How to extract X value given Y value from graph.
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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
Accepted Answer
More Answers (1)
Rik
on 8 May 2020
You can treat x as y and y as x. That way you can use normal interpolation and curve fitting tools.
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