How to extract X value given Y value from graph.

I have a graph which I ploted from X and Y values. X and Y are the data vectors. Now I wnted to find values of X corresponding to perticular value of y.

X=[0.00163629165319779	0.00919798132258265	0.0108887716090971	0.0146711509895121	0.0170727875027912	0.0183195248019579	0.0214701529053916	0.0429751397974718	0.0492808450276714	0.0535887468601313	0.0546312668441853	0.108838368209090	0.123117548287373	0.126309277779226	0.133652225605413	0.138726746967278	0.176380030642817	0.178656767960438	0.181244974266293	0.182028488504125	0.188366079912920	0.204276265010833	0.205419462095058	0.206611392842045	0.219575407312163	0.230076938132872	0.231822979086683	0.232631056161634	0.237634467232773	0.242104610234617	0.267542565734243	0.281557956744304	0.312833057094883	0.319124464220357	0.343375832711649	0.348289420385577	0.352569636089163	0.357642593625700	0.360188336139401	0.374263685881673	0.390056224874232	0.391650522675869	0.392413301822307	0.392906376647038	0.398112256101353	0.404497829354034	0.408989317941850	0.413270285981264	0.440717991244140	0.442788702297138	0.466238670974959	0.478229661633876	0.506778136096007	0.516813401433934	0.539005985379619	0.554907031467319	0.555940863634276	0.594998712090185	0.624358289172818	0.636691646673399	0.636691646673399	0.636691646673399	0.636691646673399	0.636691646673399	0.673595576899095	0.690991870967221	0.691623728204757	0.694482327270088	0.703223226864390	0.713125941582075	0.729095502236721	0.749453251696249	0.785248184904319	0.790012531082956	0.815065416646928	0.821315922015482	0.826164341954818	0.829691396788946	0.841734834718935	0.854822741928195	0.858109043192997	0.891859889006936	0.906733683367389	0.952764915345216	0.952764915345216	0.952764915345216	0.980470366913377	0.995725305844472	1.02965375760247	1.09116894860735	1.10670509719372	1.11422792999389	1.12127930245400	1.12127930245400	1.12334033118983	1.14277471093836	1.16917633244481	1.19089850966396	1.33264717043745	1.34901320716680	1.36779787738210	1.44087981077227	1.64523008957909	1.64791868775681	2.27643402946012	2.40045676947028	2.45142678398700]
Y=[0.327970660780512	0.332011235759306	0.332921494603458	0.334966833639170	0.336272047142938	0.336951613545687	0.338675071601992	0.350676523995197	0.354275568517746	0.356755561062934	0.357358328092331	0.390144287739127	0.399270422133906	0.401339342218769	0.406139930184248	0.409491004458485	0.435235365455767	0.436842919676984	0.438677608063090	0.439234531651005	0.443765372470964	0.455346819396008	0.456190529354724	0.457071870791561	0.466768441997315	0.474773779922694	0.476118038978851	0.476741456944862	0.480619720732729	0.484111312757906	0.504469097848463	0.516049003307847	0.542856789444626	0.548415510596634	0.570380400825788	0.574936780624634	0.578935485718932	0.583710823819502	0.586122035838970	0.599634471364500	0.615166607306969	0.616756803237560	0.617519073048832	0.618012319891502	0.623244115538772	0.629721998108256	0.634318708785900	0.638731195373301	0.667760657214304	0.670003487282225	0.695934458307588	0.709579482217070	0.743152374382388	0.755327344275103	0.782964775513683	0.803386803712655	0.804732868698833	0.857273337746951	0.899014193321375	0.917149276210161	0.917149276210161	0.917149276210161	0.917149276210161	0.917149276210161	0.973627548017118	1.00144444463865	1.00246961429251	1.00712073630011	1.02147703708904	1.03798894692457	1.06518092762530	1.10088015869719	1.16657580257029	1.17561063999175	1.22428363044642	1.23673808159782	1.24648599430674	1.25362549978489	1.27831360124910	1.30569436282117	1.31266115228447	1.38639837317600	1.42019536561643	1.53009968474307	1.53009968474307	1.53009968474307	1.60030840865775	1.64033208407164	1.73297416779503	1.91448899911085	1.96326411840361	1.98732634282930	2.01014827313001	2.01014827313001	2.01686820351132	2.08134856950608	2.17226025957346	2.25002870512480	2.83056628811592	2.90658351414197	2.99635450885091	3.37278249763301	4.69567928252034	4.71616719791340	13.0495320257110	15.9519723894339	17.3244414235913]

how to find value of X at Y=15? y=15 is not available in Y vector. kindly help me to write a code. Thanks

[KSSV]

interp1

Answer 1

[KSSV]

[Y,idx] = unique(Y) ; 
X = X(idx) ; 
iwant = interp1(Y,X,15)

[Govind Kumar]

Thanks

[Renu Joshi]

thanks from my side too

Answer 2

[Rik]

You can treat x as y and y as x. That way you can use normal interpolation and curve fitting tools.

[madhan ravi]

+1 , Rik you beat me up by a few seconds xD

[Govind Kumar]

No I can't use curve fitting tool. I have to use this code on loop. Can you goe me code for interploation?

[Govind Kumar]

From fig I can see for Y=8 x will be approx 2. So is their any method so I can extract excact value by code