select rows by criteria
On Thu, Mar 01, 2012 at 05:42:48PM +0100, Petr Savicky wrote:
On Thu, Mar 01, 2012 at 04:27:45AM -0800, syrvn wrote:
Hello, I am stuck with selecting the right rows from a data frame. I think the problem is rather how to select them then how to implement the R code. Consider the following data frame: df <- data.frame(ID = c(1,2,3,4,5,6,7,8,9,10), value = c(34,12,23,25,34,42,48,29,30,27)) What I want to achieve is to select 7 rows (values) so that the mean value of those rows are closest to the value of 35 and the remaining 3 rows (values) are closest to 45. However, each value is only allowed to be sampled once!
Hi. If some 3 rows have mean close to 45, then they have sum close to 3*45, so the remaining 7 rows have sum close to sum(df$value) - 3*45 # [1] 169 and they have mean close to 169/7 = 24.14286. In other words, the two criteria cannot be optimized together. For this reason, let me choose the criterion on 3 rows. The closest solution may be found as follows. # generate all triples and compute their means tripleMeans <- colMeans(combn(df$value, 3)) # select the index of the triple with mean closest to 35 indClosest <- which.min(abs(tripleMeans - 35)) # generate the indices, which form the closest triple in df$value tripleInd <- combn(1:length(df$value), 3)[, indClosest] tripleInd # [1] 1 3 7 # check the mean of the triple mean(df$value[tripleInd]) # [1] 35 This code constructs all triples. If it is used for k-tuples for a larger k and for a set of n values, its complexity will be proportional to choose(n, k), so it will be large even for moderate n, k. It is hard to provide a significant speed up, since some variants of "knapsack problem", which is NP-complete, may be reduced to your question. Consequently, it is, in general, NP-complete.
Hi. Also this statement requires a correction. It applies to the search of an exact optimum if the numbers in df$value are large. There are efficient algorithms, which find an approximate solution. Also, if the numbers in df$value are integers (or may be rounded to integers after an appropriate scaling), then there is an algorithm, whose complexity is O(k*n*max(df$value)). This may be significantly less than choose(n, k). CRAN task view Optimization and Mathematical Programming http://cran.at.r-project.org/web/views/Optimization.html may suggest also other solutions. Petr Savicky.