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How to extract X value given Y value from graph.

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Govind Kumar
Govind Kumar 2020-5-8
评论： Renu Joshi 2021-6-2
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

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采纳的回答

KSSV
KSSV 2020-5-8
[Y,idx] = unique(Y) ;
X = X(idx) ;
iwant = interp1(Y,X,15)

更多回答（1 个）

Rik
Rik 2020-5-8
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
显示 1更早的评论
Govind Kumar
Govind Kumar 2020-5-8
编辑：Govind Kumar 2020-5-8
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
Govind Kumar 2020-5-8
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

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