Python: find contour lines from matplotlib.pyplot.contour()

I’m trying to find (but not draw!) contour lines for some data:

from pprint import pprint 
import matplotlib.pyplot 
z = [[0.350087, 0.0590954, 0.002165], [0.144522, 0.885409, 0.378515], 
     [0.027956, 0.777996, 0.602663], [0.138367, 0.182499, 0.460879], 
     [0.357434, 0.297271, 0.587715]] 
cn = matplotlib.pyplot.contour(z)

I know cn contains the contour lines I want, but I can’t seem to get
to them. I’ve tried several things:

print dir(cn) 
pprint(cn.collections[0]) 
print dir(cn.collections[0]) 
pprint(cn.collections[0].figure) 
print dir(cn.collections[0].figure)

to no avail. I know cn is a ContourSet, and cn.collections is an array
of LineCollections. I would think a LineCollection is an array of line segments, but I
can’t figure out how to extract those segments.

My ultimate goal is to create a KML file that plots data on a world
map, and the contours for that data as well.

However, since some of my data points are close together, and others
are far away, I need the actual polygons (linestrings) that make up
the contours, not just a rasterized image of the contours.

I’m somewhat surprised qhull doesn’t do something like this.

Using Mathematica’s ListContourPlot and then exporting as SVG works, but I
want to use something open source.

I can’t use the well-known CONREC algorithm because my data isn’t on a
mesh (there aren’t always multiple y values for a given x value, and
vice versa).

The solution doesn’t have to python, but does have to be open source
and runnable on Linux.

Answers:

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Method 1

You can get the vertices back by looping over collections and paths and using the iter_segments() method of matplotlib.path.Path.

Here’s a function that returns the vertices as a set of nested lists of contour lines, contour sections and arrays of x,y vertices:

import numpy as np

def get_contour_verts(cn):
    contours = []
    # for each contour line
    for cc in cn.collections:
        paths = []
        # for each separate section of the contour line
        for pp in cc.get_paths():
            xy = []
            # for each segment of that section
            for vv in pp.iter_segments():
                xy.append(vv[0])
            paths.append(np.vstack(xy))
        contours.append(paths)

    return contours

Edit:

It’s also possible to compute the contours without plotting anything using the undocumented matplotlib._cntr C module:

from matplotlib import pyplot as plt
from matplotlib import _cntr as cntr

z = np.array([[0.350087, 0.0590954, 0.002165],
              [0.144522,  0.885409, 0.378515],
              [0.027956,  0.777996, 0.602663],
              [0.138367,  0.182499, 0.460879], 
              [0.357434,  0.297271, 0.587715]])

x, y = np.mgrid[:z.shape[0], :z.shape[1]]
c = cntr.Cntr(x, y, z)

# trace a contour at z == 0.5
res = c.trace(0.5)

# result is a list of arrays of vertices and path codes
# (see docs for matplotlib.path.Path)
nseg = len(res) // 2
segments, codes = res[:nseg], res[nseg:]

fig, ax = plt.subplots(1, 1)
img = ax.imshow(z.T, origin='lower')
plt.colorbar(img)
ax.hold(True)
p = plt.Polygon(segments[0], fill=False, color='w')
ax.add_artist(p)
plt.show()

Method 2

I would suggest to use scikit-image find_contours

It returns a list of contours for a given level.

matplotlib._cntr has been removed from matplotlib since v2.2 (see here).

Method 3

It seems that the contour data is in the .allsegs attribute of the QuadContourSet object returned by the plt.contour() function.

The .allseg attribute is a list of all the levels (which can be specified when calling plt.contour(X,Y,Z,V). For each level you get a list of n x 2 NumPy arrays.

plt.figure()
C = plt.contour(X, Y, Z, [0], colors='r')

plt.figure()
for ii, seg in enumerate(C.allsegs[0]):
    plt.plot(seg[:,0], seg[:,1], '.-', label=ii)
plt.legend(fontsize=9, loc='best')

In the above example, only one level is given, so len(C.allsegs) = 1. You get:

contour plot

the extracted curves

Method 4

The vertices of an all paths can be returned as a numpy array of float64 simply via:

vertices = cn.allsegs[i][j]  # for element j, in level i

with cn defines as in the original question:

import matplotlib.pyplot as plt
z = [[0.350087, 0.0590954, 0.002165], [0.144522, 0.885409, 0.378515], 
     [0.027956, 0.777996, 0.602663], [0.138367, 0.182499, 0.460879], 
     [0.357434, 0.297271, 0.587715]] 
cn = plt.contour(z)

More detailed:

Going through the collections and extracting the paths and vertices is not the most straight forward or fastest thing to do. The returned Contour object actually has attributes for the segments via cs.allsegs, which returns a nested list of shape [level][element][vertex_coord]:

num_levels = len(cn.allsegs)
num_element = len(cn.allsegs[0])  # in level 0
num_vertices = len(cn.allsegs[0][0])  # of element 0, in level 0
num_coord = len(cn.allsegs[0][0][0])  # of vertex 0, in element 0, in level 0

See reference:
https://matplotlib.org/3.1.1/api/contour_api.html

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0