o h|'@sPddlZddlZddlmZmZddlmZm Z GdddZ GdddZ dS)N) assert_equalassert_allclose) _iv_ratio _iv_ratio_cc@seZdZejdgdddZejddejdfejddfgddZ ejd d ej ej ejgejd e e j e e j ej ej ejgd d Zejd dde e jejgddZejdde e j fde e jfde e jdfde e jdfe e jee e jfgddZejdddee e je e jfgddZejde e je e jfe e jde e jfe e je e jdfgddZdS) TestIvRatiov,x,r))?UUUUUU?g.a0R#?)rUUUUUU?g<)?)rrgVS?)rUUUUUU?gP]k(?)r笪?gjD?)6Z5Z?g&R͒U?)r 窪?gZ?)r 竪?gZr!?)r ?g4e~u?)r }| @gG)ȿ?)Q@}P?g1a?)rj6i?gִN`?)r:m@g9Ƭ7?)r5T@g4+?)r翎H%@gJ]?)EdL@9L;w3@g'~V?)r^s!iFE@g/X?)rSR@g_8?)rPT`@g)X?)r>=s@g\h*?cCtt|||ddddS)aThe reference values are computed using mpmath as follows. from mpmath import mp mp.dps = 100 def iv_ratio_mp(v, x): return mp.besseli(v, x) / mp.besseli(v - 1, x) def _sample(n, *, v): '''Return n positive real numbers x such that iv_ratio(v, x) are roughly evenly spaced over (0, 1). The formula is taken from [1]. [1] Banerjee A., Dhillon, I. S., Ghosh, J., Sra, S. (2005). "Clustering on the Unit Hypersphere using von Mises-Fisher Distributions." Journal of Machine Learning Research, 6(46):1345-1382. ''' r = np.arange(1, n+1) / (n+1) return r * (2*v-r*r) / (1-r*r) for v in (0.5, 1, 2.34, 56.789): xs = _sample(5, v=v) for x in xs: print(f"({v}, {x}, {float(iv_ratio_mp(v,x))}),") 缉ؗҼ7R?)r rgL?)r rg5?)r rgZ?)rrg3zʈ?)rrgؤO??)rrgsF?)rrg1?)rrgL9Ԋ?)rrgCv`?)rrg -S?)rrg@ɣ?)rrgO?)rrg^VO?cCr)z8The reference values are one minus those of TestIvRatio.V瞯zTestIvRatioC.test_zero_xr?r@rAcCs tt||dd||dS)a=If x is much less than v, the bounds x x --------------------------- <= R <= ----------------------- v-0.5+sqrt(x**2+(v+0.5)**2) v-1+sqrt(x**2+(v+1)**2) collapses to 1-R ~= 1-x/2v. Test against this asymptotic expression. rGrNrar7r*r*r+rDs zTestIvRatioC.test_tiny_xrErFcCs"tt|||d|ddddS)aKIf x is much greater than v, the bounds x x --------------------------- <= R <= --------------------------- v-0.5+sqrt(x**2+(v+0.5)**2) v-0.5+sqrt(x**2+(v-0.5)**2) collapses to 1-R ~= (v-0.5)/x. Test against this asymptotic expression. rr^rr!Nr_r7r*r*r+rHs"zTestIvRatioC.test_huge_xrIcCs:||}d|dtd|}tt|||ddddS)aIf both x and v are very large, the bounds x x --------------------------- <= R <= ----------------------- v-0.5+sqrt(x**2+(v+0.5)**2) v-1+sqrt(x**2+(v+1)**2) collapses to 1 - R ~= 1 - x/(v+sqrt(x**2+v**2). Test against this asymptotic expression, and in particular that no numerical overflow occurs during intermediate calculations. r r rr!N)r5rJrr`rKr*r*r+rNszTestIvRatioC.test_huge_v_xNrOr*r*r*r+r]sJ           r]) rSnumpyr5 numpy.testingrrscipy.special._ufuncsrr$rr`rr]r*r*r*r+s