What is the purpose of meshgrid in Python / NumPy?
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
)
What is the purpose of meshgrid in Python / NumPy?
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
)
Numpy meshgrid in 3D
Here is the source code of meshgrid:
def meshgrid(x,y):
"""
Return coordinate matrices from two coordinate vectors.
Parameters
----------
x, y : ndarray
Two 1-D arrays representing the x and y coordinates of a grid.
Returns
-------
X, Y : ndarray
For vectors `x`, `y` with lengths ``Nx=len(x)`` and ``Ny=len(y)``,
return `X`, `Y` where `X` and `Y` are ``(Ny, Nx)`` shaped arrays
with the elements of `x` and y repeated to fill the matrix along
the first dimension for `x`, the second for `y`.
See Also
--------
index_tricks.mgrid : Construct a multi-dimensional "meshgrid"
using indexing notation.
index_tricks.ogrid : Construct an open multi-dimensional "meshgrid"
using indexing notation.
Examples
--------
>>> X, Y = np.meshgrid([1,2,3], [4,5,6,7])
>>> X
array([[1, 2, 3],
[1, 2, 3],
[1, 2, 3],
[1, 2, 3]])
>>> Y
array([[4, 4, 4],
[5, 5, 5],
[6, 6, 6],
[7, 7, 7]])
`meshgrid` is very useful to evaluate functions on a grid.
>>> x = np.arange(-5, 5, 0.1)
>>> y = np.arange(-5, 5, 0.1)
>>> xx, yy = np.meshgrid(x, y)
>>> z = np.sin(xx**2+yy**2)/(xx**2+yy**2)
"""
x = asarray(x)
y = asarray(y)
numRows, numCols = len(y), len(x) # yes, reversed
x = x.reshape(1,numCols)
X = x.repeat(numRows, axis=0)
y = y.reshape(numRows,1)
Y = y.repeat(numCols, axis=1)
return X, Y
It is fairly simple to understand. I extended the pattern to an arbitrary number of dimensions, but this code is by no means optimized (and not thoroughly error-checked either), but you get what you pay for. Hope it helps:def meshgrid2(*arrs):
arrs = tuple(reversed(arrs)) #edit
lens = map(len, arrs)
dim = len(arrs)
sz = 1
for s in lens:
sz*=s
ans = []
for i, arr in enumerate(arrs):
slc = [1]*dim
slc[i] = lens[i]
arr2 = asarray(arr).reshape(slc)
for j, sz in enumerate(lens):
if j!=i:
arr2 = arr2.repeat(sz, axis=j)
ans.append(arr2)
return tuple(ans)
numpy.meshgrid explanation
In [214]: nx, ny = (3, 2)
In [215]: x = np.linspace(0, 1, nx)
In [216]: x
Out[216]: array([ 0. , 0.5, 1. ])
In [217]: y = np.linspace(0, 1, ny)
In [218]: y
Out[218]: array([ 0., 1.])
Using unpacking to better see the 2 arrays produced by meshgrid
:In [225]: X,Y = np.meshgrid(x, y)
In [226]: X
Out[226]:
array([[ 0. , 0.5, 1. ],
[ 0. , 0.5, 1. ]])
In [227]: Y
Out[227]:
array([[ 0., 0., 0.],
[ 1., 1., 1.]])
and for the sparse version. Notice that X1
looks like one row of X
(but 2d). and Y1
like one column of Y
.In [228]: X1,Y1 = np.meshgrid(x, y, sparse=True)
In [229]: X1
Out[229]: array([[ 0. , 0.5, 1. ]])
In [230]: Y1
Out[230]:
array([[ 0.],
[ 1.]])
When used in calculations like plus and times, both forms behave the same. That's because of numpy's
broadcasting.In [231]: X+Y
Out[231]:
array([[ 0. , 0.5, 1. ],
[ 1. , 1.5, 2. ]])
In [232]: X1+Y1
Out[232]:
array([[ 0. , 0.5, 1. ],
[ 1. , 1.5, 2. ]])
The shapes might also help:In [235]: X.shape, Y.shape
Out[235]: ((2, 3), (2, 3))
In [236]: X1.shape, Y1.shape
Out[236]: ((1, 3), (2, 1))
The X
and Y
have more values than are actually needed for most uses. But usually there isn't much of penalty for using them instead the sparse versions. Function of meshgrid numpy
You could use np.frompyfunc
:
import numpy as np
import matplotlib.pyplot as plt
def f(x, y):
DD = np.matrix([[0., 0.],[0., 0.]]) + 0.j
omega = x + 1.j * y
# set up dispersion matrix
DD[0,0] = 1 + omega
DD[1,0] = omega
DD[0,1] = omega
DD[1,1] = 1 - omega
metric = np.linalg.det(DD)
return metric
f = np.frompyfunc(f, 2, 1)
xx = np.arange(1., 2., 0.1)
yy = np.arange(1., 2., 0.1)
x, y = np.meshgrid(xx, yy)
FPlot = f(x, y)
plt.contourf(x, y, FPlot) # Note that this is only using the real part of FPlot
plt.show()
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