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3

Perhaps PairedHistogram[Sequence@@#, {15,45,5}]&/@ Subsets[Normal@(Values@ds[GroupBy[Key["Value1"]],All,"Value3"]),{2}]


0

For discussion purposes, let's use some data from Mathematica's example databases: $connection = OpenSQLConnection["publisher"]; $dbTest = SQLSelect[$connection, "SALESDETAILS", {"TITLE_ID", "QTY_ORDERED"}] (* {{"PY2002", 75}, {"PY2002", 10}, <<17>> , {"CK4017", 20}, {"CK4016", 20}} *) We can create a histogram directly from this data ...


2

Guessing/hoping that the following is not too far off from what you have in mind: lld = {{49.5671, 15.34, 50.0033, 17.9569, 100}, {48.7635, 16.0479, 49.5671, 15.34, 35}, {50.0033, 17.9569, 48.7635, 16.0479, 75}}; vertices = Join @@ (Partition[#, 2] & /@ lld[[All, ;; -2]]); edges = Property[UndirectedEdge[{#, #2}, {#3, #4}], EdgeStyle ...


5

The error I get is that the integrals do not converge. I expect that this is because the empirical PDFs are zero far from the data, and Mathematica does not automatically take $0\log 0$ to be $0$. You can do it explicitly as follows: NIntegrate[With[{f = PDF[smoothDistribution1D, x]}, If[f > 0, f Log[f], 0]], {x, -∞, ∞}] (* -1.4238 *) Compared to the ...


2

Perhaps Histogram[A[[All, All, -1]]]



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