# Ratio between first max. and first min. of g[r] vs time from Coordination analysis

Quote from Prashant Dwivedi on August 26, 2020, 4:41 pmHello Constanze,

I’m trying to calculate the ratio between first max. and the first min. value of g[r] .

I did a coordination analysis(for calculating RDF) and then using a python script modifier implemented max() and min() function.

But the problem with this is, it’s not finding the 1st max. and 1min. value. For min. it takes 0. Also if there is two-particle type in the system then it can not take value for rdf.y and return nan.So, is there any possible way to do it in Ovito by using python modifier.

I tried something like this:

from ovito.data import * import numpy as np def modify(frame, data, output): rdf = data.tables[‘coordination-rdf’] rdf_max = np.max(rdf.y) rdf_min = np.min(rdf.y) rdf_ratio = (rdf_min/rdf_max) print(rdf_max) print(rdf_min) print(rdf_ratio)

Thanks,

Best

Prash

Hello Constanze,

I’m trying to calculate the ratio between first max. and the first min. value of g[r] .

I did a coordination analysis(for calculating RDF) and then using a python script modifier implemented max() and min() function.

But the problem with this is, it’s not finding the 1st max. and 1min. value. For min. it takes 0. Also if there is two-particle type in the system then it can not take value for rdf.y and return nan.

So, is there any possible way to do it in Ovito by using python modifier.

I tried something like this:

from ovito.data import * import numpy as np def modify(frame, data, output): rdf = data.tables[‘coordination-rdf’] rdf_max = np.max(rdf.y) rdf_min = np.min(rdf.y) rdf_ratio = (rdf_min/rdf_max) print(rdf_max) print(rdf_min) print(rdf_ratio)

Thanks,

Best

Prash

**Uploaded files:**

Quote from Constanze Kalcher on August 27, 2020, 1:13 pmHi Prashant,

yes, that is the expected behavior of the

`numpy.max()`

and`numpy.min()`

functions, please refer to the numpy documentation https://numpy.org/doc/stable/.There are a lot of different ways of finding local maxima or minima in a 1-d numpy array with python, see e.g. https://stackoverflow.com/questions/4624970/finding-local-maxima-minima-with-numpy-in-a-1d-numpy-array.

Concerning the shape of

`rdf.y`

: Please note that when you activate the option "Compute partial RDFs",`data.tables[‘coordination-rdf’].y`

will be a numpy array of shape (n, X), where n is the number of bins and X is the number of possible particle type pair-combinations. If your system consists of 2 particle types, you will thus get three columns, i.e. 1-1, 1-2 and 2-2, and you can access them like this:#1-1 print(rdf.y[:,0]) #1-2 print(rdf.y[:,1]) #2-2 print(rdf.y[:,2])

Hi Prashant,

yes, that is the expected behavior of the `numpy.max()`

and `numpy.min()`

functions, please refer to the numpy documentation https://numpy.org/doc/stable/.

There are a lot of different ways of finding local maxima or minima in a 1-d numpy array with python, see e.g. https://stackoverflow.com/questions/4624970/finding-local-maxima-minima-with-numpy-in-a-1d-numpy-array.

Concerning the shape of `rdf.y`

: Please note that when you activate the option "Compute partial RDFs", `data.tables[‘coordination-rdf’].y`

will be a numpy array of shape (n, X), where n is the number of bins and X is the number of possible particle type pair-combinations. If your system consists of 2 particle types, you will thus get three columns, i.e. 1-1, 1-2 and 2-2, and you can access them like this:

#1-1 print(rdf.y[:,0]) #1-2 print(rdf.y[:,1]) #2-2 print(rdf.y[:,2])

Quote from Prashant Dwivedi on August 27, 2020, 2:19 pmThank you, Constanze.

It works.

Have a nice day.

Thank you, Constanze.

It works.

Have a nice day.