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How to find the rows having values between -1 and 1 in a given numpy 2D-array?

I have a np.array of shape (15,3).
final_vals = array([[ 37, -84, -143],
[ 29, 2, -2],
[ -18, -2, 0],
[ -3, 6, 0],
[ 361, -5, 2],
[ -23, 4, 8],
[ 0, -1, 0],
[ -1, 1, 0],
[ 62, 181, 83],
[-193, -14, -2],
[ 42, -154, -92],
[ 16, -13, 1],
[ -10, -3, 0],
[-299, 244, 110],
[ 223, -237, -110]])
am trying to find the rows whose element values are between -1 and 1.In the array printed above ROW-6 and ROW-7 are target/result rows.
I tried,
result_idx = np.where(np.logical_and(final_vals>=-1, final_vals<=1))
which returns,
result_idx = (array([ 2, 3, 6, 6, 6, 7, 7, 7, 11, 12], dtype=int64),
array([2, 2, 0, 1, 2, 0, 1, 2, 2, 2], dtype=int64))
I want my program to return only row numbers

You could take the absolute value of all elements, and check which rows's elements are smaller or equal to 1. Then use np.flatnonzero to find the indices where all columns fullfil the condition:
np.flatnonzero((np.abs(final_vals) <= 1).all(axis=1))
Output
array([6, 7], dtype=int64)

Another way to do this based on your approach is to find the truth value of each element and then use numpy.all for each row. Then numpy.where gets you what you want.
mask = (final_vals <= 1) * (final_vals >= -1)
np.where(np.all(mask, axis=1))

How about
np.where(np.all((-1<=final_vals) & (final_vals<=1),axis=1))

You could use np.argwhere:
r = np.logical_and(final_vals <= 1, final_vals >=-1)
result = np.argwhere(r.all(1)).flatten()
print(result)
Output
[6 7]

Another way is using pandas,
you can achieve the row with following code:
df = pd.DataFrame(final_vals)
temp= ((df>=-1) & (df<=1 )).product(axis=1)
rows = temp[temp!=0].keys()
rows
At first it check numbers that are between -1 and +1 and then check rows(with axis=1) that all values accept the condition.
and the result is:
Int64Index([ 6, 7], dtype='int64')

Just a simple list comprehension:
[ i for i, row in enumerate(final_vals) if all([ e >= -1 and e <= 1 for e in row ]) ]
#=> [6, 7]

Sliced numpy array does not modify original array

I've run into this interaction with arrays that I'm a little confused. I can work around it, but for my own understanding, I'd like to know what is going on.
Essentially, I have a datafile that I'm trying to tailor so I can run this as an input for some code I've already written. This involves some calculations on some columns, rows, etc. In particular, I also need to rearrange some elements, where the original array isn't being modified as I expect it would.
import numpy as np
ex_data = np.arange(12).reshape(4,3)
ex_data[2,0] = 0 #Constructing some fake data
ex_data[ex_data[:,0] == 0][:,1] = 3
print ex_data
Basically, I look in a column of interest, collect all the rows where that column contains a parameter value of interest and just reassigning values.
With the snippet of code above, I would expect ex_data to have it's column 1 elements, conditional if it's column 0 element is equal to 0, to be assigned a value of 3. However what I'm seeing is that there is no effect at all.
>>> ex_data
array([[ 0, 1, 2],
[ 3, 4, 5],
[ 0, 7, 8],
[ 9, 10, 11]])
In another case, if I don't 'slice', my 'sliced' data file, then the reassignment goes on as normal.
ex_data[ex_data[:,0] == 0] = 3
print ex_data
Here I'd expect my entire row, conditional to where column 0 is equal to 0, be populated with 3. This is what you see.
>>> ex_data
array([[ 3, 3, 3],
[ 3, 4, 5],
[ 3, 3, 3],
[ 9, 10, 11]])
Can anyone explain the interaction?

In [368]: ex_data
Out[368]:
array([[ 0, 1, 2],
[ 3, 4, 5],
[ 0, 7, 8],
[ 9, 10, 11]])
The column 0 test:
In [369]: ex_data[:,0]==0
Out[369]: array([ True, False, True, False])
That boolean mask can be applied to the rows as:
In [370]: ex_data[ex_data[:,0]==0,0]
Out[370]: array([0, 0]) # the 0's you expected
In [371]: ex_data[ex_data[:,0]==0,1]
Out[371]: array([1, 7]) # the col 1 values you want to replace
In [372]: ex_data[ex_data[:,0]==0,1] = 3
In [373]: ex_data
Out[373]:
array([[ 0, 3, 2],
[ 3, 4, 5],
[ 0, 3, 8],
[ 9, 10, 11]])
The indexing you tried:
In [374]: ex_data[ex_data[:,0]==0]
Out[374]:
array([[0, 3, 2],
[0, 3, 8]])
produces a copy. Assigning ...[:,1]=3 just changes that copy, not the original array. Fortunately in this case, it is easy to use
ex_data[ex_data[:,0]==0,1]
instead of
ex_data[ex_data[:,0]==0][:,1]

Numpy assignment like 'numpy.take'

Is it possible to assign to a numpy array along the lines of how the take functionality works?
E.g. if I have a an array a, a list of indices inds, and a desired axis, I can use take as follows:
import numpy as np
a = np.arange(12).reshape((3, -1))
inds = np.array([1, 2])
print(np.take(a, inds, axis=1))
[[ 1 2]
[ 5 6]
[ 9 10]]
This is extremely useful when the indices / axis needed may change at runtime.
However, numpy does not let you do this:
np.take(a, inds, axis=1) = 0
print(a)
It looks like there is some limited (1-D) support for this via numpy.put, but I was wondering if there was a cleaner way to do this?

In [222]: a = np.arange(12).reshape((3, -1))
...: inds = np.array([1, 2])
...:
In [223]: np.take(a, inds, axis=1)
Out[223]:
array([[ 1, 2],
[ 5, 6],
[ 9, 10]])
In [225]: a[:,inds]
Out[225]:
array([[ 1, 2],
[ 5, 6],
[ 9, 10]])
construct an indexing tuple
In [226]: idx=[slice(None)]*a.ndim
In [227]: axis=1
In [228]: idx[axis]=inds
In [229]: a[tuple(idx)]
Out[229]:
array([[ 1, 2],
[ 5, 6],
[ 9, 10]])
In [230]: a[tuple(idx)] = 0
In [231]: a
Out[231]:
array([[ 0, 0, 0, 3],
[ 4, 0, 0, 7],
[ 8, 0, 0, 11]])
Or for a[inds,:]:
In [232]: idx=[slice(None)]*a.ndim
In [233]: idx[0]=inds
In [234]: a[tuple(idx)]
Out[234]:
array([[ 4, 0, 0, 7],
[ 8, 0, 0, 11]])
In [235]: a[tuple(idx)]=1
In [236]: a
Out[236]:
array([[0, 0, 0, 3],
[1, 1, 1, 1],
[1, 1, 1, 1]])
PP's suggestion:
def put_at(inds, axis=-1, slc=(slice(None),)):
return (axis<0)*(Ellipsis,) + axis*slc + (inds,) + (-1-axis)*slc
To be used as in a[put_at(ind_list,axis=axis)]
I've seen both styles on numpy functions. This looks like one used for extend_dims, mine was used in apply_along/over_axis.
earlier thoughts
In a recent take question I/we figured out that it was equivalent to arr.flat[ind] for some some raveled index. I'll have to look that up.
There is an np.put that is equivalent to assignment to the flat:
Signature: np.put(a, ind, v, mode='raise')
Docstring:
Replaces specified elements of an array with given values.
The indexing works on the flattened target array. `put` is roughly
equivalent to:
a.flat[ind] = v
Its docs also mention place and putmask (and copyto).
numpy multidimensional indexing and the function 'take'
I commented take (without axis) is equivalent to:
lut.flat[np.ravel_multi_index(arr.T, lut.shape)].T
with ravel:
In [257]: a = np.arange(12).reshape((3, -1))
In [258]: IJ=np.ix_(np.arange(a.shape[0]), inds)
In [259]: np.ravel_multi_index(IJ, a.shape)
Out[259]:
array([[ 1, 2],
[ 5, 6],
[ 9, 10]], dtype=int32)
In [260]: np.take(a,np.ravel_multi_index(IJ, a.shape))
Out[260]:
array([[ 1, 2],
[ 5, 6],
[ 9, 10]])
In [261]: a.flat[np.ravel_multi_index(IJ, a.shape)] = 100
In [262]: a
Out[262]:
array([[ 0, 100, 100, 3],
[ 4, 100, 100, 7],
[ 8, 100, 100, 11]])
and to use put:
In [264]: np.put(a, np.ravel_multi_index(IJ, a.shape), np.arange(1,7))
In [265]: a
Out[265]:
array([[ 0, 1, 2, 3],
[ 4, 3, 4, 7],
[ 8, 5, 6, 11]])
Use of ravel is unecessary in this case but might useful in others.

I have given an example for use of
numpy.take in 2 dimensions. Perhaps you can adapt that to your problem

You can juste use indexing in this way :
a[:,[1,2]]=0

What is the purpose of meshgrid in Python / NumPy?

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 3 x 2 kilometer area (horizontal x vertical). 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 scipy.interpolate.interp2d,
scipy.interpolate.griddata 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.

Short answer
The purpose of meshgrid is to help replace slow Python loops by faster vectorized operations available in NumPy library. meshgrid role is to prepare 2D arrays required by the vectorized operation.
Basic example showing the principle
Let's say we have two sequences of values,
a = [2,7,9,20]
b = [1,6,7,9] ​
and we want to perform an operation on each possible pair of values, one taken from the first list, one taken from the second list. We also want to store the result. For example, let's say we want to get the sum of the values for each possible pair.
Slow and laborious method
c = []
for i in range(len(b)):
row = []
for j in range(len(a)):
row.append (a[j] + b[i])
c.append (row)
print (c)
Result:
[[3, 8, 10, 21],
[8, 13, 15, 26],
[9, 14, 16, 27],
[11, 16, 18, 29]]
Fast and easy method
i,j = np.meshgrid (a,b)
c = i + j
print (c)
Result:
[[ 3 8 10 21]
[ 8 13 15 26]
[ 9 14 16 27]
[11 16 18 29]]
You can see from this basic illustration how the explicit slow Python loops have been replaced by hidden faster C loops in Numpy library. This principle is widely used for 3D operations, included colored pixel maps. The common example is a 3D plot.
Common use: 3D plot
x = np.arange(-4, 4, 0.25)
y = np.arange(-4, 4, 0.25)
X, Y = np.meshgrid(x, y)
R = np.sqrt(X**2 + Y**2)
Z = np.sin(R)
(Borrowed from this site)
meshgrid is used to create pairs of coordinates between -4 and +4 with .25 increments in each direction X and Y. Each pair is then used to find R, and Z from it. This way of preparing "a grid" of coordinates is frequently used in plotting 3D surfaces, or coloring 2D surfaces.
Meshgrid under the hood
The two arrays prepared by meshgrid are:
(array([[ 2, 7, 9, 20],
[ 2, 7, 9, 20],
[ 2, 7, 9, 20],
[ 2, 7, 9, 20]]),
array([[1, 1, 1, 1],
[6, 6, 6, 6],
[7, 7, 7, 7],
[9, 9, 9, 9]]))
These arrays are created by repeating the values provided, either horizontally or vertically. The two arrays are shape compatible for a vector operation.
Origin
numpy.meshgrid comes from MATLAB, like many other NumPy functions. So you can also study the examples from MATLAB to see meshgrid in use, the code for the 3D plotting looks the same in MATLAB.

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.

Behind the scenes:
import numpy as np
def meshgrid(x , y):
XX = []
YY = []
for colm in range(len(y)):
XX.append([])
YY.append([])
for row in range(len(x)):
XX[colm].append(x[row])
YY[colm].append(y[colm])
return np.asarray(XX), np.asarray(YY)
Lets take dataset of #Sarsaparilla's answer as example:
y = [7, 6, 5]
x = [1, 2, 3, 4]
xx, yy = meshgrid(x , y)
and it outputs:
>>> xx
array([[1, 2, 3, 4],
[1, 2, 3, 4],
[1, 2, 3, 4]])
>>> yy
array([[7, 7, 7, 7],
[6, 6, 6, 6],
[5, 5, 5, 5]])

Numpy: get index of smallest value based on conditions

I have an array as such:
array([[ 10, -1],
[ 3, 1],
[ 5, -1],
[ 7, 1]])
What I want is to get the index of row with the smallest value in the first column and -1 in the second.
So basically, np.argmin() with a condition for the second column to be equal to -1 (or any other value for that matter).
In my example, I would like to get 2 which is the index of [ 5, -1].
I'm pretty sure there's a simple way, but I can't find it.

import numpy as np
a = np.array([
[10, -1],
[ 3, 1],
[ 5, -1],
[ 7, 1]])
mask = (a[:, 1] == -1)
arg = np.argmin(a[mask][:, 0])
result = np.arange(a.shape[0])[mask][arg]
print result

np.argwhere(a[:,1] == -1)[np.argmin(a[a[:, 1] == -1, 0])]

This is not efficient but if you have a relatively small array and want a one-line solution:
>>> a = np.array([[ 10, -1],
... [ 3, 1],
... [ 5, -1],
... [ 7, 1]])
>>> [i for i in np.argsort(a[:, 0]) if a[i, 1] == -1][0]
2