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 // Copyright 2015 The Go Authors. All rights reserved. // Use of this source code is governed by a BSD-style // license that can be found in the LICENSE file. package stats import ( "math" ) // A UDist is the discrete probability distribution of the // Mann-Whitney U statistic for a pair of samples of sizes N1 and N2. // // The details of computing this distribution with no ties can be // found in Mann, Henry B.; Whitney, Donald R. (1947). "On a Test of // Whether one of Two Random Variables is Stochastically Larger than // the Other". Annals of Mathematical Statistics 18 (1): 50–60. // Computing this distribution in the presence of ties is described in // Klotz, J. H. (1966). "The Wilcoxon, Ties, and the Computer". // Journal of the American Statistical Association 61 (315): 772-787 // and Cheung, Ying Kuen; Klotz, Jerome H. (1997). "The Mann Whitney // Wilcoxon Distribution Using Linked Lists". Statistica Sinica 7: // 805-813 (the former paper contains details that are glossed over in // the latter paper but has mathematical typesetting issues, so it's // easiest to get the context from the former paper and the details // from the latter). type UDist struct { N1, N2 int // T is the count of the number of ties at each rank in the // input distributions. T may be nil, in which case it is // assumed there are no ties (which is equivalent to an M+N // slice of 1s). It must be the case that Sum(T) == M+N. T []int } // hasTies returns true if d has any tied samples. func (d UDist) hasTies() bool { for _, t := range d.T { if t > 1 { return true } } return false } // p returns the p_{d.N1,d.N2} function defined by Mann, Whitney 1947 // for values of U from 0 up to and including the U argument. // // This algorithm runs in Θ(N1*N2*U) = O(N1²N2²) time and is quite // fast for small values of N1 and N2. However, it does not handle ties. func (d UDist) p(U int) []float64 { // This is a dynamic programming implementation of the // recursive recurrence definition given by Mann and Whitney: // // p_{n,m}(U) = (n * p_{n-1,m}(U-m) + m * p_{n,m-1}(U)) / (n+m) // p_{n,m}(U) = 0 if U < 0 // p_{0,m}(U) = p{n,0}(U) = 1 / nCr(m+n, n) if U = 0 // = 0 if U > 0 // // (Note that there is a typo in the original paper. The first // recursive application of p should be for U-m, not U-M.) // // Since p{n,m} only depends on p{n-1,m} and p{n,m-1}, we only // need to store one "plane" of the three dimensional space at // a time. // // Furthermore, p_{n,m} = p_{m,n}, so we only construct values // for n <= m and obtain the rest through symmetry. // // We organize the computed values of p as followed: // // n → N // m * // ↓ * * // * * * // * * * * // * * * * // M * * * * // // where each * is a slice indexed by U. The code below // computes these left-to-right, top-to-bottom, so it only // stores one row of this matrix at a time. Furthermore, // computing an element in a given U slice only depends on the // same and smaller values of U, so we can overwrite the U // slice we're computing in place as long as we start with the // largest value of U. Finally, even though the recurrence // depends on (n,m) above the diagonal and we use symmetry to // mirror those across the diagonal to (m,n), the mirrored // indexes are always available in the current row, so this // mirroring does not interfere with our ability to recycle // state. N, M := d.N1, d.N2 if N > M { N, M = M, N } memo := make([][]float64, N+1) for n := range memo { memo[n] = make([]float64, U+1) } for m := 0; m <= M; m++ { // Compute p_{0,m}. This is zero except for U=0. memo[0][0] = 1 // Compute the remainder of this row. nlim := N if m < nlim { nlim = m } for n := 1; n <= nlim; n++ { lp := memo[n-1] // p_{n-1,m} var rp []float64 if n <= m-1 { rp = memo[n] // p_{n,m-1} } else { rp = memo[m-1] // p{m-1,n} and m==n } // For a given n,m, U is at most n*m. // // TODO: Actually, it's at most ⌈n*m/2⌉, but // then we need to use more complex symmetries // in the inner loop below. ulim := n * m if U < ulim { ulim = U } out := memo[n] // p_{n,m} nplusm := float64(n + m) for U1 := ulim; U1 >= 0; U1-- { l := 0.0 if U1-m >= 0 { l = float64(n) * lp[U1-m] } r := float64(m) * rp[U1] out[U1] = (l + r) / nplusm } } } return memo[N] } type ukey struct { n1 int // size of first sample twoU int // 2*U statistic for this permutation } // This computes the cumulative counts of the Mann-Whitney U // distribution in the presence of ties using the computation from // Cheung, Ying Kuen; Klotz, Jerome H. (1997). "The Mann Whitney // Wilcoxon Distribution Using Linked Lists". Statistica Sinica 7: // 805-813, with much guidance from appendix L of Klotz, A // Computational Approach to Statistics. // // makeUmemo constructs a table memo[K][ukey{n1, 2*U}], where K is the // number of ranks (up to len(t)), n1 is the size of the first sample // (up to the n1 argument), and U is the U statistic (up to the // argument twoU/2). The value of an entry in the memo table is the // number of permutations of a sample of size n1 in a ranking with tie // vector t[:K] having a U statistic <= U. func makeUmemo(twoU, n1 int, t []int) []map[ukey]float64 { // Another candidate for a fast implementation is van de Wiel, // "The split-up algorithm: a fast symbolic method for // computing p-values of distribution-free statistics". This // is what's used by R's coin package. It's a comparatively // recent publication, so it's presumably faster (or perhaps // just more general) than previous techniques, but I can't // get my hands on the paper. // // TODO: ~40% of this function's time is spent in mapassign on // the assignment lines in the two loops and another ~20% in // map access and iteration. Improving map behavior or // replacing the maps altogether with some other constant-time // structure could double performance. // // TODO: The worst case for this function is when there are // few ties. Yet the best case overall is when there are *no* // ties. Can we get the best of both worlds? Use the fast // algorithm for the most part when there are few ties and mix // in the general algorithm just where we need it? That's // certainly possible for sub-problems where t[:k] has no // ties, but that doesn't help if t[0] has a tie but nothing // else does. Is it possible to rearrange the ranks without // messing up our computation of the U statistic for // sub-problems? K := len(t) // Compute a coefficients. The a slice is indexed by k (a[0] // is unused). a := make([]int, K+1) a[1] = t[0] for k := 2; k <= K; k++ { a[k] = a[k-1] + t[k-2] + t[k-1] } // Create the memo table for the counts function, A. The A // slice is indexed by k (A[0] is unused). // // In "The Mann Whitney Distribution Using Linked Lists", they // use linked lists (*gasp*) for this, but within each K it's // really just a memoization table, so it's faster to use a // map. The outer structure is a slice indexed by k because we // need to find all memo entries with certain values of k. // // TODO: The n1 and twoU values in the ukeys follow strict // patterns. For each K value, the n1 values are every integer // between two bounds. For each (K, n1) value, the twoU values // are every integer multiple of a certain base between two // bounds. It might be worth turning these into directly // indexible slices. A := make([]map[ukey]float64, K+1) A[K] = map[ukey]float64{ukey{n1: n1, twoU: twoU}: 0} // Compute memo table (k, n1, twoU) triples from high K values // to low K values. This drives the recurrence relation // downward to figure out all of the needed argument triples. // // TODO: Is it possible to generate this table bottom-up? If // so, this could be a pure dynamic programming algorithm and // we could discard the K dimension. We could at least store // the inputs in a more compact representation that replaces // the twoU dimension with an interval and a step size (as // suggested by Cheung, Klotz, not that they make it at all // clear *why* they're suggesting this). tsum := sumint(t) // always ∑ t[0:k] for k := K - 1; k >= 2; k-- { tsum -= t[k] A[k] = make(map[ukey]float64) // Construct A[k] from A[k+1]. for A_kplus1 := range A[k+1] { rkLow := maxint(0, A_kplus1.n1-tsum) rkHigh := minint(A_kplus1.n1, t[k]) for rk := rkLow; rk <= rkHigh; rk++ { twoU_k := A_kplus1.twoU - rk*(a[k+1]-2*A_kplus1.n1+rk) n1_k := A_kplus1.n1 - rk if twoUmin(n1_k, t[:k], a) <= twoU_k && twoU_k <= twoUmax(n1_k, t[:k], a) { key := ukey{n1: n1_k, twoU: twoU_k} A[k][key] = 0 } } } } // Fill counts in memo table from low K values to high K // values. This unwinds the recurrence relation. // Start with K==2 base case. // // TODO: Later computations depend on these, but these don't // depend on anything (including each other), so if K==2, we // can skip the memo table altogether. if K < 2 { panic("K < 2") } N_2 := t[0] + t[1] for A_2i := range A[2] { Asum := 0.0 r2Low := maxint(0, A_2i.n1-t[0]) r2High := (A_2i.twoU - A_2i.n1*(t[0]-A_2i.n1)) / N_2 for r2 := r2Low; r2 <= r2High; r2++ { Asum += mathChoose(t[0], A_2i.n1-r2) * mathChoose(t[1], r2) } A[2][A_2i] = Asum } // Derive counts for the rest of the memo table. tsum = t[0] // always ∑ t[0:k-1] for k := 3; k <= K; k++ { tsum += t[k-2] // Compute A[k] counts from A[k-1] counts. for A_ki := range A[k] { Asum := 0.0 rkLow := maxint(0, A_ki.n1-tsum) rkHigh := minint(A_ki.n1, t[k-1]) for rk := rkLow; rk <= rkHigh; rk++ { twoU_kminus1 := A_ki.twoU - rk*(a[k]-2*A_ki.n1+rk) n1_kminus1 := A_ki.n1 - rk x, ok := A[k-1][ukey{n1: n1_kminus1, twoU: twoU_kminus1}] if !ok && twoUmax(n1_kminus1, t[:k-1], a) < twoU_kminus1 { x = mathChoose(tsum, n1_kminus1) } Asum += x * mathChoose(t[k-1], rk) } A[k][A_ki] = Asum } } return A } func twoUmin(n1 int, t, a []int) int { K := len(t) twoU := -n1 * n1 n1_k := n1 for k := 1; k <= K; k++ { twoU_k := minint(n1_k, t[k-1]) twoU += twoU_k * a[k] n1_k -= twoU_k } return twoU } func twoUmax(n1 int, t, a []int) int { K := len(t) twoU := -n1 * n1 n1_k := n1 for k := K; k > 0; k-- { twoU_k := minint(n1_k, t[k-1]) twoU += twoU_k * a[k] n1_k -= twoU_k } return twoU } func (d UDist) PMF(U float64) float64 { if U < 0 || U >= 0.5+float64(d.N1*d.N2) { return 0 } if d.hasTies() { // makeUmemo computes the CDF directly. Take its // difference to get the PMF. p1, ok1 := makeUmemo(int(2*U)-1, d.N1, d.T)[len(d.T)][ukey{d.N1, int(2*U) - 1}] p2, ok2 := makeUmemo(int(2*U), d.N1, d.T)[len(d.T)][ukey{d.N1, int(2 * U)}] if !ok1 || !ok2 { panic("makeUmemo did not return expected memoization table") } return (p2 - p1) / mathChoose(d.N1+d.N2, d.N1) } // There are no ties. Use the fast algorithm. U must be integral. Ui := int(math.Floor(U)) // TODO: Use symmetry to minimize U return d.p(Ui)[Ui] } func (d UDist) CDF(U float64) float64 { if U < 0 { return 0 } else if U >= float64(d.N1*d.N2) { return 1 } if d.hasTies() { // TODO: Minimize U? p, ok := makeUmemo(int(2*U), d.N1, d.T)[len(d.T)][ukey{d.N1, int(2 * U)}] if !ok { panic("makeUmemo did not return expected memoization table") } return p / mathChoose(d.N1+d.N2, d.N1) } // There are no ties. Use the fast algorithm. U must be integral. Ui := int(math.Floor(U)) // The distribution is symmetric around U = m * n / 2. Sum up // whichever tail is smaller. flip := Ui >= (d.N1*d.N2+1)/2 if flip { Ui = d.N1*d.N2 - Ui - 1 } pdfs := d.p(Ui) p := 0.0 for _, pdf := range pdfs[:Ui+1] { p += pdf } if flip { p = 1 - p } return p } func (d UDist) Step() float64 { return 0.5 } func (d UDist) Bounds() (float64, float64) { // TODO: More precise bounds when there are ties. return 0, float64(d.N1 * d.N2) }