RAD 2 25 0 32 23.1 6.5
23.1SFNSW_DIC:C
5 BENALLA_1 1 10 YE=1985 1 20.25 slope=14 SPP:P.RAD 2 25 0 32 23.1 23.1 100
SFNSW_DIC:C
6 BENALLA_1 1 10 YE=1985 1 20.25 slope=14 SPP:P.RAD 3 25 0 39.5 22.2407 0
4.7 SFNSW_DIC:A
7 BENALLA_1 1 10 YE=1985 1 20.25 slope=14 SPP:P.RAD 3 25 0 39.5 22.2407 4.7
6.7 SFNSW_DIC:C
8 BENALLA_1 1 10 YE=1985 2 20.25 slope=13 SPP:P.RAD 1 25 0 38 22.1474 0 1
SFNSW_DIC:G
9 BENALLA_1 1 10 YE=1985 2 20.25 slope=13 SPP:P.RAD 1 25 0 38 22.1474
1 2.3SFNSW_DIC:A
10 BENALLA_1 1 10 YE=1985 2 20.25 slope=13 SPP:P.RAD 1001 25 0 38 22.1474 0
1 SFNSW_DIC:G
11 BENALLA_1 1 10 YE=1985 2 20.25 slope=13 SPP:P.RAD 1001 25 0 38 22.1474 1
2.3 SFNSW_DIC:A
12 BENALLA_1 1 10 YE=1985 2 20.25 slope=13 SPP:P.RAD 2 25 0 32.5 21.7386 0 2
SFNSW_DIC:A
13 BENALLA_1 1 10 YE=1985 2 20.25 slope=13 SPP:P.RAD 2 25 0 32.5 21.7386 2
3.3 SFNSW_DIC:G
14 BENALLA_1 1 10 YE=1985 2 20.25 slope=13 SPP:P.RAD 2 25 0 32.5 21.7386 3.3
10.4 SFNSW_DIC:C
15 BENALLA_1 2 10 YE=1985 1 20.25 slope=14 SPP:P.RAD 1 25 0 28.5 21.3528 0
21.3528 SFNSW_DIC:P
16 BENALLA_1 2 10 YE=1985 1 20.25 slope=14 SPP:P.RAD 1 25 0 28.5 21.3528
21.3528 100 SFNSW_DIC:P
17 BENALLA_1 2 10 YE=1985 1 20.25 slope=14 SPP:P.RAD 2 25 0 32 23.1 0
6.5SFNSW_DIC:A
18 BENALLA_1 2 10 YE=1985 1 20.25 slope=14 SPP:P.RAD 2 25 0 32 23.1
6.5 23.1SFNSW_DIC:C
19 BENALLA_1 2 10 YE=1985 1 20.25 slope=14 SPP:P.RAD 2 25 0 32 23.1 23.1 100
SFNSW_DIC:C
20 BENALLA_1 2 10 YE=1985 1 20.25 slope=14 SPP:P.RAD 3 25 0 39.5 22.2407 0
4.7 SFNSW_DIC:A
21 BENALLA_1 2 10 YE=1985 1 20.25 slope=14 SPP:P.RAD 3 25 0 39.5 22.2407 4.7
6.7 SFNSW_DIC:C
22 BENALLA_1 2 10 YE=1985 2 20.25 slope=13 SPP:P.RAD 1 25 0 38 22.1474 0 1
SFNSW_DIC:G
23 BENALLA_1 2 10 YE=1985 2 20.25 slope=13 SPP:P.RAD 1 25 0 38 22.1474
1 2.3SFNSW_DIC:A
24 BENALLA_1 2 10 YE=1985 2 20.25 slope=13 SPP:P.RAD 1001 25 0 38 22.1474 0
1 SFNSW_DIC:G
25 BENALLA_1 2 10 YE=1985 2 20.25 slope=13 SPP:P.RAD 1001 25 0 38 22.1474 1
2.3 SFNSW_DIC:A
26 BENALLA_1 2 10 YE=1985 2 20.25 slope=13 SPP:P.RAD 2 25 0 32.5 21.7386 0 2
SFNSW_DIC:A
27 BENALLA_1 2 10 YE=1985 2 20.25 slope=13 SPP:P.RAD 2 25 0 32.5 21.7386 2
3.3 SFNSW_DIC:G
28 BENALLA_1 2 10 YE=1985 2 20.25 slope=13 SPP:P.RAD 2 25 0 32.5 21.7386 3.3
10.4 SFNSW_DIC:C
>
>
>
On 10/3/05, [EMAIL PROTECTED] <[EMAIL PROTECTED]>
wrote:
>
> Dear List,
>
> Im new to R - making a transition from SAS. I have a space delimited file
> with the following structure. Each line in the datafile is identified by
> the first letter.
>
> A = Inventory (Inventory)
> X = Stratum (Stratum_no Total Ye=year established)
> P = Plot (Plot_no age slope= species)
> T = Tree (tree_no frequency)
> L = Leader (leader diameter height)
> F = Feature (start_height finish_height feature)
>
> On each of these lines there are some 'line specific' variables (in
> brackets). The data is hierarchical in nature - A feature belongs to a
> leader, a leader belongs to a tree, a tree belongs to a plot, a plot
> belongs to a stratum, a stratum belongs to inventory. There are many
> features in a tree. Many trees in a plot etc.
>
> In SAS I would read in the data in a procedural way using first. and last.
> variables to work out where inventories/stratums/plots/trees finished and
> started so I could create summary statistics for each of them. For
> example, how many plots in a stratum? How many trees in a plot? An example
> of the sas code I would (not checked for errors!!!). If anybody could give
> me some idea on what the right approach in R would be for a similar
> analysis it would be greatly appreciated.
>
> regards Andrew
>
>
> Data datafile;
> infile 'test.txt';
> input @1 tag $1. @@;
> retain inventory stratum plot tree leader;
> if tag = 'A' then input @3 inventory $.;
> if tag = 'X' then input @3 stratum_no $. total $. yearest $. ;
> if tag = 'P' then input @3 plot_no $. age $. slope $. species $;
> if tag = 'T' then input @3 tree_no $. frequency ;
> if tag = 'L' then input @3 leader_no $ diameter height ;
> if tag = 'F' then input @3 start $ finish $ feature $;
> if tag = 'F' then output;
> run;
> proc sort data = datafile;
> by inventory stratum_no plot_no tree_no leader_no;
>
> * calculate mean dbh in each plot
> data dbh
> set datafile;
> by inventory stratum_no plot_no tree_no leader_no
> if first.leader_no then output;
>
> proc summary data = diameter;
> by inventory stratum plot tree;
> var diameter;
> output out = mean mean=;
> run;
>
> A BENALLA_1
> X 1 10 YE=1985
> P 1 20.25 slope=14 SPP:P.RAD
> T 1 25
> L 0 28.5 21.3528
> F 0 21.3528 SFNSW_DIC:P
> F 21.3528 100 SFNSW_DIC:P
> T 2 25
> L 0 32 23.1
> F 0 6.5 SFNSW_DIC:A
> F 6.5 23.1 SFNSW_DIC:C
> F 23.1 100 SFNSW_DIC:C
> T 3 25
> L 0 39.5 22.2407
> F 0 4.7 SFNSW_DIC:A
> F 4.7 6.7 SFNSW_DIC:C
> P 2 20.25 slope=13 SPP:P.RAD
> T 1 25
> L 0 38 22.1474
> F 0 1 SFNSW_DIC:G
> F 1 2.3 SFNSW_DIC:A
> T 1001 25
> L 0 38 22.1474
> F 0 1 SFNSW_DIC:G
> F 1 2.3 SFNSW_DIC:A
> T 2 25
> L 0 32.5 21.7386
> F 0 2 SFNSW_DIC:A
> F 2 3.3 SFNSW_DIC:G
> F 3.3 10.4 SFNSW_DIC:C
> X 2 10 YE=1985
> P 1 20.25 slope=14 SPP:P.RAD
> T 1 25
> L 0 28.5 21.3528
> F 0 21.3528 SFNSW_DIC:P
> F 21.3528 100 SFNSW_DIC:P
> T 2 25
> L 0 32 23.1
> F 0 6.5 SFNSW_DIC:A
> F 6.5 23.1 SFNSW_DIC:C
> F 23.1 100 SFNSW_DIC:C
> T 3 25
> L 0 39.5 22.2407
> F 0 4.7 SFNSW_DIC:A
> F 4.7 6.7 SFNSW_DIC:C
> P 2 20.25 slope=13 SPP:P.RAD
> T 1 25
> L 0 38 22.1474
> F 0 1 SFNSW_DIC:G
> F 1 2.3 SFNSW_DIC:A
> T 1001 25
> L 0 38 22.1474
> F 0 1 SFNSW_DIC:G
> F 1 2.3 SFNSW_DIC:A
> T 2 25
> L 0 32.5 21.7386
> F 0 2 SFNSW_DIC:A
> F 2 3.3 SFNSW_DIC:G
> F 3.3 10.4 SFNSW_DIC:C
>
>
>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
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>
--
Jim Holtman
Cincinnati, OH
+1 513 247 0281
What the problem you are trying to solve?
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