132485adb0
Followed the Migration Guide at https://cli.urfave.org/migrate-v1-to-v2/ The major changes not pointed out in the migration guide are: - context.Args() no longer produces a []slice, so context.Args().Slice() in substitued - All cli.Global***** are deprecated (the migration guide is somewhat unclear on this) Signed-off-by: Derek Nola <derek.nola@suse.com> Vendor in urfave cli/v2 Signed-off-by: Derek Nola <derek.nola@suse.com> Fix NewStringSlice calls Signed-off-by: Derek Nola <derek.nola@suse.com>
95 lines
2.3 KiB
Go
95 lines
2.3 KiB
Go
package smetrics
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import (
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"math"
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)
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// The Ukkonen algorithm for calculating the Levenshtein distance. The algorithm is described in http://www.cs.helsinki.fi/u/ukkonen/InfCont85.PDF, or in docs/InfCont85.PDF. It runs on O(t . min(m, n)) where t is the actual distance between strings a and b. It needs O(min(t, m, n)) space. This function might be preferred over WagnerFischer() for *very* similar strings. But test it out yourself.
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// The first two parameters are the two strings to be compared. The last three parameters are the insertion cost, the deletion cost and the substitution cost. These are normally defined as 1, 1 and 2 respectively.
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func Ukkonen(a, b string, icost, dcost, scost int) int {
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var lowerCost int
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if icost < dcost && icost < scost {
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lowerCost = icost
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} else if dcost < scost {
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lowerCost = dcost
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} else {
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lowerCost = scost
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}
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infinite := math.MaxInt32 / 2
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var r []int
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var k, kprime, p, t int
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var ins, del, sub int
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if len(a) > len(b) {
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t = (len(a) - len(b) + 1) * lowerCost
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} else {
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t = (len(b) - len(a) + 1) * lowerCost
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}
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for {
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if (t / lowerCost) < (len(b) - len(a)) {
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continue
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}
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// This is the right damn thing since the original Ukkonen
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// paper minimizes the expression result only, but the uncommented version
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// doesn't need to deal with floats so it's faster.
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// p = int(math.Floor(0.5*((float64(t)/float64(lowerCost)) - float64(len(b) - len(a)))))
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p = ((t / lowerCost) - (len(b) - len(a))) / 2
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k = -p
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kprime = k
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rowlength := (len(b) - len(a)) + (2 * p)
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r = make([]int, rowlength+2)
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for i := 0; i < rowlength+2; i++ {
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r[i] = infinite
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}
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for i := 0; i <= len(a); i++ {
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for j := 0; j <= rowlength; j++ {
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if i == j+k && i == 0 {
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r[j] = 0
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} else {
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if j-1 < 0 {
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ins = infinite
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} else {
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ins = r[j-1] + icost
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}
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del = r[j+1] + dcost
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sub = r[j] + scost
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if i-1 < 0 || i-1 >= len(a) || j+k-1 >= len(b) || j+k-1 < 0 {
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sub = infinite
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} else if a[i-1] == b[j+k-1] {
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sub = r[j]
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}
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if ins < del && ins < sub {
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r[j] = ins
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} else if del < sub {
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r[j] = del
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} else {
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r[j] = sub
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}
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}
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}
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k++
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}
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if r[(len(b)-len(a))+(2*p)+kprime] <= t {
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break
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} else {
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t *= 2
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}
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}
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return r[(len(b)-len(a))+(2*p)+kprime]
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}
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