3 #shorthand: map 1->1, 2->2, 3->3, 4->1, ..., 149->2, 150->3, ... (is base==3)
7 test_that("computeClusters1&2 behave as expected",
9 require("MASS", quietly=TRUE)
10 if (!require("clue", quietly=TRUE))
11 skip("'clue' package not available")
13 # 3 gaussian clusters, 300 items; and then 7 gaussian clusters, 490 items
17 for (ndK in list( c(300,5,3), c(490,10,7) ))
19 n = ndK[1] ; d = ndK[2] ; K = ndK[3]
20 cs = n/K #cluster size
22 coefs = do.call(rbind,
23 lapply(1:K, function(i) MASS::mvrnorm(cs, c(rep(0,(i-1)),5,rep(0,d-i)), Id)))
24 indices_medoids1 = computeClusters1(coefs, K, verbose=TRUE)
25 indices_medoids2 = computeClusters2(dist(coefs), K, verbose=TRUE)
26 # Get coefs assignments (to medoids)
27 assignment1 = sapply(seq_len(n), function(i)
28 which.min( rowSums( sweep(coefs[indices_medoids1,],2,coefs[i,],'-')^2 ) ) )
29 assignment2 = sapply(seq_len(n), function(i)
30 which.min( rowSums( sweep(coefs[indices_medoids2,],2,coefs[i,],'-')^2 ) ) )
33 expect_equal(sum(assignment1==i), cs, tolerance=5)
34 expect_equal(sum(assignment2==i), cs, tolerance=5)
37 costs_matrix1 = matrix(nrow=K,ncol=K)
38 costs_matrix2 = matrix(nrow=K,ncol=K)
43 # assign i (in result) to j (order 1,2,3)
44 costs_matrix1[i,j] = abs( mean(assignment1[((i-1)*cs+1):(i*cs)]) - j )
45 costs_matrix2[i,j] = abs( mean(assignment2[((i-1)*cs+1):(i*cs)]) - j )
48 permutation1 = as.integer( clue::solve_LSAP(costs_matrix1) )
49 permutation2 = as.integer( clue::solve_LSAP(costs_matrix2) )
53 mean(assignment1[((i-1)*cs+1):(i*cs)]), permutation1[i], tolerance=0.05)
55 mean(assignment2[((i-1)*cs+1):(i*cs)]), permutation2[i], tolerance=0.05)
60 test_that("computeSynchrones behave as expected",
64 L = length(x) #96 1/4h
68 s3 = c( s1[1:(L%/%2)] , s2[(L%/%2+1):L] )
73 series = matrix(nrow=n, ncol=L)
75 series[i,] = s[[I(i,K)]] + rnorm(L,sd=0.01)
76 getRefSeries = function(indices) {
77 indices = indices[indices <= n]
78 if (length(indices)>0) series[indices,] else NULL
80 synchrones = computeSynchrones(bigmemory::as.big.matrix(rbind(s1,s2,s3)), getRefSeries,
81 n, 100, verbose=TRUE, parll=FALSE)
83 expect_equal(dim(synchrones), c(K,L))
85 expect_equal(synchrones[i,], s[[i]], tolerance=0.01)
88 # NOTE: medoids can be a big.matrix
89 computeDistortion = function(series, medoids)
91 n = nrow(series) ; L = ncol(series)
93 if (bigmemory::is.big.matrix(medoids))
96 distortion = distortion + min( rowSums( sweep(medoids,2,series[i,],'-')^2 ) / L )
100 test_that("clusteringTask1 behave as expected",
104 L = length(x) #96 1/4h
106 s = lapply( seq_len(K1), function(i) x^(1+i/30)*cos(x+i) )
107 series = matrix(nrow=n, ncol=L)
108 for (i in seq_len(n))
109 series[i,] = s[[I(i,K1)]] + rnorm(L,sd=0.01)
110 getSeries = function(indices) {
111 indices = indices[indices <= n]
112 if (length(indices)>0) series[indices,] else NULL
116 getContribs = function(indices) curvesToContribs(series[indices,],wf,ctype)
117 indices1 = clusteringTask1(1:n, getContribs, K1, 75, verbose=TRUE, parll=FALSE)
118 medoids_K1 = getSeries(indices1)
120 expect_equal(dim(medoids_K1), c(K1,L))
121 # Not easy to evaluate result: at least we expect it to be better than random selection of
122 # medoids within initial series
123 distorGood = computeDistortion(series, medoids_K1)
125 expect_lte( distorGood, computeDistortion(series,series[sample(1:n, K1),]) )
128 test_that("clusteringTask2 behave as expected",
132 L = length(x) #96 1/4h
135 #for (i in 1:60) {plot(x^(1+i/30)*cos(x+i),type="l",col=i,ylim=c(-50,50)); par(new=TRUE)}
136 s = lapply( seq_len(K1), function(i) x^(1+i/30)*cos(x+i) )
137 series = matrix(nrow=n, ncol=L)
138 for (i in seq_len(n))
139 series[i,] = s[[I(i,K1)]] + rnorm(L,sd=0.01)
140 getRefSeries = function(indices) {
141 indices = indices[indices <= n]
142 if (length(indices)>0) series[indices,] else NULL
144 # Artificially simulate 60 medoids - perfect situation, all equal to one of the refs
145 medoids_K1 = bigmemory::as.big.matrix(
146 do.call(rbind, lapply( 1:K1, function(i) s[[I(i,K1)]] ) ) )
147 medoids_K2 = clusteringTask2(medoids_K1, K2, getRefSeries, n, 75, verbose=TRUE, parll=FALSE)
149 expect_equal(dim(medoids_K2), c(K2,L))
150 # Not easy to evaluate result: at least we expect it to be better than random selection of
151 # medoids within 1...K1 (among references)
152 distorGood = computeDistortion(series, medoids_K2)
154 expect_lte( distorGood, computeDistortion(series,medoids_K1[sample(1:K1, K2),]) )
157 #NOTE: rather redundant test
158 #test_that("clusteringTask1 + clusteringTask2 behave as expected",
162 # L = length(x) #96 1/4h
165 # s = lapply( seq_len(K1), function(i) x^(1+i/30)*cos(x+i) )
166 # series = matrix(nrow=n, ncol=L)
167 # for (i in seq_len(n))
168 # series[i,] = s[[I(i,K1)]] + rnorm(L,sd=0.01)
169 # getSeries = function(indices) {
170 # indices = indices[indices <= n]
171 # if (length(indices)>0) series[indices,] else NULL
175 # getContribs = function(indices) curvesToContribs(series[indices,],wf,ctype)
176 # require("bigmemory", quietly=TRUE)
177 # indices1 = clusteringTask1(1:n, getContribs, K1, 75, verbose=TRUE, parll=FALSE)
178 # medoids_K1 = bigmemory::as.big.matrix( getSeries(indices1) )
179 # medoids_K2 = clusteringTask2(medoids_K1, K2, getSeries, n, 120, verbose=TRUE, parll=FALSE)
181 # expect_equal(dim(medoids_K1), c(K1,L))
182 # expect_equal(dim(medoids_K2), c(K2,L))
183 # # Not easy to evaluate result: at least we expect it to be better than random selection of
184 # # medoids within 1...K1 (among references)
185 # distorGood = computeDistortion(series, medoids_K2)
187 # expect_lte( distorGood, computeDistortion(series,medoids_K1[sample(1:K1, K2),]) )