R help - Feb 2007 - Is there better alternative to this loop?

Dear List,

Thanks to those who helped with my enquiry a few days ago.

I have a another question on loops, in this case I am trying to print out 
the row of a data frame if the previous 3 values (daily values) in col5 are 
in descending order. I have this loop which works, but ask whether this can 
be done differently (without conventional loop) in R:

flag="T"
d= 3 # d represents previous down days
for(i in (d+1): 100)
{
for( j in (i-d):(i-1))
{
if(x[j,5]<x[i,5]){flag="F"}
}
if( flag == "T"){ print(x[i,1])}
flag="T";
}

Any help appreciated,

Regards,
Alf Sammassimo
Melbourne, Australia.

You can use 'rle' to find the 'runs' of decreasing values by
looking at the
'diff'. You can change the values to get the different lengths of runs.
> # generate some test data
> set.seed(3)
> x <- runif(100)
> # use 'diff' to determine if decreasing
> x.dec <- diff(x) < 0
> # use rle to find 'runs' of 3 TRUEs
> x.rle <- rle(x.dec)
> sum(x.rle$length == 3 & x.rle$values) # how many occurances
[1] 6
> # determine the indices in the x.rle vector
> x.loc <- which(x.rle$lengths == 3 & x.rle$values)
> # print out the location of occurance
> cumsum(x.rle$lengths)[x.loc]
[1] 21 32 46 58 69 83
> x  # print out data for comparison
  [1] 0.168041526 0.807516399 0.384942351 0.327734317 0.602100675
0.604394054 0.124633444 0.294600924
  [9] 0.577609919 0.630979274 0.512015898 0.505023914 0.534035353
0.557249436 0.867919488 0.829708693
 [17] 0.111449153 0.703688359 0.897488264 0.279732554 0.228201881
0.015329893 0.128981559 0.093381929
 [25] 0.236885007 0.791147409 0.599731566 0.910147711 0.560424554
0.755704769 0.379171891 0.373280977
 [33] 0.170290643 0.453307324 0.258413960 0.336265952 0.889583035
0.201946299 0.579186043 0.207632030
 [41] 0.281468792 0.786281204 0.173019349 0.570747518 0.419282963
0.267622165 0.047809442 0.103493054
 [49] 0.314031456 0.800641062 0.229324695 0.212998441 0.877100906
0.993221963 0.844247020 0.910436549
 [57] 0.471269732 0.224418408 0.127814657 0.279683512 0.816106060
0.057612993 0.802829250 0.104387835
 [65] 0.766606019 0.304810566 0.769287421 0.540655726 0.362370532
0.092556916 0.759755255 0.760819947
 [73] 0.903260845 0.966282786 0.515256623 0.549480674 0.163719897
0.164596954 0.786332777 0.751113420
 [81] 0.784214704 0.654354915 0.378104397 0.008566967 0.955329554
0.838616148 0.213424938 0.494713556
 [89] 0.636244256 0.921091393 0.011744041 0.267402917 0.435572102
0.829467095 0.870944038 0.251068959
 [97] 0.324358248 0.306237812 0.184282035 0.679977400
>


On 2/21/07, Alfonso Sammassimo <cincinattikid@bigpond.com>
wrote:
>
> Dear List,
>
> Thanks to those who helped with my enquiry a few days ago.
>
> I have a another question on loops, in this case I am trying to print out
> the row of a data frame if the previous 3 values (daily values) in col5
> are
> in descending order. I have this loop which works, but ask whether this
> can
> be done differently (without conventional loop) in R:
>
> flag="T"
> d= 3 # d represents previous down days
> for(i in (d+1): 100)
> {
> for( j in (i-d):(i-1))
> {
> if(x[j,5]<x[i,5]){flag="F"}
> }
> if( flag == "T"){ print(x[i,1])}
> flag="T";
> }
>
> Any help appreciated,
>
> Regards,
> Alf Sammassimo
> Melbourne, Australia.
>
> ______________________________________________
> R-help@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.
>


-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem you are trying to solve?

	[[alternative HTML version deleted]]

On Thu, 2007-02-22 at 12:33 +1100, Alfonso Sammassimo
wrote:
> Dear List,
> 
> Thanks to those who helped with my enquiry a few days ago.
> 
> I have a another question on loops, in this case I am trying to print out 
> the row of a data frame if the previous 3 values (daily values) in col5 are
> in descending order. I have this loop which works, but ask whether this can
> be done differently (without conventional loop) in R:
> 
> flag="T"
> d= 3 # d represents previous down days
> for(i in (d+1): 100)
> {
> for( j in (i-d):(i-1))
> {
> if(x[j,5]<x[i,5]){flag="F"}
> }
> if( flag == "T"){ print(x[i,1])}
> flag="T";
> }
> 
> Any help appreciated,
> 
> Regards,
> Alf Sammassimo
> Melbourne, Australia.
How about this. We'll take it one step at a time.

Get the indices of the rows in the data frame, where a column value in
the prior three rows is in decreasing order.  Let's use the iris data
set as an example:
> iris$Sepal.Length
  [1] 5.1 4.9 4.7 4.6 5.0 5.4 4.6 5.0 4.4 4.9 5.4 4.8 4.8 4.3 5.8 5.7
 [17] 5.4 5.1 5.7 5.1 5.4 5.1 4.6 5.1 4.8 5.0 5.0 5.2 5.2 4.7 4.8 5.4
 [33] 5.2 5.5 4.9 5.0 5.5 4.9 4.4 5.1 5.0 4.5 4.4 5.0 5.1 4.8 5.1 4.6
 [49] 5.3 5.0 7.0 6.4 6.9 5.5 6.5 5.7 6.3 4.9 6.6 5.2 5.0 5.9 6.0 6.1
 [65] 5.6 6.7 5.6 5.8 6.2 5.6 5.9 6.1 6.3 6.1 6.4 6.6 6.8 6.7 6.0 5.7
 [81] 5.5 5.5 5.8 6.0 5.4 6.0 6.7 6.3 5.6 5.5 5.5 6.1 5.8 5.0 5.6 5.7
 [97] 5.7 6.2 5.1 5.7 6.3 5.8 7.1 6.3 6.5 7.6 4.9 7.3 6.7 7.2 6.5 6.4
[113] 6.8 5.7 5.8 6.4 6.5 7.7 7.7 6.0 6.9 5.6 7.7 6.3 6.7 7.2 6.2 6.1
[129] 6.4 7.2 7.4 7.9 6.4 6.3 6.1 7.7 6.3 6.4 6.0 6.9 6.7 6.9 5.8 6.8
[145] 6.7 6.7 6.3 6.5 6.2 5.9


Use diff() to get the differences between successive values in the
vector:
> diff(iris$Sepal.Length, 1)
  [1] -0.2 -0.2 -0.1  0.4  0.4 -0.8  0.4 -0.6  0.5  0.5 -0.6  0.0 -0.5
 [14]  1.5 -0.1 -0.3 -0.3  0.6 -0.6  0.3 -0.3 -0.5  0.5 -0.3  0.2  0.0
 [27]  0.2  0.0 -0.5  0.1  0.6 -0.2  0.3 -0.6  0.1  0.5 -0.6 -0.5  0.7
 [40] -0.1 -0.5 -0.1  0.6  0.1 -0.3  0.3 -0.5  0.7 -0.3  2.0 -0.6  0.5
 [53] -1.4  1.0 -0.8  0.6 -1.4  1.7 -1.4 -0.2  0.9  0.1  0.1 -0.5  1.1
 [66] -1.1  0.2  0.4 -0.6  0.3  0.2  0.2 -0.2  0.3  0.2  0.2 -0.1 -0.7
 [79] -0.3 -0.2  0.0  0.3  0.2 -0.6  0.6  0.7 -0.4 -0.7 -0.1  0.0  0.6
 [92] -0.3 -0.8  0.6  0.1  0.0  0.5 -1.1  0.6  0.6 -0.5  1.3 -0.8  0.2
[105]  1.1 -2.7  2.4 -0.6  0.5 -0.7 -0.1  0.4 -1.1  0.1  0.6  0.1  1.2
[118]  0.0 -1.7  0.9 -1.3  2.1 -1.4  0.4  0.5 -1.0 -0.1  0.3  0.8  0.2
[131]  0.5 -1.5 -0.1 -0.2  1.6 -1.4  0.1 -0.4  0.9 -0.2  0.2 -1.1  1.0
[144] -0.1  0.0 -0.4  0.2 -0.3 -0.3


Then use sign() to get the sign of the differences, positive, negative
or zero:
> sign(diff(iris$Sepal.Length, 1))
  [1] -1 -1 -1  1  1 -1  1 -1  1  1 -1  0 -1  1 -1 -1 -1  1 -1  1 -1 -1
 [23]  1 -1  1  0  1  0 -1  1  1 -1  1 -1  1  1 -1 -1  1 -1 -1 -1  1  1
 [45] -1  1 -1  1 -1  1 -1  1 -1  1 -1  1 -1  1 -1 -1  1  1  1 -1  1 -1
 [67]  1  1 -1  1  1  1 -1  1  1  1 -1 -1 -1 -1  0  1  1 -1  1  1 -1 -1
 [89] -1  0  1 -1 -1  1  1  0  1 -1  1  1 -1  1 -1  1  1 -1  1 -1  1 -1
[111] -1  1 -1  1  1  1  1  0 -1  1 -1  1 -1  1  1 -1 -1  1  1  1  1 -1
[133] -1 -1  1 -1  1 -1  1 -1  1 -1  1 -1  0 -1  1 -1 -1


So, now we essentially want to get the index for a value where it and
the previous value are -1, indicating a decreasing sequence of 3 in the
original data vector. The easiest way to do this may be to actually look
for a moving average of -1 over two successive values in moving windows.
In this case, we use filter():
> which(filter(sign(diff(iris$Sepal.Length)), rep(1/2, 2), 
               sides = 1) == -1)
 [1]   2   3  16  17  22  38  41  42  60  78  79  80  88  89  93 111 127
[18] 133 134 149


That gives us the index of the second value in the sign() sequence.
Since we want the index of the 4th value in the original data vector, we
add 2:
> which(filter(sign(diff(iris$Sepal.Length)), rep(1/2, 2), 
               sides = 1) == -1) + 2
 [1]   4   5  18  19  24  40  43  44  62  80  81  82  90  91  95 113 129
[18] 135 136 151


So these values now should correspond to the indices of the values in
iris$Sepal.Length, where the three preceding values are in a decreasing
sequence.

So to apply that to your dataset, something like the following should
work:

Ind <- which(filter(sign(diff(DF$Col5)), rep(1/2, 2), 
                    sides = 1) == -1) + 2

Then print the rows with:

  DF[Ind, ]

HTH,

Marc Schwartz