New answers tagged filtering
2
First, let's do it with datasets having unnamed columns, because they are simpler. We use Intersection to find common elements in the first column. We use a simple replacement rule that replaces a list, which represents a row in the dataset, with Nothing.
{nest1, nest2} = {{{"A", 1}, {"B", 2}, {"C", 3}, {"D", 4}}, {{"A",
111}, {"C", 333}, {"E", 555}}...
5
GeneralUtilities`PrintDefinitions@SavitzkyGolayMatrix
3
We can use Complement to find the elements in set1 whose ID values are not contained within set2:
Complement[set1, set2, SameTest -> (#["ID"] === #2["ID"] &)]
Update for the revised question
Complement provides a nice notational convenience, but for large (15,000) sets it runs orders of magnitude slower than the solution proposed in the question. ...
5
This should work:
Select[ll, AllTrue[#, NumericQ] &]
{{1, 2, 3}, {[Pi], 4, 5}}
1
Using replacement rules:
SomeData /. x : {a_, b_, c_} :> If[b - c < 0.2, Most@x, {a, 999999}]
(* {{0., 999999}, {0.5, 999999}, {1., 1.538}, {1.5, 999999},
{2., 0.6}, {2.5, 999999}, {3., 0.149}, {3.5, 0.408},
{4., 1.141}, {4.5, 1.696}, {5., 1.833}, {5.5, 999999},
{6., 2.1}, {6.5, 1.773}, {7., 999999}, {7.5, 0.889},
{8., 999999}, {8.5, 0....
3
ClearAll[clip];
clip[δ_, repl_: 9999][{a_, b_, c_}] :=
{a, Clip[b, {-∞, c + δ - 2 $MachineEpsilon}, {-∞, repl}]};
Map[clip[.2]] @ SomeData
{{0., 9999}, {0.5, 9999}, {1., 1.538}, {1.5, 9999}, {2., 0.6}, {2.5,
9999}, {3., 0.149}, {3.5, 0.408}, {4., 1.141}, {4.5, 1.696}, {5.,
1.833}, {5.5, 9999}, {6., 2.1}, {6.5, 1.773}, {7., 9999}, {...
2
Would something like this do what you want:
Block[{outlierPositions =
Flatten[Position[SomeData, #] & /@
Select[SomeData, ! (#[[2]] - #[[3]] < 0.2) &], 1],
replacements},
replacements = Append[#, 2] & /@ outlierPositions;
ReplacePart[SomeData, replacements -> 999999][[;; , {1, 2}]]
]
(*{{0., 999999}, {0.5, 999999}, {1., 1....
7
Is this ok?
Select[GatherBy[l, #[[{-2, -1}]] &], Length[#] > 1 &]
{{{1, 2, 3, 6, 6}, {1, 2, 4, 6, 6}}, {{1, 2, 5, 8, 8}, {1, 2, 6, 8,
8}}}
3
Try Binomial:
prb[tot_, sub_] := 1 - Binomial[tot - sub, 5]/Binomial[tot, 5]
7
You can use GatherBy, SplitBy or GroupBy as follows:
min1 = First /@ GatherBy[Sort@list, Most]
{{0, 2, 1}, {0, 3, 0}, {1, 2, 0}, {2, 2, 0}}
max1 = Last /@ GatherBy[Sort@list, Most]
{{0, 2, 3}, {0, 3, 0}, {1, 2, 0}, {2, 2, 1}}
Alternatively,
min2 = First /@ SplitBy[Sort@list, Most];
max2 = Last /@ SplitBy[Sort@list, Most];
min3 = First@*MinimalBy[...
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