The problem is, that numpy can't give you the derivatives directly and you have two options:

With NUMPY

What you essentially have to do, is to define a grid in three dimension and to evaluate the function on this grid. Afterwards you feed this table of function values to numpy.gradient to get an array with the numerical derivative for every dimension (variable).

Example from here:

from numpy import *

x,y,z = mgrid[-100:101:25., -100:101:25., -100:101:25.]

V = 2*x**2 + 3*y**2 - 4*z # just a random function for the potential

Ex,Ey,Ez = gradient(V)

Without NUMPY

You could also calculate the derivative yourself by using the centered difference quotient.

This is essentially, what numpy.gradient is doing for every point of your predefined grid.

You need to give gradient a matrix that describes your angular frequency values for your (x,y) points. e.g.

def f(x,y):
    return np.sin((x + y))
x = y = np.arange(-5, 5, 0.05)
X, Y = np.meshgrid(x, y)
zs = np.array([f(x,y) for x,y in zip(np.ravel(X), np.ravel(Y))])
Z = zs.reshape(X.shape)

gx,gy = np.gradient(Z,0.05,0.05)

You can see that plotting Z as a surface gives:

Here is how to interpret your gradient:

gx is a matrix that gives the change dz/dx at all points. e.g. gx[0][0] is dz/dx at (x0,y0). Visualizing gx helps in understanding:

Since my data was generated from f(x,y) = sin(x+y) gy looks the same.

Here is a more obvious example using f(x,y) = sin(x)...

f(x,y)

and the gradients

update Let's take a look at the xy pairs.

This is the code I used:

def f(x,y):
    return np.sin(x)
x = y = np.arange(-3,3,.05)
X, Y = np.meshgrid(x, y)
zs = np.array([f(x,y) for x,y in zip(np.ravel(X), np.ravel(Y))])
xy_pairs = np.array([str(x)+','+str(y) for x,y in zip(np.ravel(X), np.ravel(Y))])
Z = zs.reshape(X.shape)
xy_pairs = xy_pairs.reshape(X.shape)

gy,gx = np.gradient(Z,.05,.05)

Now we can look and see exactly what is happening. Say we wanted to know what point was associated with the value atZ[20][30]? Then...

>>> Z[20][30]
-0.99749498660405478

And the point is

>>> xy_pairs[20][30]
'-1.5,-2.0'

Is that right? Let's check.

>>> np.sin(-1.5)
-0.99749498660405445

Yes.

And what are our gradient components at that point?

>>> gy[20][30]
0.0
>>> gx[20][30]
0.070707731517679617

Do those check out?

dz/dy always 0 check. dz/dx = cos(x) and...

>>> np.cos(-1.5)
0.070737201667702906

Looks good.

You'll notice they aren't exactly correct, that is because my Z data isn't continuous, there is a step size of 0.05 and gradient can only approximate the rate of change.

The Numpy documentation indicates that gradient works for any dimensions:

numpy.gradient(f, *varargs)

Return the gradient of an N-dimensional array.

The gradient is computed using central differences in the interior and first differences at the boundaries. The returned gradient hence has the same shape as the input array.

Parameters :

f: array_like. An N-dimensional array containing samples of a scalar function.

*varargs: 0, 1, or N scalars specifying the sample distances in each direction, that is: dx, dy, dz, ... The default distance is 1.

Returns :

g: ndarray. N arrays of the same shape as f giving the derivative of f with respect to each dimension.

Seems like you should be able to extend your 2-dimensional code to 3D like you would expect.

pd.Series.diff() only takes the differences. It doesn't divide by the delta of the index as well.

This gets you the answer

recv.diff() / recv.index.to_series().diff().dt.total_seconds()

2017-01-20 20:00:00            NaN
2017-01-20 20:05:00    4521.493333
2017-01-20 20:10:00    4533.760000
2017-01-20 20:15:00    4557.493333
2017-01-20 20:20:00    4536.053333
2017-01-20 20:25:00    4567.813333
2017-01-20 20:30:00    4406.160000
2017-01-20 20:35:00    4366.720000
2017-01-20 20:40:00    4407.520000
2017-01-20 20:45:00    4421.173333
Freq: 300S, dtype: float64

You could also use numpy.gradient passing the bytes_in and the delta you expect to have. This will not reduce the length by one, instead making assumptions about the edges.

np.gradient(bytes_in, 300) * 8

array([ 4521.49333333,  4527.62666667,  4545.62666667,  4546.77333333,
        4551.93333333,  4486.98666667,  4386.44      ,  4387.12      ,
        4414.34666667,  4421.17333333])