I went back and reviewed the merge.data.frame method and then (re-)reviewed the data.table documents. I was disappointed to find the there were relatively few worked examples of data.table merges, but finally got something that seems to make sense:

dftxt <- as.data.frame(dtxt)
# The dataframe method
> merge(dftxt[dftxt$YEAR==2008, c(1,2,4)], dftxt[dftxt$YEAR==2009, c(1,2,4)], by="STUDENT_ID")
   STUDENT_ID SCHOOL_NUMBER.x SCORE.x SCHOOL_NUMBER.y SCORE.y
1           1             100      39             200      48
2           2             100      64             200      73
3           3             100      35             200      35
4           4             100      52             200      61
5           5             100      51             200      58
6           6             300      45             400      55
7           7             300      69             400      77
8           8             300      47             400      47
9           9             300      57             400      58
10         10             300      47             400      53

#The datatable method:

> setkey(dtxt) <- "STUDENT_ID" #as far as I can tell this is needed to repalce the by= argument in merge.data.frame()

#( I tried several failed efforts at incorporating the key= statement within the data.table calls.)

> merge(dtxt[YEAR==2008,  ], dtxt[YEAR==2009,  ])
STUDENT_ID SCHOOL_NUMBER YEAR SCORE SCHOOL_NUMBER.1 YEAR.1 SCORE.1 [1,] 1 100 2008 39 200 2009 48 [2,] 2 100 2008 64 200 2009 73 [3,] 3 100 2008 35 200 2009 35 [4,] 4 100 2008 52 200 2009 61 [5,] 5 100 2008 51 200 2009 58 [6,] 6 300 2008 45 400 2009 55 [7,] 7 300 2008 69 400 2009 77 [8,] 8 300 2008 47 400 2009 47 [9,] 9 300 2008 57 400 2009 58
[10,]         10           300 2008    47             400   2009      53


> mtxt <- merge(dtxt[YEAR==2008,  ], dtxt[YEAR==2009,  ])

> mtxt[, mean(SCORE) , by=list("2009_SCH"=SCHOOL_NUMBER.1)]
     X2009_SCH   V1
[1,]       200 48.2
[2,]       400 53.0


On Sep 5, 2010, at 4:43 PM, David Winsemius wrote:


On Sep 5, 2010, at 3:02 PM, Damian Betebenner wrote:

Hi David,

Thanks for the quick and thoughtful reply.

Sorry for not being clearer.

There are two schools that the students attended in 2009 (200 and 400). I'd like to break on those, and calculate the mean for all the students in those two schools but for their 2008 scores.

Thus, the output would have 2 rows:

YEAR, 2009_SCHOOL_NUMBER, 2008_SCORE_MEAN

2009  200    54.4
2009  400    53

I dont see what sort of linkage you have between the 2008 and 2009 school-numbers but see if this satisfies:

> dtxt[YEAR==2008 & STUDENT_ID %in% dtxt[YEAR==2009, STUDENT_ID], mean(SCORE), by=SCHOOL_NUMBER]
    SCHOOL_NUMBER   V1
[1,]           100 48.2
[2,]           300 53.0


Thanks for considering this,

Damian


Damian Betebenner
Center for Assessment
PO Box 351
Dover, NH   03821-0351

Phone (office): (603) 516-7900
Phone (cell): (857) 234-2474
Fax: (603) 516-7910

[email protected]
www.nciea.org




-----Original Message-----
From: David Winsemius [mailto:[email protected]]
Sent: Sunday, September 05, 2010 1:03 PM
To: David Winsemius
Cc: Damian Betebenner; [email protected]
Subject: Re: [datatable-help] Data table syntax


On Sep 5, 2010, at 12:43 PM, David Winsemius wrote:


On Sep 5, 2010, at 11:38 AM, Damian Betebenner wrote:

Thanks for the invaluable help on my previous questions. The speed
up in create summary tables has been immense and I'm enthused about
all the possibilities going forward.

I'm currently stuck in trying to put together syntax for a "long"
for table. In the example below, each case is a unique Student by
Year combination. What I'm trying to do is take
such a table, aggregate on the student's  current year (i.e., 2009
in this data) SCHOOL_NUMBER, and calculate their mean score in the
previous year (i.e., 2008 in this data).

If the file were "wide", with each case representing a unique
student with separate variables for the year, then it would be easy
to break on the 2009 SCHOOL_NUMBER and take the
mean of the 2008 SCORE.

But there is only one 2008 SCORE for each student???


Is conversion of long to wide necessary to do this?

Probably not. Are you familiar with the "ave" function in base R?

I am having some difficulty understanding the structure of the desired
output. I initially thought it might be something like:
rd.txt <-
function(txt, header=TRUE, ...) {
    rd <- read.table(textConnection(txt), header=header, ...)
      closeAllConnections()
    rd }
txt <- rd.txt("STUDENT_ID SCHOOL_NUMBER YEAR SCORE
        1           100 2008    39
        1           200 2009    48
        2           100 2008    64
        2           200 2009    73
        3           100 2008    35
        3           200 2009    35
        4           100 2008    52
        4           200 2009    61
        5           100 2008    51
        5           200 2009    58
        6           300 2008    45
        6           400 2009    55
        7           300 2008    69
        7           400 2009    77
        8           300 2008    47
        8           400 2009    47
        9           300 2008    57
        9           400 2009    58
       10           300 2008    47
       10           400 2009    53")
dtxt <- data.table(txt)

dtxt$avScr <- dtxt[ , ave(SCORE, list(STUDENT_ID))]  # returns a
vector as long as its input
dtxt

But now I am wondering if you wanted:

dtxt[ , tapply(SCORE, list(STUDENT_ID), mean)]  # returns vector
only as long as product of category levels.
  1    2    3    4    5    6    7    8    9   10
43.5 68.5 35.0 56.5 54.5 50.0 73.0 47.0 57.5 50.0




   STUDENT_ID SCHOOL_NUMBER YEAR SCORE
[1,]          1           100 2008    39
[2,]          1           200 2009    48
[3,]          2           100 2008    64
[4,]          2           200 2009    73
[5,]          3           100 2008    35
[6,]          3           200 2009    35
[7,]          4           100 2008    52
[8,]          4           200 2009    61
[9,]          5           100 2008    51
[10,]          5           200 2009    58
[11,]          6           300 2008    45
[12,]          6           400 2009    55
[13,]          7           300 2008    69
[14,]          7           400 2009    77
[15,]          8           300 2008    47
[16,]          8           400 2009    47
[17,]          9           300 2008    57
[18,]          9           400 2009    58
[19,]         10           300 2008    47
[20,]         10           400 2009    53


Thanks,

Damian



David Winsemius, MD
West Hartford, CT


David Winsemius, MD
West Hartford, CT

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West Hartford, CT

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