How to extract X value given Y value from graph.
154 visualizzazioni (ultimi 30 giorni)
Mostra commenti meno recenti
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
Risposta accettata
Più risposte (1)
Rik
il 8 Mag 2020
You can treat x as y and y as x. That way you can use normal interpolation and curve fitting tools.
3 Commenti
Vedere anche
Categorie
Scopri di più su Graph and Network Algorithms in Help Center e File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!