R help - Sep 2013 - how to do this trimming/selecting in canonical R?

This is better explained by example:
> A <- data.frame(force = sort(runif(10, 0, 1)), condition =  
> sort(sample(0:100, 10)))
> B <- data.frame(counterforce = sort(runif(15, 0, 1), decreasing=T),  
> condition = sort(sample(0:100, 15)))
So we have:
> A
         force condition
1  0.03515542         1
2  0.13267882        13
3  0.26155689        24
4  0.37453142        38
5  0.39360520        45 <--- trim everything after this
6  0.43924737        48
7  0.47669800        50
8  0.57044795        51
9  0.81177499        61
10 0.98860450        94
> B
    counterforce condition
1   0.965769548         2
2   0.965266255         5
3   0.846941244         7
4   0.818013029        11
5   0.813139978        22
6   0.730599939        34
7   0.715985436        39
8   0.658073895        40
9   0.421264948        42 <--- trim everything after this
10  0.373774505        52
11  0.242191461        62
12  0.090584590        63
13  0.070020635        68
14  0.067366062        83
15  0.001585313        84

I need to trim away rows after No. 5, from A, trim away rows after No.  
9, from B.

Because

A[5, condition] > max(B[1:9, condition] && A[5, force] > B[9+1,
counterforce]

In a general way, I am looking for x and y, where:

A[x, condition] > max(B[1:y, condition] && A[x, force] > B[y+1,
counterforce]

and I will select A[1:x,] and B[1:y,], or trim away the rest, because  
they are irrelevent for the calculation onwards.

This is easy to do it in C, and I actually have done it in C-like R  
script, by looping through all rows of A, and breaking from the loop  
when finding the matching trim-point in B. But I am learning R, so  
what is the native way to do it in R?


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On Sat, Sep 14, 2013 at 12:35 PM,  <gildororonar at mail-on.us>
wrote:
> This is better explained by example:
>
>> A <- data.frame(force = sort(runif(10, 0, 1)), condition >>
sort(sample(0:100, 10)))
>> B <- data.frame(counterforce = sort(runif(15, 0, 1), decreasing=T),
>> condition = sort(sample(0:100, 15)))
>
>
> So we have:
>
>> A
>
>         force condition
> 1  0.03515542         1
> 2 0.13267882        13
> 3 0.26155689        24
> 4 0.37453142        38
> 5 0.39360520        45 <--- trim everything after this
> 6 0.43924737        48
> 7 0.47669800        50
> 8 0.57044795        51
> 9  0.81177499        61
> 10 0.98860450        94
>
>> B
>
>    counterforce condition
> 1   0.965769548         2
> 2   0.965266255         5
> 3   0.846941244         7
> 4   0.818013029        11
> 5   0.813139978        22
> 6   0.730599939        34
> 7   0.715985436        39
> 8   0.658073895        40
> 9   0.421264948        42 <--- trim everything after this
> 10  0.373774505        52
> 11  0.242191461        62
> 12  0.090584590        63
> 13  0.070020635        68
> 14  0.067366062        83
> 15  0.001585313        84
>
> I need to trim away rows after No. 5, from A, trim away rows after No. 9,
> from B.
>
> Because
>
> A[5, condition] > max(B[1:9, condition] && A[5, force] >
B[9+1,
> counterforce]
>
> In a general way, I am looking for x and y, where:
>
> A[x, condition] > max(B[1:y, condition] && A[x, force] >
B[y+1,
> counterforce]
>
> and I will select A[1:x,] and B[1:y,], or trim away the rest, because they
> are irrelevent for the calculation onwards.
Try this:

library(sqldf)

r <- sqldf("select a.condition,  b1.condition
   from A a, B b1, B b2
   where a.condition > b1.condition
   and a.force > b2.counterforce
   and b2.rowid = b1.rowid + 1
   order by a.condition, b1.condition
   limit 1")

Atrim <- fn$sqldf("select * from A where condition <= `r[1]` ")
Btrim <- fn$sqldf("select * from B where condition <= `r[2]` ")

giving:
> Atrim
       force condition
1 0.03515542         1
2 0.13267882        13
3 0.26155689        24
4 0.37453142        38
5 0.39360520        45
> Btrim
  counterforce condition
1    0.9657695         2
2    0.9652663         5
3    0.8469412         7
4    0.8180130        11
5    0.8131400        22
6    0.7305999        34
7    0.7159854        39
8    0.6580739        40
9    0.4212649        42

On 14-09-2013, at 18:35, gildororonar at mail-on.us wrote:
> This is better explained by example:
> 
>> A <- data.frame(force = sort(runif(10, 0, 1)), condition =
sort(sample(0:100, 10)))
>> B <- data.frame(counterforce = sort(runif(15, 0, 1), decreasing=T),
condition = sort(sample(0:100, 15)))
> 
> So we have:
> 
>> A
>        force condition
> 1  0.03515542         1
> 2  0.13267882        13
> 3  0.26155689        24
> 4  0.37453142        38
> 5  0.39360520        45 <--- trim everything after this
> 6  0.43924737        48
> 7  0.47669800        50
> 8  0.57044795        51
> 9  0.81177499        61
> 10 0.98860450        94
> 
>> B
>   counterforce condition
> 1   0.965769548         2
> 2   0.965266255         5
> 3   0.846941244         7
> 4   0.818013029        11
> 5   0.813139978        22
> 6   0.730599939        34
> 7   0.715985436        39
> 8   0.658073895        40
> 9   0.421264948        42 <--- trim everything after this
> 10  0.373774505        52
> 11  0.242191461        62
> 12  0.090584590        63
> 13  0.070020635        68
> 14  0.067366062        83
> 15  0.001585313        84
> 
> I need to trim away rows after No. 5, from A, trim away rows after No. 9,
from B.
> 
> Because
> 
> A[5, condition] > max(B[1:9, condition] && A[5, force] >
B[9+1, counterforce]
> 
> In a general way, I am looking for x and y, where:
> 
> A[x, condition] > max(B[1:y, condition] && A[x, force] >
B[y+1, counterforce]
> 
> and I will select A[1:x,] and B[1:y,], or trim away the rest, because they
are irrelevent for the calculation onwards.
> 
> This is easy to do it in C, and I actually have done it in C-like R script,
by looping through all rows of A, and breaking from the loop when finding the
matching trim-point in B. But I am learning R, so what is the native way to do
it in R?
Your trim-point in B is not unique (at least for the data you provided).
Use a loop in R like this

for( x in seq_len(nrow(A)) ) {
    for( y in seq_len(nrow(B)-1) ) {
        res1 <- A[x,"condition"] >
max(B[1:y,"condition"])
        res2 <- A[x,"force"] > B[y+1,"counterforce"]
        res <- res1 && res2 
        if(res)
cat("x=",x,"y=",y,"res=",res,"\n")
    }
}

Result is:

# x= 5 y= 9 res= TRUE 
# x= 6 y= 8 res= TRUE 
# x= 6 y= 9 res= TRUE 
# x= 7 y= 8 res= TRUE 
# x= 7 y= 9 res= TRUE 
# x= 8 y= 8 res= TRUE 
# x= 8 y= 9 res= TRUE 
# x= 9 y= 5 res= TRUE 
# x= 9 y= 6 res= TRUE 
# x= 9 y= 7 res= TRUE 
# x= 9 y= 8 res= TRUE 
# x= 9 y= 9 res= TRUE 
# x= 9 y= 10 res= TRUE 
# x= 10 y= 1 res= TRUE 
# x= 10 y= 2 res= TRUE 
# x= 10 y= 3 res= TRUE 
# x= 10 y= 4 res= TRUE 
# x= 10 y= 5 res= TRUE 
# x= 10 y= 6 res= TRUE 
# x= 10 y= 7 res= TRUE 
# x= 10 y= 8 res= TRUE 
# x= 10 y= 9 res= TRUE 
# x= 10 y= 10 res= TRUE 
# x= 10 y= 11 res= TRUE 
# x= 10 y= 12 res= TRUE 
# x= 10 y= 13 res= TRUE 
# x= 10 y= 14 res= TRUE 

If you want a unique answer you'll need additional restrictions.

Berend