I looked at all the questions/answers relating to placement of ContourLabels but could not find an answer to this. Some of the answers before are based on user giving before-hand, the Contours levels to use, using the option Contours -> {v1, v2, ....}.

But if one just want to use Contours->number, then it gets hard to figure how to solve this.

The labels that show up are not in the nice locations and hard to read. I'd like them to be in the middle of the contour lines. This is how Matlab actually does it by default.

Here is a MWE, and I also show the Matlab result.

f = (11 - x - y)^2 + (1 + x + 10 y - x y)^2;
ContourPlot[f, {x, 0, 20}, {y, 0, 15}, 
 Contours -> 10, ContourShading -> None,
 ContourLabels -> Function[{x, y, z},
   Text[z, {x, y}]], AspectRatio -> Automatic, PlotRangePadding -> 2]

Mathematica graphics

I am trying to get closer output to Matlab's

clear; close all;
x = 0:0.05:20;
y = 0:0.05:15;
[x,y] = meshgrid(x,y);
z = (11-x-y).^2 + (1+x+10*y-x.*y).^2;
[C,h] =contour(x,y,z,10);

Mathematica graphics

I also noticed that Matlab seems to have additional heuristics in placing labels. It does not label all the lines if the space is tight. Notice the top levels are not labeled in Matlab. Also the text is not written over the lines, but there is space in the line where the text goes. This makes it easier to read.

But my main question I have is how to put the labels in the middle of the level lines, where it is more pleasing to see than on the edges.

If someone knows of a package to use for this, I will try it. I looked at few here http://packagedata.net/ but did not see anything yet. Looked at sciDraw package also. I think Contour label placement in Mathematica should be improved more.

Version 10.3.1. windows.

Related questions

How to control the positions of contour labels?

Custom contour labels in ContourPlot

How do I add contour labels to contour plot?


FYI, I found the book Mathematica graphics which has an function called LabelContourLines[], which from the example shown, seems to do what I want. Page 348 in the book:

Mathematica graphics

Mathematica graphics

Here is how Mathematica 10.3.1. gives, for the same example in the book. Notice the difference :

ContourPlot[x y, {x, -2, 2}, {y, -2, 2}, ContourShading -> False, 
 Contours -> 10, ContourLabels -> All]

Mathematica graphics

I do not have this book around myself, but will try to see if the author has the software on the net to use or I can look for it at the university library.

It looks like the above function is what I am looking for. But the book is old (1994), and do not know if book software will even work on current Mathematica even if I find it.


This is a good question which I had tried to work on several years ago, and I just revisited it to see if the code could be useful.

I had initially only worked on rotating the contour labels to be tangential with the contours, but in the code it turned out to be quite easy to add a line that also attempts to center the labels along the contour. My simple approach is to estimate the center of the contour by just choosing the midpoint in the Line specification defining that particular contour.

The code comes with comments, but it's quite lengthy:

rotateContourLabels::usage = 
  "The function rotateContourLabels accepts the output of ContourPlot \
or ListContourPlot, assuming they were made with the option \
ContourLabel-> All. The plot is passed via the required first \
argument. The function rotates the labels of all contours to be \
approximately parallel to the iso-lines. The optional LabelFunction \
can specify a custom label style in the form of a function f[#1, #2, \
#3], where {#1, #2} is the 2D location vector of the label, and #3 \
the value of the plotted function at that location. For examples see \
the documentation on ContourLabels. The default label style is given \
by the function Text[#3,{#1,#2}]&. A second option, Alignment, \
influences the placement of the labels along the contour. The default \
Alignment -> Automatic leaves the original placement intact, any \
other value will shift the labels to an estimated center position \
along the contour.";

Options[rotateContourLabels] = {LabelingFunction -> (Text[#3, {#1, \
#2}] &), Alignment -> Automatic};

rotateContourLabels[plot_, OptionsPattern[]] := 
 Module[{gc = plot[[1]] , valueList, pointList, rotatePoint, 
   labelCoordinatesIndexed, directionVector},
  (******* Created by Jens U. Nöckel, August 17, 2009, 
  last modified May 16, 2017  *******)
   labelCoordinatesIndexed = 
    Quiet@Check[gc[[-1, -1, All, 2]], Throw[plot]];
   valueList = Quiet@Check[gc[[-1, -1, All, 1]], Throw[plot]];
     pointList = labelCoordinatesIndexed, 
     pointList = Quiet@Check[gc[[1, labelCoordinatesIndexed]],
      Throw[{"The plot must have the option ContourLabels->All !!!", 
   (** gc is the GraphicComplex, gc[[1]] a list of 2D points. 
   The list of contour labels is contained in plot[[1,-1,-1]] = 
   gc[[-1,-1]] and we want to replace this list. **)

   (* Local function definition: *)

   rotatePoint[labelPoint_] := 
    Module[{pt, extractSecant , allContourLines, contourPoints2D, 
      contourPointsIndexed , closestContourIndexed, 
     pt = labelPoint;
     allContourLines = 
       The First extracts the list of line points from the Line \
expressions collected in the next line: **)

         MapAll[If[SameQ[Head[#], Line], #, Apply[List, #]] &, 
          gc[[-1]]], _Line]];
     (** Line can be nested inside other non-
     list expressions such as Tooltip, 
     so the above flattens all levels of the expression except for \
subexpressions with head Line, 
     before it proceeds to collect the Lines using Cases. **) 

     {contourPoints2D, contourPointsIndexed} =
      (** Given a point pt, 
      find out for each contour what is its closest point to pt. 
      Return this information both as 2D points and as a list of \
indices counted within each contour. **)
           # -> Transpose[{#, Range[Length[#]]}],
           ] (** End First@Nearest **)&,
        Map[gc[[1, #]] &, allContourLines]
        ](* End Map *)
       ] (* End Transpose *);

     {closestContourIndexed, closestPointIndexed} =
      (** Given a point pt, find closest contour. 
      Give its index in the list of contours, 
      and the index of the closest point within this contour. **)

        contourPoints2D -> 
        ]];(* End First@Nearest *)

     If[OptionValue[Alignment] =!= Automatic,
      (* Alignment refers to the placement of labels on contours. 
      If not Automatic, 
      try to approximately center the labels according to contour \
length, measured by points in their line: *)
      closestPointIndexed = 
      pt = Part[gc[[1]], 
     extractSecant = (** Given the closest point to pt, 
      record it and its next neighbor on the same contour line. 
      The two points are returned as real 2D points, 
      determined from their indices. **)
       allContourLines[[closestContourIndexed, closestPointIndexed]] (* 
       The first point *),
           closestPointIndexed + 1]], 
           closestPointIndexed - 1]]] (* 
       The second sequential point - 
       here I want to avoid falling outside the range of the contour \
point list. *)

     (* The rotation aims to make this vector parallel to the \
contour. We approximate the contour direction by taking two points on \
the contour and forming their difference: *)

     directionVector = Apply[Subtract,
       (** The two required points are found from the list gc[[1]] : **)

        The Part of gc[[1]] we're looking for is a set of two points \
that define the secant closest to the contour label: *)


     (** Here starts the body of rotatePoint: **)
      (* The composition sandwiches a rotation around the origin \
between a translation to the origin and its inverse: *)

          {1, 0},  (* 
          The reference direction for the rotation is the horizontal: \
          (Sign[directionVector[[1]]] directionVector)
          To avoid rotations outside [-Pi,Pi] the above \
statement places the direction vector into the first or fourth \
quadrant. *) 
          }] (* End RotationTransform *)
        , Identity (* 
        No rotation happens if for some reason we can't get a valid \
direction vector *)],
      ] (* End Composition *) 
     ] ;(** End function rotatePoint **)

   (**** Main function body: *****)

   ReplacePart[plot, {1, -1, -1} -> 
     (** the replacement is a geometric transformation of the \
original list gc[[-1,-1]]: **)

        Append[Transpose[pointList], valueList]],
       What follows is the list of transformations for each contour \
label: **)
       Map[rotatePoint, pointList]}]]

This defines a function rotateContourLabels which takes an existing plot and post-processes it. As always with post-processing, I have to make some assumptions about how the input was prepared:

In the ContourPlot or ListContourPlot that serves as the input, you must set either the option ContourLabels -> True or ContourLabels -> All.

With this assumption, you can then use the function as follows:

a = ContourPlot[Im[(x + I y)^(1/2)], {x, -1, 1}, {y, -1, 1},
   Contours -> 20, ContourLabels -> True]


This is the original output in Mathematica 10.2.

Now the usage example:

rotateContourLabels[a, Alignment -> Center, 
 LabelingFunction -> (Text[#3, {#1, #2}, 
     Background -> Directive[Opacity[.5], White]] &)]


I think this result is an improvement (given that it's fully automatic), although of course the centering is not perfect.


For completeness, here is the OP's minimum working example:

Clear[x, y]
f = (11 - x - y)^2 + (1 + x + 10 y - x y)^2;
b = ContourPlot[f, {x, 0, 20}, {y, 0, 15}, Contours -> 7, 
   ContourShading -> None, ContourLabels -> True, 
   AspectRatio -> Automatic, PlotRangePadding -> 2];

rotateContourLabels[b, Alignment -> Center, 
 LabelingFunction -> Function[{x, y, z}, Text[z, {x, y}]]]

second example

Note that I added the label function as an option to rotateContourLabels instead of the original plot (which I called b). I also decreased the number of contours to 7 so the labels are more readable. The placement of the labels is obviously much better than the original. The background I chose in the first example would probably also make this plot look a little better.

Finally, here is my automatic version of the book example in the question:

c = ContourPlot[x y, {x, -2, 2}, {y, -2, 2}, ContourShading -> False, 
   Contours -> 10, ContourLabels -> All];

rotateContourLabels[c, Alignment -> Center, 
 LabelingFunction -> (Text[#3, {#1, #2}, Background -> White] &)]

third example

One could probably also add an option to place the labels at the ends of the contours instead of the center, but I haven't done that yet (because it usually looks ugly, even though Mathematica seems to like doing that by default...).

  • $\begingroup$ i tried to run this last example but i get the following message "The plot must have the option ContourLabels->All !!!". I am using Mathematica 11.1.1 $\endgroup$ – Valerio May 16 '17 at 12:33
  • $\begingroup$ @Valerio Thanks for pointing this out. It appears to be a change in the way label coordinates are stored internally. My code makes assumptions about this internal storage. I think it's fixed now, and should also still work with older Mathematica versions. $\endgroup$ – Jens May 16 '17 at 16:27
  • $\begingroup$ @Nasser: Have you improved Jens's rotateContourLabels function? It seems that it does not label properly when the space is tight. Also, sometimes the directions of the text are inappropriate! :( I have some sample pics but cannot show them in a comment. $\endgroup$ – H. R. Jul 25 '17 at 21:23
  • $\begingroup$ @H.R. maybe you can post code to reproduce the plots? I can't guarantee that I can fix it, but I'd like to know what you're seeing. I did a small fix recently to keep up with changes in version 11... maybe go back to the earlier version in the revision history and see if that works better (i.e. perhaps I broke something in my May edit of the answer). $\endgroup$ – Jens Jul 25 '17 at 21:45
  • 1
    $\begingroup$ @H.R. Thanks, I see what's going on. But it looks like for those scrunched-up contours you may be better off with this more manual approach. $\endgroup$ – Jens Jul 25 '17 at 23:00

The software is located at http://library.wolfram.com/infocenter/Books/3753/ as a .tar and a .zip file.


How about a shameless lifting of one of the references you mention and adding a Rotate function?

ContourPlot[x y, {x, -2, 2}, {y, -2, 2}, ContourShading -> False, 
 Contours -> 10, 
 ContourLabels -> (Rotate[Text[" " <> ToString[#3] <> " ", {#1, #2}, 
      Background -> White], -ArcTan[#2/#1]] &)]


For other curves one would need to determine the tangent to the curve at point {#1, #2}.

  • $\begingroup$ Thanks for the link. I downloaded and installed it. But as expected, it does not work with 10.3.1. This is an old package from version 3.0 ! I get lots of errors: !Mathematica graphics but what I will do next, is locate this specific function in the package, and try to make it work for 10.3.1. I've always had problems running old packages in new Mathematica versions. Something always changes. The LabelContour function is very complicated 450 lines long ! $\endgroup$ – Nasser Jan 4 '16 at 0:43
  • $\begingroup$ @Nasser. Bummer! $\endgroup$ – JimB Jan 4 '16 at 1:23
  • 1
    $\begingroup$ Ok, I tried for the last 40 minutes, but too many problems to resolve. This package was done using old graphics and uses too many old options and such, and it will need much longer time to fix. But I think it will be worth it. I am really surprised that WRI did not integrate some of the functions in this package in Mathematica. This package has been around for only 20 years! I guess WRI is busy with cloud computing now and has no time to work on such basic things like contour labels location :) $\endgroup$ – Nasser Jan 4 '16 at 1:37
  • $\begingroup$ @H.R. sorry, do not think I ever did. $\endgroup$ – Nasser Jun 14 '17 at 21:01
  • $\begingroup$ In general, determining the tangent to a curve is not really applicable in closed form and a numerical algorithm should be employed. $\endgroup$ – H. R. Jul 25 '17 at 20:49

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