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58

answers:

4

Hello,

I have a data frame holding information on options like this

> chData
myIdx strike_price       date     exdate cp_flag strike_price    return
1 8355342       605000 1996-04-02 1996-05-18       P       605000  0.002340
2 8355433       605000 1996-04-02 1996-05-18       C       605000  0.002340
3 8356541       605000 1996-04-09 1996-05-18       P       605000 -0.003182
4 8356629       605000 1996-04-09 1996-05-18       C       605000 -0.003182
5 8358033       605000 1996-04-16 1996-05-18       P       605000  0.003907
6 8358119       605000 1996-04-16 1996-05-18       C       605000  0.003907
7 8359391       605000 1996-04-23 1996-05-18       P       605000  0.005695

where cp_flag means that a certain option is either a call or a put. What is a way to make sure that for each date, there is a both a call and a put, and drop the rows for which this does not exist? I can do it with a for loop, but is there a more clever way?

A: 

Here's one way using split and lapply:

> tmp <- lapply(split(d, list(d$date)), function(x) if(all(c('P', 'C') %in% x[, 5])) x)
> do.call(rbind, tmp)
             myIdx strike_price       date     exdate cp_flag strike_price    return
1996-05-18.1 8355342       605000 1996-04-02 1996-05-18       P       605000  0.002340
1996-05-18.2 8355433       605000 1996-04-02 1996-05-18       C       605000  0.002340
1996-05-18.3 8356541       605000 1996-04-09 1996-05-18       P       605000 -0.003182
1996-05-18.4 8356629       605000 1996-04-09 1996-05-18       C       605000 -0.003182
1996-05-18.5 8358033       605000 1996-04-16 1996-05-18       P       605000  0.003907
1996-05-18.6 8358119       605000 1996-04-16 1996-05-18       C       605000  0.003907
1996-05-18.7 8359391       605000 1996-04-23 1996-05-18       P       605000  0.005695

Edit: Here's the full version implied by my last answer. I tend to think in base functions rather than plyr or reshape... but these answers seem good too.

Vince
I must be taking crazy pills... `lapply` + `split` is better done with just `tapply`. But wch's solution seems *much* cleaner.
Vince
+1  A: 

Using the plyr package:

> ddply(chData, "date", function(x) if(all(c("P","C") %in% x$cp_flag)) x)
    myIdx strike_price       date     exdate cp_flag strike_price.1    return
1 8355342       605000 1996-04-02 1996-05-18       P         605000  0.002340
2 8355433       605000 1996-04-02 1996-05-18       C         605000  0.002340
3 8356541       605000 1996-04-09 1996-05-18       P         605000 -0.003182
4 8356629       605000 1996-04-09 1996-05-18       C         605000 -0.003182
5 8358033       605000 1996-04-16 1996-05-18       P         605000  0.003907
6 8358119       605000 1996-04-16 1996-05-18       C         605000  0.003907
Joshua Ulrich
This language keeps getting core cryptic and non-intuitive the more I read about it. What's a ddply plyr?
Karl
@Karl, that's a package, not the "core" language.
Vince
It just looks cryptic because of the function in there. `plyr` and its functions really *are* wonderful.
JoFrhwld
+7  A: 

Get the dates that have P's and those that have C's, and use intersect to find the dates that have both.

keep_dates <- with(x, intersect(date[cp_flag=='P'], date[cp_flag=='C']) )
# "1996-04-02" "1996-04-09" "1996-04-16"

Keep only the rows that have dates appearing in keep_dates.

x[ x$date %in% keep_dates, ]
#   myIdx strike_price       date     exdate cp_flag strike_price.1
# 8355342       605000 1996-04-02 1996-05-18       P         605000
# 8355433       605000 1996-04-02 1996-05-18       C         605000
# 8356541       605000 1996-04-09 1996-05-18       P         605000
# 8356629       605000 1996-04-09 1996-05-18       C         605000
# 8358033       605000 1996-04-16 1996-05-18       P         605000
# 8358119       605000 1996-04-16 1996-05-18       C         605000
wch
Elegant! I like this one a lot.
Vince
A: 

Here's a reshape approach.

library(reshape)
#Add a dummy value
df$value <- 1
check <- cast(df, myIdx + strike_price + date + exdate + strike_price + return ~ cp_flag)

#take stock of what just happened
summary(check)

#use only complete cases. If you have NAs elsewhere, this will knock out those obs too
check <- check[complete.cases(check),]

#back to original form
df.clean <- melt(check, id = 1:6)
JoFrhwld