doc/v614/df012__DefinesAndFiltersAsStrings_8py.html

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).
npoints = 10000000
tdf = ROOT.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
pidf = tdf.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 4 times the number of points inside
## the unit circle over the total number of points:
incircle = pidf.Filter("r <= 1.0").Count().GetValue()
pi_approx = 4.0 * incircle / npoints
print("pi is approximately equal to %g" % (pi_approx))