Python / NumPy中的meshgrid的用途是什么？

Can someone explain to me what is the purpose of meshgrid function in Numpy? I know it creates some kind of grid of coordinates for plotting, but I can’t really see the direct benefit of it.

I am studying “Python Machine Learning” from Sebastian Raschka, and he is using it for plotting the decision borders. See input 11 here.

I have also tried this code from official documentation, but, again, the output doesn’t really make sense to me.

x = np.arange(-5, 5, 1)
y = np.arange(-5, 5, 1)
xx, yy = np.meshgrid(x, y, sparse=True)
z = np.sin(xx**2 + yy**2) / (xx**2 + yy**2)
h = plt.contourf(x,y,z)

Please, if possible, also show me a lot of real-world examples.

The purpose of meshgrid is to create a rectangular grid out of an array of x values and an array of y values.

So, for example, if we want to create a grid where we have a point at each integer value between 0 and 4 in both the x and y directions. To create a rectangular grid, we need every combination of the x and y points.

This is going to be 25 points, right? So if we wanted to create an x and y array for all of these points, we could do the following.

x[0,0] = 0    y[0,0] = 0
x[0,1] = 1    y[0,1] = 0
x[0,2] = 2    y[0,2] = 0
x[0,3] = 3    y[0,3] = 0
x[0,4] = 4    y[0,4] = 0
x[1,0] = 0    y[1,0] = 1
x[1,1] = 1    y[1,1] = 1
...
x[4,3] = 3    y[4,3] = 4
x[4,4] = 4    y[4,4] = 4

This would result in the following x and y matrices, such that the pairing of the corresponding element in each matrix gives the x and y coordinates of a point in the grid.

x =   0 1 2 3 4        y =   0 0 0 0 0
      0 1 2 3 4              1 1 1 1 1
      0 1 2 3 4              2 2 2 2 2
      0 1 2 3 4              3 3 3 3 3
      0 1 2 3 4              4 4 4 4 4

We can then plot these to verify that they are a grid:

plt.plot(x,y, marker='.', color='k', linestyle='none')

Obviously, this gets very tedious especially for large ranges of x and y. Instead, meshgrid can actually generate this for us: all we have to specify are the unique x and y values.

xvalues = np.array([0, 1, 2, 3, 4]);
yvalues = np.array([0, 1, 2, 3, 4]);

Now, when we call meshgrid, we get the previous output automatically.

xx, yy = np.meshgrid(xvalues, yvalues)

plt.plot(xx, yy, marker='.', color='k', linestyle='none')

Creation of these rectangular grids is useful for a number of tasks. In the example that you have provided in your post, it is simply a way to sample a function (sin(x**2 + y**2) / (x**2 + y**2)) over a range of values for x and y.

Because this function has been sampled on a rectangular grid, the function can now be visualized as an “image”.

Additionally, the result can now be passed to functions which expect data on rectangular grid (i.e. contourf)

Courtesy of Microsoft Excel: 

Actually the purpose of np.meshgrid is already mentioned in the documentation:

np.meshgrid

Return coordinate matrices from coordinate vectors.

Make N-D coordinate arrays for vectorized evaluations of N-D scalar/vector fields over N-D grids, given one-dimensional coordinate arrays x1, x2,…, xn.

So it’s primary purpose is to create a coordinates matrices.

You probably just asked yourself:

Why do we need to create coordinate matrices?

The reason you need coordinate matrices with Python/NumPy is that there is no direct relation from coordinates to values, except when your coordinates start with zero and are purely positive integers. Then you can just use the indices of an array as the index. However when that’s not the case you somehow need to store coordinates alongside your data. That’s where grids come in.

Suppose your data is:

1  2  1
2  5  2
1  2  1

However, each value represents a 2 kilometers wide region horizontally and 3 kilometers vertically. Suppose your origin is the upper left corner and you want arrays that represent the distance you could use:

import numpy as np
h, v = np.meshgrid(np.arange(3)*3, np.arange(3)*2)

where v is:

array([[0, 0, 0],
       [2, 2, 2],
       [4, 4, 4]])

and h:

array([[0, 3, 6],
       [0, 3, 6],
       [0, 3, 6]])

So if you have two indices, let’s say x and y (that’s why the return value of meshgrid is usually xx or xs instead of x in this case I chose h for horizontally!) then you can get the x coordinate of the point, the y coordinate of the point and the value at that point by using:

h[x, y]    # horizontal coordinate
v[x, y]    # vertical coordinate
data[x, y]  # value

That makes it much easier to keep track of coordinates and (even more importantly) you can pass them to functions that need to know the coordinates.

A slightly longer explanation

However, np.meshgrid itself isn’t often used directly, mostly one just uses one of similar objects np.mgrid or np.ogrid. Here np.mgrid represents the sparse=False and np.ogrid the sparse=True case (I refer to the sparse argument of np.meshgrid). Note that there is a significant difference between np.meshgrid and np.ogrid and np.mgrid: The first two returned values (if there are two or more) are reversed. Often this doesn’t matter but you should give meaningful variable names depending on the context.

For example, in case of a 2D grid and matplotlib.pyplot.imshow it makes sense to name the first returned item of np.meshgrid x and the second one y while it’s the other way around for np.mgrid and np.ogrid.

np.ogrid and sparse grids

>>> import numpy as np
>>> yy, xx = np.ogrid[-5:6, -5:6]
>>> xx
array([[-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5]])
>>> yy
array([[-5],
       [-4],
       [-3],
       [-2],
       [-1],
       [ 0],
       [ 1],
       [ 2],
       [ 3],
       [ 4],
       [ 5]])
       

As already said the output is reversed when compared to np.meshgrid, that’s why I unpacked it as yy, xx instead of xx, yy:

>>> xx, yy = np.meshgrid(np.arange(-5, 6), np.arange(-5, 6), sparse=True)
>>> xx
array([[-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5]])
>>> yy
array([[-5],
       [-4],
       [-3],
       [-2],
       [-1],
       [ 0],
       [ 1],
       [ 2],
       [ 3],
       [ 4],
       [ 5]])

This already looks like coordinates, specifically the x and y lines for 2D plots.

Visualized:

yy, xx = np.ogrid[-5:6, -5:6]
plt.figure()
plt.title('ogrid (sparse meshgrid)')
plt.grid()
plt.xticks(xx.ravel())
plt.yticks(yy.ravel())
plt.scatter(xx, np.zeros_like(xx), color="blue", marker="*")
plt.scatter(np.zeros_like(yy), yy, color="red", marker="x")

np.mgrid and dense/fleshed out grids

>>> yy, xx = np.mgrid[-5:6, -5:6]
>>> xx
array([[-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5]])
>>> yy
array([[-5, -5, -5, -5, -5, -5, -5, -5, -5, -5, -5],
       [-4, -4, -4, -4, -4, -4, -4, -4, -4, -4, -4],
       [-3, -3, -3, -3, -3, -3, -3, -3, -3, -3, -3],
       [-2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2],
       [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1],
       [ 0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0],
       [ 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1],
       [ 2,  2,  2,  2,  2,  2,  2,  2,  2,  2,  2],
       [ 3,  3,  3,  3,  3,  3,  3,  3,  3,  3,  3],
       [ 4,  4,  4,  4,  4,  4,  4,  4,  4,  4,  4],
       [ 5,  5,  5,  5,  5,  5,  5,  5,  5,  5,  5]])
       

The same applies here: The output is reversed compared to np.meshgrid:

>>> xx, yy = np.meshgrid(np.arange(-5, 6), np.arange(-5, 6))
>>> xx
array([[-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5],
       [-5, -4, -3, -2, -1,  0,  1,  2,  3,  4,  5]])
>>> yy
array([[-5, -5, -5, -5, -5, -5, -5, -5, -5, -5, -5],
       [-4, -4, -4, -4, -4, -4, -4, -4, -4, -4, -4],
       [-3, -3, -3, -3, -3, -3, -3, -3, -3, -3, -3],
       [-2, -2, -2, -2, -2, -2, -2, -2, -2, -2, -2],
       [-1, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1],
       [ 0,  0,  0,  0,  0,  0,  0,  0,  0,  0,  0],
       [ 1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1],
       [ 2,  2,  2,  2,  2,  2,  2,  2,  2,  2,  2],
       [ 3,  3,  3,  3,  3,  3,  3,  3,  3,  3,  3],
       [ 4,  4,  4,  4,  4,  4,  4,  4,  4,  4,  4],
       [ 5,  5,  5,  5,  5,  5,  5,  5,  5,  5,  5]])
       

Unlike ogrid these arrays contain all xx and yy coordinates in the -5 <= xx <= 5; -5 <= yy <= 5 grid.

yy, xx = np.mgrid[-5:6, -5:6]
plt.figure()
plt.title('mgrid (dense meshgrid)')
plt.grid()
plt.xticks(xx[0])
plt.yticks(yy[:, 0])
plt.scatter(xx, yy, color="red", marker="x")

Functionality

It’s not only limited to 2D, these functions work for arbitrary dimensions (well, there is a maximum number of arguments given to function in Python and a maximum number of dimensions that NumPy allows):

>>> x1, x2, x3, x4 = np.ogrid[:3, 1:4, 2:5, 3:6]
>>> for i, x in enumerate([x1, x2, x3, x4]):
...     print('x{}'.format(i+1))
...     print(repr(x))
x1
array([[[[0]]],


       [[[1]]],


       [[[2]]]])
x2
array([[[[1]],

        [[2]],

        [[3]]]])
x3
array([[[[2],
         [3],
         [4]]]])
x4
array([[[[3, 4, 5]]]])

>>> # equivalent meshgrid output, note how the first two arguments are reversed and the unpacking
>>> x2, x1, x3, x4 = np.meshgrid(np.arange(1,4), np.arange(3), np.arange(2, 5), np.arange(3, 6), sparse=True)
>>> for i, x in enumerate([x1, x2, x3, x4]):
...     print('x{}'.format(i+1))
...     print(repr(x))
# Identical output so it's omitted here.

Even if these also work for 1D there are two (much more common) 1D grid creation functions:

• np.arange
• np.linspace

Besides the start and stop argument it also supports the step argument (even complex steps that represent the number of steps):

>>> x1, x2 = np.mgrid[1:10:2, 1:10:4j]
>>> x1  # The dimension with the explicit step width of 2
array([[1., 1., 1., 1.],
       [3., 3., 3., 3.],
       [5., 5., 5., 5.],
       [7., 7., 7., 7.],
       [9., 9., 9., 9.]])
>>> x2  # The dimension with the "number of steps"
array([[ 1.,  4.,  7., 10.],
       [ 1.,  4.,  7., 10.],
       [ 1.,  4.,  7., 10.],
       [ 1.,  4.,  7., 10.],
       [ 1.,  4.,  7., 10.]])
       

Applications

You specifically asked about the purpose and in fact, these grids are extremely useful if you need a coordinate system.

For example if you have a NumPy function that calculates the distance in two dimensions:

def distance_2d(x_point, y_point, x, y):
    return np.hypot(x-x_point, y-y_point)
    

And you want to know the distance of each point:

>>> ys, xs = np.ogrid[-5:5, -5:5]
>>> distances = distance_2d(1, 2, xs, ys)  # distance to point (1, 2)
>>> distances
array([[9.21954446, 8.60232527, 8.06225775, 7.61577311, 7.28010989,
        7.07106781, 7.        , 7.07106781, 7.28010989, 7.61577311],
       [8.48528137, 7.81024968, 7.21110255, 6.70820393, 6.32455532,
        6.08276253, 6.        , 6.08276253, 6.32455532, 6.70820393],
       [7.81024968, 7.07106781, 6.40312424, 5.83095189, 5.38516481,
        5.09901951, 5.        , 5.09901951, 5.38516481, 5.83095189],
       [7.21110255, 6.40312424, 5.65685425, 5.        , 4.47213595,
        4.12310563, 4.        , 4.12310563, 4.47213595, 5.        ],
       [6.70820393, 5.83095189, 5.        , 4.24264069, 3.60555128,
        3.16227766, 3.        , 3.16227766, 3.60555128, 4.24264069],
       [6.32455532, 5.38516481, 4.47213595, 3.60555128, 2.82842712,
        2.23606798, 2.        , 2.23606798, 2.82842712, 3.60555128],
       [6.08276253, 5.09901951, 4.12310563, 3.16227766, 2.23606798,
        1.41421356, 1.        , 1.41421356, 2.23606798, 3.16227766],
       [6.        , 5.        , 4.        , 3.        , 2.        ,
        1.        , 0.        , 1.        , 2.        , 3.        ],
       [6.08276253, 5.09901951, 4.12310563, 3.16227766, 2.23606798,
        1.41421356, 1.        , 1.41421356, 2.23606798, 3.16227766],
       [6.32455532, 5.38516481, 4.47213595, 3.60555128, 2.82842712,
        2.23606798, 2.        , 2.23606798, 2.82842712, 3.60555128]])
        

The output would be identical if one passed in a dense grid instead of an open grid. NumPys broadcasting makes it possible!

Let’s visualize the result:

plt.figure()
plt.title('distance to point (1, 2)')
plt.imshow(distances, origin='lower', interpolation="none")
plt.xticks(np.arange(xs.shape[1]), xs.ravel())  # need to set the ticks manually
plt.yticks(np.arange(ys.shape[0]), ys.ravel())
plt.colorbar()

And this is also when NumPys mgrid and ogrid become very convenient because it allows you to easily change the resolution of your grids:

ys, xs = np.ogrid[-5:5:200j, -5:5:200j]
# otherwise same code as above

However, since imshow doesn’t support x and y inputs one has to change the ticks by hand. It would be really convenient if it would accept the x and y coordinates, right?

It’s easy to write functions with NumPy that deal naturally with grids. Furthermore, there are several functions in NumPy, SciPy, matplotlib that expect you to pass in the grid.

I like images so let’s explore matplotlib.pyplot.contour:

ys, xs = np.mgrid[-5:5:200j, -5:5:200j]
density = np.sin(ys)-np.cos(xs)
plt.figure()
plt.contour(xs, ys, density)

Note how the coordinates are already correctly set! That wouldn’t be the case if you just passed in the density.

Or to give another fun example using astropy models (this time I don’t care much about the coordinates, I just use them to create some grid):

from astropy.modeling import models
z = np.zeros((100, 100))
y, x = np.mgrid[0:100, 0:100]
for _ in range(10):
    g2d = models.Gaussian2D(amplitude=100, 
                           x_mean=np.random.randint(0, 100), 
                           y_mean=np.random.randint(0, 100), 
                           x_stddev=3, 
                           y_stddev=3)
    z += g2d(x, y)
    a2d = models.AiryDisk2D(amplitude=70, 
                            x_0=np.random.randint(0, 100), 
                            y_0=np.random.randint(0, 100), 
                            radius=5)
    z += a2d(x, y)
    

Although that’s just “for the looks” several functions related to functional models and fitting (for example , even show examples using np.mgrid) in Scipy, etc. require grids. Most of these work with open grids and dense grids, however some only work with one of them.

Suppose you have a function:

def sinus2d(x, y):
    return np.sin(x) + np.sin(y)

and you want, for example, to see what it looks like in the range 0 to 2*pi. How would you do it? There np.meshgrid comes in:

xx, yy = np.meshgrid(np.linspace(0,2*np.pi,100), np.linspace(0,2*np.pi,100))
z = sinus2d(xx, yy) # Create the image on this grid

and such a plot would look like:

import matplotlib.pyplot as plt
plt.imshow(z, origin='lower', interpolation='none')
plt.show()

So np.meshgrid is just a convenience. In principle the same could be done by:

z2 = sinus2d(np.linspace(0,2*np.pi,100)[:,None], np.linspace(0,2*np.pi,100)[None,:])

but there you need to be aware of your dimensions (suppose you have more than two …) and the right broadcasting. np.meshgrid does all of this for you.

Also meshgrid allows you to delete coordinates together with the data if you, for example, want to do an interpolation but exclude certain values:

condition = z>0.6
z_new = z[condition] # This will make your array 1D

so how would you do the interpolation now? You can give x and y to an interpolation function like scipy.interpolate.interp2d so you need a way to know which coordinates were deleted:

x_new = xx[condition]
y_new = yy[condition]

and then you can still interpolate with the “right” coordinates (try it without the meshgrid and you will have a lot of extra code):

from scipy.interpolate import interp2d
interpolated = interp2d(x_new, y_new, z_new)

and the original meshgrid allows you to get the interpolation on the original grid again:

interpolated_grid = interpolated(xx[0], yy[:, 0]).reshape(xx.shape)

These are just some examples where I used the meshgrid there might be a lot more.

meshgrid helps in creating a rectangular grid from two 1-D arrays of all pairs of points from the two arrays.

x = np.array([0, 1, 2, 3, 4])
y = np.array([0, 1, 2, 3, 4])

Now, if you have defined a function f(x,y) and you wanna apply this function to all the possible combination of points from the arrays ‘x’ and ‘y’, then you can do this:

f(*np.meshgrid(x, y))

Say, if your function just produces the product of two elements, then this is how a cartesian product can be achieved, efficiently for large arrays.

Referred from here

Basic Idea

Given possible x values, xs, (think of them as the tick-marks on the x-axis of a plot) and possible y values, ys, meshgrid generates the corresponding set of (x, y) grid points—analogous to set((x, y) for x in xs for y in yx). For example, if xs=[1,2,3] and ys=[4,5,6], we’d get the set of coordinates {(1,4), (2,4), (3,4), (1,5), (2,5), (3,5), (1,6), (2,6), (3,6)}.

Form of the Return Value

However, the representation that meshgrid returns is different from the above expression in two ways:

First, meshgrid lays out the grid points in a 2d array: rows correspond to different y-values, columns correspond to different x-values—as in list(list((x, y) for x in xs) for y in ys), which would give the following array:

   [[(1,4), (2,4), (3,4)],
    [(1,5), (2,5), (3,5)],
    [(1,6), (2,6), (3,6)]]

Second, meshgrid returns the x and y coordinates separately (i.e. in two different numpy 2d arrays):

   xcoords, ycoords = (
       array([[1, 2, 3],
              [1, 2, 3],
              [1, 2, 3]]),
       array([[4, 4, 4],
              [5, 5, 5],
              [6, 6, 6]]))
   # same thing using np.meshgrid:
   xcoords, ycoords = np.meshgrid([1,2,3], [4,5,6])
   # same thing without meshgrid:
   xcoords = np.array([xs] * len(ys)
   ycoords = np.array([ys] * len(xs)).T

Note, np.meshgrid can also generate grids for higher dimensions. Given xs, ys, and zs, you’d get back xcoords, ycoords, zcoords as 3d arrays. meshgrid also supports reverse ordering of the dimensions as well as sparse representation of the result.

Applications

Why would we want this form of output?

Apply a function at every point on a grid: One motivation is that binary operators like (+, -, *, /, **) are overloaded for numpy arrays as elementwise operations. This means that if I have a function def f(x, y): return (x - y) ** 2 that works on two scalars, I can also apply it on two numpy arrays to get an array of elementwise results: e.g. f(xcoords, ycoords) or f(*np.meshgrid(xs, ys)) gives the following on the above example:

array([[ 9,  4,  1],
       [16,  9,  4],
       [25, 16,  9]])

Higher dimensional outer product: I’m not sure how efficient this is, but you can get high-dimensional outer products this way: np.prod(np.meshgrid([1,2,3], [1,2], [1,2,3,4]), axis=0).

Contour plots in matplotlib: I came across meshgrid when investigating drawing contour plots with matplotlib for plotting decision boundaries. For this, you generate a grid with meshgrid, evaluate the function at each grid point (e.g. as shown above), and then pass the xcoords, ycoords, and computed f-values (i.e. zcoords) into the contourf function.