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"source": [
"# vo005_Combinations\n",
"In this tutorial we learn how combinations of RVecs can be built.\n",
"\n",
"\n",
"\n",
"\n",
"**Author:** Stefan Wunsch \n",
"This notebook tutorial was automatically generated with ROOTBOOK-izer from the macro found in the ROOT repository on Tuesday, May 19, 2026 at 08:28 PM."
]
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"source": [
"import ROOT\n",
"from ROOT.VecOps import Take, Combinations"
]
},
{
"cell_type": "markdown",
"id": "63433a4e",
"metadata": {},
"source": [
"RVec can be sorted in Python with the inbuilt sorting function because\n",
"PyROOT implements a Python iterator"
]
},
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"source": [
"v1 = ROOT.RVecD(3)\n",
"v1[0], v1[1], v1[2] = 1, 2, 3\n",
"v2 = ROOT.RVecD(2)\n",
"v2[0], v2[1] = -4, -5"
]
},
{
"cell_type": "markdown",
"id": "74b03caf",
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"source": [
"To get the indices, which result in all combinations, you can call the\n",
"following helper.\n",
"Note that you can also pass the size of the vectors directly."
]
},
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"outputs": [],
"source": [
"idx = Combinations(v1, v2)"
]
},
{
"cell_type": "markdown",
"id": "21abbe62",
"metadata": {},
"source": [
"Next, the respective elements can be taken via the computed indices."
]
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"outputs": [],
"source": [
"c1 = Take(v1, idx[0])\n",
"c2 = Take(v2, idx[1])"
]
},
{
"cell_type": "markdown",
"id": "3f19c7d1",
"metadata": {},
"source": [
"Finally, you can perform any set of operations conveniently."
]
},
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"name": "stdout",
"output_type": "stream",
"text": [
"Combinations of { 1.0000000, 2.0000000, 3.0000000 } and { -4.0000000, -5.0000000 }:\n",
"1.0 * -4.0 = -4.0\n",
"1.0 * -5.0 = -5.0\n",
"2.0 * -4.0 = -8.0\n",
"2.0 * -5.0 = -10.0\n",
"3.0 * -4.0 = -12.0\n",
"3.0 * -5.0 = -15.0\n"
]
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""
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"source": [
"v3 = c1 * c2\n",
"\n",
"print(\"Combinations of {} and {}:\".format(v1, v2))\n",
"for i in range(len(v3)):\n",
" print(\"{} * {} = {}\".format(c1[i], c2[i], v3[i]))\n",
"print"
]
},
{
"cell_type": "markdown",
"id": "46b7d994",
"metadata": {},
"source": [
"However, if you want to compute operations on unique combinations of a\n",
"single RVec, you can perform this as follows."
]
},
{
"cell_type": "markdown",
"id": "ad192f9c",
"metadata": {},
"source": [
"Get the indices of unique triples for the given vector."
]
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},
"outputs": [],
"source": [
"v4 = ROOT.RVecD(4)\n",
"v4[0], v4[1], v4[2], v4[3] = 1, 2, 3, 4\n",
"idx2 = Combinations(v4, 3)"
]
},
{
"cell_type": "markdown",
"id": "81c2fdab",
"metadata": {},
"source": [
"Take the elements and compute any operation on the returned collections."
]
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"outputs": [
{
"name": "stdout",
"output_type": "stream",
"text": [
"Unique triples of { 1.0000000, 2.0000000, 3.0000000, 4.0000000 }:\n",
"1.0 * 2.0 * 3.0 = 6.0\n",
"1.0 * 2.0 * 4.0 = 8.0\n",
"1.0 * 3.0 * 4.0 = 12.0\n",
"2.0 * 3.0 * 4.0 = 24.0\n"
]
}
],
"source": [
"c3 = Take(v4, idx2[0])\n",
"c4 = Take(v4, idx2[1])\n",
"c5 = Take(v4, idx2[2])\n",
"\n",
"v5 = c3 * c4 * c5\n",
"\n",
"print(\"Unique triples of {}:\".format(v4))\n",
"for i in range(len(v5)):\n",
" print(\"{} * {} * {} = {}\".format(c3[i], c4[i], c5[i], v5[i]))"
]
}
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