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

采纳的回答
更多回答(1 个)
Rik
2020-5-8
1 个投票
You can treat x as y and y as x. That way you can use normal interpolation and curve fitting tools.
3 个评论
madhan ravi
2020-5-8
+1 , Rik you beat me up by a few seconds xD
Govind Kumar
2020-5-8
编辑:Govind Kumar
2020-5-8
Govind Kumar
2020-5-8
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