epclust/tests/testthat/test.clustering.R

   1 context("clustering")
   2
   3 test_that("computeSynchrones behave as expected",
   4 {
   5     n = 300
   6     x = seq(0,9.5,0.1)
   7     L = length(x) #96 1/4h
   8     K = 3
   9     s1 = cos(x)
  10     s2 = sin(x)
  11     s3 = c( s1[1:(L%/%2)] , s2[(L%/%2+1):L] )
  12     #sum((s1-s2)^2) == 96
  13     #sum((s1-s3)^2) == 58
  14     #sum((s2-s3)^2) == 38
  15     s = list(s1, s2, s3)
  16     series = matrix(nrow=L, ncol=n)
  17     for (i in seq_len(n))
  18         series[,i] = s[[I(i,K)]] + rnorm(L,sd=0.01)
  19     getRefSeries = function(indices) {
  20         indices = indices[indices <= n]
  21         if (length(indices)>0) series[,indices] else NULL
  22     }
  23     synchrones = computeSynchrones(bigmemory::as.big.matrix(cbind(s1,s2,s3)), getRefSeries,
  24         n, 100, sync_mean=TRUE, verbose=TRUE, parll=FALSE)
  25
  26     expect_equal(dim(synchrones), c(L,K))
  27     for (i in 1:K)
  28         expect_equal(synchrones[,i], s[[i]], tolerance=0.01)
  29 })
  30
  31 # Helper function to divide indices into balanced sets
  32 test_that("Helper function to spread indices work properly",
  33 {
  34     indices <- 1:400
  35
  36     # bigger nb_per_set than length(indices)
  37     expect_equal(epclust:::.spreadIndices(indices,500), list(indices))
  38
  39     # nb_per_set == length(indices)
  40     expect_equal(epclust:::.spreadIndices(indices,400), list(indices))
  41
  42     # length(indices) %% nb_per_set == 0
  43     expect_equal(epclust:::.spreadIndices(indices,200),
  44         c( list(indices[1:200]), list(indices[201:400]) ))
  45     expect_equal(epclust:::.spreadIndices(indices,100),
  46         c( list(indices[1:100]), list(indices[101:200]),
  47             list(indices[201:300]), list(indices[301:400]) ))
  48
  49     # length(indices) / nb_per_set == 1, length(indices) %% nb_per_set == 100
  50     expect_equal(epclust:::.spreadIndices(indices,300), list(indices))
  51     # length(indices) / nb_per_set == 2, length(indices) %% nb_per_set == 42
  52     repartition <- epclust:::.spreadIndices(indices,179)
  53     expect_equal(length(repartition), 2)
  54     expect_equal(length(repartition[[1]]), 179 + 21)
  55     expect_equal(length(repartition[[1]]), 179 + 21)
  56 })
  57
  58 test_that("clusteringTask1 behave as expected",
  59 {
  60     n = 900
  61     x = seq(0,9.5,0.1)
  62     L = length(x) #96 1/4h
  63     K1 = 60
  64     s = lapply( seq_len(K1), function(i) x^(1+i/30)*cos(x+i) )
  65     series = matrix(nrow=n, ncol=L)
  66     for (i in seq_len(n))
  67         series[,i] = s[[I(i,K1)]] + rnorm(L,sd=0.01)
  68     getSeries = function(indices) {
  69         indices = indices[indices <= n]
  70         if (length(indices)>0) series[,indices] else NULL
  71     }
  72     wf = "haar"
  73     ctype = "absolute"
  74     getContribs = function(indices) curvesToContribs(series[,indices],wf,ctype)
  75     indices1 = clusteringTask1(1:n, getContribs, K1, 75, verbose=TRUE, parll=FALSE)
  76     medoids_K1 = getSeries(indices1)
  77
  78     expect_equal(dim(medoids_K1), c(L,K1))
  79     # Not easy to evaluate result: at least we expect it to be better than random selection of
  80     # medoids within initial series
  81     distorGood = computeDistortion(series, medoids_K1)
  82     for (i in 1:3)
  83         expect_lte( distorGood, computeDistortion(series,series[,sample(1:n, K1)]) )
  84 })
  85
  86 test_that("clusteringTask2 behave as expected",
  87 {
  88     n = 900
  89     x = seq(0,9.5,0.1)
  90     L = length(x) #96 1/4h
  91     K1 = 60
  92     K2 = 3
  93     #for (i in 1:60) {plot(x^(1+i/30)*cos(x+i),type="l",col=i,ylim=c(-50,50)); par(new=TRUE)}
  94     s = lapply( seq_len(K1), function(i) x^(1+i/30)*cos(x+i) )
  95     series = matrix(nrow=n, ncol=L)
  96     for (i in seq_len(n))
  97         series[i,] = s[[I(i,K1)]] + rnorm(L,sd=0.01)
  98     getRefSeries = function(indices) {
  99         indices = indices[indices <= n]
 100         if (length(indices)>0) series[,indices] else NULL
 101     }
 102     # Artificially simulate 60 medoids - perfect situation, all equal to one of the refs
 103     medoids_K1 = bigmemory::as.big.matrix( sapply( 1:K1, function(i) s[[I(i,K1)]] ) )
 104     medoids_K2 = clusteringTask2(medoids_K1, K2, getRefSeries, n, 75, verbose=TRUE, parll=FALSE)
 105
 106     expect_equal(dim(medoids_K2), c(L,K2))
 107     # Not easy to evaluate result: at least we expect it to be better than random selection of
 108     # medoids within 1...K1 (among references)
 109     distorGood = computeDistortion(series, medoids_K2)
 110     for (i in 1:3)
 111         expect_lte( distorGood, computeDistortion(series,medoids_K1[,sample(1:K1, K2)]) )
 112 })