#!/usr/bin/env python
# coding: utf-8

# # df012_DefinesAndFiltersAsStrings
# Use just-in-time-compiled Filters and Defines for quick prototyping.
# 
# This tutorial illustrates how to use jit-compiling features of RDataFrame
# to define data using C++ code in a Python script.
# 
# 
# 
# 
# **Author:** Guilherme Amadio (CERN)  
# <i><small>This notebook tutorial was automatically generated with <a href= "https://github.com/root-project/root/blob/master/documentation/doxygen/converttonotebook.py">ROOTBOOK-izer</a> from the macro found in the ROOT repository  on Wednesday, April 17, 2024 at 11:07 AM.</small></i>

# In[1]:


import ROOT


#  We will inefficiently calculate an approximation of pi by generating
#  some data and doing very simple filtering and analysis on it.

#  We start by creating an empty dataframe where we will insert 10 million
#  random points in a square of side 2.0 (that is, with an inscribed unit
#  circle).

# In[2]:


npoints = 10000000
df = ROOT.RDataFrame(npoints)


#  Define what data we want inside the dataframe. We do not need to define p
#  as an array, but we do it here to demonstrate how to use jitting with RDataFrame.

# In[3]:


pidf = df.Define("x", "gRandom->Uniform(-1.0, 1.0)") \
         .Define("y", "gRandom->Uniform(-1.0, 1.0)") \
         .Define("p", "std::array<double, 2> v{x, y}; return v;") \
         .Define("r", "double r2 = 0.0; for (auto&& w : p) r2 += w*w; return sqrt(r2);")


#  Now we have a dataframe with columns x, y, p (which is a point based on x
#  and y), and the radius r = sqrt(x*x + y*y). In order to approximate pi, we
#  need to know how many of our data points fall inside the circle of radius
#  one compared with the total number of points. The ratio of the areas is
# 
#      A_circle / A_square = pi r*r / l * l, where r = 1.0, and l = 2.0
# 
#  Therefore, we can approximate pi with four times the number of points inside
#  the unit circle over the total number of points:

# In[4]:


incircle = pidf.Filter("r <= 1.0").Count().GetValue()

pi_approx = 4.0 * incircle / npoints

print("pi is approximately equal to %g" % (pi_approx))