Skip to main content

plotting - How can I get the interpolation line from Listplot?


My discrete data is:


data={{1., 5827.}, {3., 6242.}, {15., 6066.}, {60., 5972.}}


I want to get the interpolation line from ListPlot:


pl=ListPlot[data, Joined -> True,InterpolationOrder -> 2, PlotStyle -> Dashed, PlotRange -> {5900, 6300}]

enter image description here


But I can't get the same dash line from InterPolation:


Plot[Interpolation[data, InterpolationOrder -> 2][x], {x, 1, 60}, 
     PlotStyle -> Dashed]

enter image description here



How can I get the curve like 1st one?


Or can I get the dash line points automatically?



Answer



More of an extended comment here, but it is clear that ListPlot must use a different interpolation function than is available via Interpolation. We can get a very close approximation by Janus's answer here and amending the Method option. The idea is that ListPlot performs an parametric interpolation on the x and y axes separately.


data = {{1., 5827.}, {3., 6242.}, {15., 6066.}, {60., 5972.}};
pl = ListPlot[data, Joined -> True, InterpolationOrder -> 2, 
   PlotStyle -> Dashed, PlotRange -> {All, {5900, 6300}}];
pl2 = With[{xyInterpolation = 
     Interpolation[#, InterpolationOrder -> 2, Method -> "Spline"] & /@
       Transpose[data]},

   ParametricPlot[Through[xyInterpolation[i]], {i, 1, Length[data]}, 
    AspectRatio -> 2/3, PlotStyle -> Red]];
Show[pl, pl2]

Mathematica graphics


But those curves only match up approximately.


To answer the last question,



"Or can I get the dash line points automatically?"




Yes you can,


Cases[pl, Line[pts_] :> pts, Infinity]
(* {{{1.20269, 5900.}, {1.23947, 5913.25}, {1.28811, 
   5930.77}, {1.33925, 5949.19}, {1.39288, 5968.5}, {1.44901, 
   5988.72}, {1.50763, 6009.83}, {1.56874, 6031.84}, {1.63235, 
   6054.75}, {1.70031, 6077.42}, {1.77448, 6098.71}, {1.85486, 
   6118.63}, {1.94144, 6137.16}, {2.03423, 6154.32}, {2.13322, 
   6170.11}, {2.23843, 6184.51}, {2.34984, 6197.54}, {2.46746, 
   6209.18}, {2.59128, 6219.45}, {2.72131, 6228.35}, {2.85755, 
   6235.86}, {3., 6242.}, {3.14865, 6246.76}, {3.30352, 

   6250.14}, {3.46458, 6252.15}, {3.63186, 6252.77}, {3.80534, 
   6252.02}, {3.98503, 6249.89}, {4.17093, 6246.39}, {4.36304, 
   6241.5}, {4.56135, 6235.24}, {4.76587, 6227.6}, {4.97659, 
   6218.58}, {5.19353, 6208.19}, {5.41667, 6196.42}, {5.68221, 
   6184.14}, {6.02637, 6172.24}, {6.44912, 6160.71}, {6.95049, 
   6149.56}, {7.53046, 6138.78}, {8.18903, 6128.38}, {8.92621, 
   6118.34}, {9.74199, 6108.68}, {10.6364, 6099.4}, {11.6094, 
   6090.49}, {12.661, 6081.95}, {13.7912, 6073.79}, {15., 
   6066.}, {16.2874, 6058.58}, {17.6534, 6051.54}, {19.0981, 
   6044.87}, {20.6213, 6038.58}, {22.2231, 6032.66}, {23.9036, 

   6027.11}, {25.6626, 6021.94}, {27.5003, 6017.14}, {29.4166, 
   6012.71}, {31.4114, 6008.66}, {33.4849, 6004.98}, {35.637, 
   6001.68}, {37.8676, 5998.75}, {40.094, 5996.06}, {42.233, 
   5993.47}, {44.2847, 5990.99}, {46.2491, 5988.62}, {48.1262, 
   5986.35}, {49.916, 5984.19}, {51.6185, 5982.13}, {53.2337, 
   5980.18}, {54.7616, 5978.33}, {56.2021, 5976.59}, {57.5554, 
   5974.95}, {58.8214, 5973.42}, {60., 5972.}}} *)

 ListPlot@%


Mathematica graphics


Comments

Post a Comment

Popular posts from this blog

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
Post a Comment
Read more

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more