From: <gyadav_at_ccilindia.co.in>

Date: Mon, 14 May 2007 18:29:29 +0530

*[19,] 38 72 66 78 81
*

*[20,] 78 41 41 96 60
*

>

> for ( j in 1:length(x[1,]) - 1) {

*[20,] 0 0 0 0 0
*

>

price/mcap data for a portfolio of stocks, and the date. To get the total value of the portfolio on a daily basis, I calculate rowSums of the data.frame.

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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. Received on Mon 14 May 2007 - 13:18:46 GMT

Date: Mon, 14 May 2007 18:29:29 +0530

Hi Tirthankar

this will help you

ind is a matrix which indicates the start of any new stock. ind[i,j] means that in j + 1 column all the values from 1st row to i - 1 row are all NAs.

> x

V2 V3 V4 V5 V6

[1,] 27 93 82 NA NA [2,] 37 21 65 NA NA [3,] 57 65 78 NA NA [4,] 91 13 55 NA NA [5,] 20 27 53 NA NA [6,] 90 39 79 26 NA [7,] 94 NA NA 48 NA [8,] 66 38 48 77 NA [9,] 63 87 73 8 NA

[10,] NA 34 69 88 NA

[11,] 21 48 48 34 24[12,] 18 60 86 84 NA[13,] 69 49 44 35 64[14,] 38 19 24 33 88[15,] 77 83 NA 48 78[16,] 50 67 10 89 80[17,] 72 79 32 86 46[18,] 99 11 52 39 41

>

> for ( j in 1:length(x[1,]) - 1) {

+ for ( i in 2:length(x[,1])) { + indicator<-TRUE + for (k in 1: i - 1){ + indicator <- indicator && is.na(x[k,j+1]) + } + ind[i,j]<-indicator + + } + } > ind V2 V3 V4 V5 V6 [1,] NA NA NA NA NA [2,] 0 0 NA NA 0 [3,] 0 0 NA NA 0 [4,] 0 0 NA NA 0 [5,] 0 0 NA NA 0 [6,] 0 0 NA NA 0 [7,] 0 0 0 NA 0 [8,] 0 0 0 NA 0 [9,] 0 0 0 NA 0[19,] 0 0 0 0 0

[10,] 0 0 0 NA 0[11,] 0 0 0 NA 0[12,] 0 0 0 0 0[13,] 0 0 0 0 0[14,] 0 0 0 0 0[15,] 0 0 0 0 0[16,] 0 0 0 0 0[17,] 0 0 0 0 0[18,] 0 0 0 0 0

>

Regards,

Gaurav Yadav

+++++++++++

Assistant Manager, CCIL, Mumbai (India)

Mob: +919821286118 Email: mailtogauravyadav_at_gmail.com
Bhagavad Gita: Man is made by his Belief, as He believes, so He is

"Patnaik, Tirthankar " <tirthankar.patnaik_at_citi.com> Sent by: r-help-bounces_at_stat.math.ethz.ch 05/14/2007 11:53 AM

To

<r-help_at_stat.math.ethz.ch>

cc

Subject

[R] Conditional Sums for Index creation

Hi,

Apologies for the long mail. I have a data.frame withcolumns of

price/mcap data for a portfolio of stocks, and the date. To get the total value of the portfolio on a daily basis, I calculate rowSums of the data.frame.

> set.seed(1) > ab <- matrix(round(runif(100)*100),nrow=20,ncol=5) > ab[1:5,4:5] <- NA > ab[6:10,5] <- NA > ac <- as.data.frame(ifelse(ab <= 7,NA,ab)) > ac

V1 V2 V3 V4 V5

1 27 93 82 NA NA

2 37 21 65 NA NA

3 57 65 78 NA NA

4 91 13 55 NA NA

5 20 27 53 NA NA

6 90 39 79 26 NA

7 94 NA NA 48 NA

8 66 38 48 77 NA

9 63 87 73 8 NA

10 NA 34 69 88 NA

11 21 48 48 34 24

12 18 60 86 84 NA

13 69 49 44 35 64

14 38 19 24 33 88

15 77 83 NA 48 78

16 50 67 10 89 80

17 72 79 32 86 46

18 99 11 52 39 41

19 38 72 66 78 81

20 78 41 41 96 60

>

Here the rows 1:20 are dates (also in my data.frame).

Since some of the prices have NA, the rowSums is made to ignore these entries.

> rowSums(ac,na.rm=TRUE)

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18
19 20

202 123 200 159 100 234 142 229 231 191 175 248 261 202 286 296 315 242
335 316

>

Stocks are being added to the portfolio too. So from date=6 (or row=6) we have the 4th stock V4, and from date=11, we have the 5th stock V5. My problem is that I need to calculate the rowSums for row=6 (When a new stock was added), _with_ and _without_ the new stock. So my answer for row=6 would be 234 for the plain row-sum, and 234 - 26 = 208 for the original set of stocks (without V4). Similarly, my answer for row=11 would be 175 for the plain sum, and 175 - 24 = 151 for the original sum (without V5).

Basically I'm interested in finding out the value of the portfolio with and without the new stock for the purposes of creating an index. It's possible that some stocks my get dropped later, in which case there would be an NA series starting for say V1 at row=18 and so on. In that case, the aim would be to find the sum at row=18 with and without the value of V1.

Is there any way I can get the sum over columns, deleting specific colums? To get the columns that are NA in any row, I tried (shown for the first 12 rows):

> apply(ac[1:12,],1,function(y)which(is.na(y)))

Which correctly gives

$`1`

V4 V5

4 5

$`2`

V4 V5

4 5

$`3`

V4 V5

4 5

$`4`

V4 V5

4 5

$`5`

V4 V5

4 5

$`6`

V5

5

$`7`

V2 V3 V5

2 3 5

$`8`

V5

5

$`9`

V5

5

$`10`

V1 V5

1 5

$`11`

integer(0)

$`12`

V5

5

>

But now I'm stuck. I don't how to use this list of indices at each row to exclude my columns.

Any pointers please? Would such an exercise be easier if I use a time-series based object, like a zoo.

TIA and best,

-Tir

Tirthankar Patnaik

India Strategy

Citigroup Investment Research

+91-22-6631 9887

R-help_at_stat.math.ethz.ch 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.

DISCLAIMER AND CONFIDENTIALITY CAUTION:\ \ This message and ...{{dropped}}

R-help_at_stat.math.ethz.ch 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. Received on Mon 14 May 2007 - 13:18:46 GMT

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