Thank you for sharing. I like the way you described the plot in your prompt. It played a significant role in getting such a good prediction. So the basic rule here is to try to provide as much as we observe to the model.

Charlie-san,

I'm sorry I've not built an abnormal route in my tool. I don't know if the following is the problem you're facing, but an endless loop may occur when locator2coordinate function does not have unique solution from given two red points.

You may avoid the loop by changing the function to the following. Please change the location of two red points and try again.

locator2coordinate[list_, sample_] := 
 Module[{l1x, l1y, l2x, l2y, s1x, s1y, s2x, s2y, d, c, ysolve, xlist, 
   ylist},
  {l1x, l1y} = list[[1]]; {l2x, l2y} = list[[2]];
  If[l1x === l2x || l1y === l2y, Beep[]; Return[{}]];
  {s1x, s1y} = sample[[1]]; {s2x, s2y} = sample[[2]];
  ysolve = Solve[d*l1y + c == s1y && d*l2y + c == s2y, {d, c}];
  {{d, c}} = {d, c} /. ysolve;
  ({Exp[(#[[1]] - l1x)/(l2x - l1x)*(Log[s2x] - Log[s1x]) + Log[s1x]], 
       d*#[[2]] + c} & /@ list) // Sort
  ]

Hello Kataro,

thank you so much for your method. It is really great.

Charlie-san,

Thanks for your comment. To work for graphs that have a logarithmic scaling on the x-axis, change the locator2coordinate function below.

locator2coordinate[list_, sample_] := 
 Module[{l1x, l1y, l2x, l2y, s1x, s1y, s2x, s2y, d, c, ysolve, xlist, 
   ylist},
  {l1x, l1y} = list[[1]]; {l2x, l2y} = list[[2]];
  {s1x, s1y} = sample[[1]]; {s2x, s2y} = sample[[2]];
  ysolve = Solve[d*l1y + c == s1y && d*l2y + c == s2y, {d, c}];
  {{d, c}} = {d, c} /. ysolve;
  ({Exp[(#[[1]] - l1x)/(l2x - l1x)*(Log[s2x] - Log[s1x]) + Log[s1x]], 
       d*#[[2]] + c} & /@ list) // Sort
  ]

This is my result of your example.

For your reference:

• GraphImage2List for multi-colored graph-image-data
• GraphImage2List : more automatically
• presentation(Japanese) at Wolfram Conference Japan 2018

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