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df012_DefinesAndFiltersAsStrings.C
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1/// \file
2/// \ingroup tutorial_dataframe
3/// \notebook -nodraw
4///
5/// This tutorial illustrates how to save some typing when using RDataFrame
6/// by invoking functions that perform jit-compiling at runtime.
7///
8/// \macro_code
9/// \macro_output
10///
11/// \date October 2017
12/// \author Guilherme Amadio
13
15{
16 // We will inefficiently calculate an approximation of pi by generating
17 // some data and doing very simple filtering and analysis on it
18
19 // We start by creating an empty dataframe where we will insert 10 million
20 // random points in a square of side 2.0 (that is, with an inscribed circle
21 // of radius 1.0)
22
23 size_t npoints = 10000000;
24 ROOT::RDataFrame df(npoints);
25
26 // Define what we want inside the dataframe. We do not need to define p as an array,
27 // but we do it here to demonstrate how to use jitting with RDataFrame
28
29 // NOTE: Although it's possible to use "for (auto&& x : p)" below, it will
30 // shadow the name of the data column "x", and may cause compilation failures
31 // if the local variable and the data column are of different types or the
32 // local x variable is declared in the global scope of the lambda function
33
34 auto pidf = df.Define("x", "gRandom->Uniform(-1.0, 1.0)")
35 .Define("y", "gRandom->Uniform(-1.0, 1.0)")
36 .Define("p", "std::array<double, 2> v{x, y}; return v;")
37 .Define("r", "double r2 = 0.0; for (auto&& x : p) r2 += x*x; return sqrt(r2);");
38
39 // Now we have a dataframe with columns x, y, p (which is a point based on x
40 // and y), and the radius r = sqrt(x*x + y*y). In order to approximate pi, we
41 // need to know how many of our data points fall inside the unit circle compared
42 // with the total number of points. The ratio of the areas is
43 //
44 // A_circle / A_square = pi r*r / l * l, where r = 1.0, and l = 2.0
45 //
46 // Therefore, we can approximate pi with 4 times the number of points inside the
47 // unit circle over the total number of points in our dataframe:
48
49 auto incircle = *(pidf.Filter("r <= 1.0").Count());
50
51 double pi_approx = 4.0 * incircle / npoints;
52
53 std::cout << "pi is approximately equal to " << pi_approx << std::endl;
54}
ROOT's RDataFrame offers a high level interface for analyses of data stored in TTrees,...
Definition: RDataFrame.hxx:42
REAL incircle(struct mesh *m, struct behavior *b, vertex pa, vertex pb, vertex pc, vertex pd)
Definition: triangle.c:5863