New answers tagged pattern-matching
7
The problem arises from naming the inner sub-patterns in your two pattern definitions with the same identifier. When you combine the patterns with Alternatives, you then have a pattern with two sub-patterns tagged with the same name, which is bad.
Consider this simpler case.
pattern1 = StartOfString ~~ "(" ~~ t : WordCharacter .. ~~ ")1" ~...
0
Yet another approach is to take the Union of 2-subsets that satisfy the condition:
highlighted = Union @@ Select[Length[Intersection @@ #] >= 2 &] @ Subsets[list, {2}]
{{a, b, c}, {b, c, d}, {c, a, m}, {c, d, n}}
list /. x : Alternatives @@ highlighted :> Style[x, Gray]
1
list /. x : {__Symbol} /;
Max[Length[Intersection[x, #]] & /@ DeleteCases[list, x]] >= 2 :>
Style[x, Gray]
A method using GatherBy:
gb = Join @@ Select[Length@# > 1 &]@
GatherBy[list, Function[x, Max[Length[Intersection[x, #]] & /@ DeleteCases[x][list]]]]
{{a, b, c}, {b, c, d}, {c, a, m}, {c, d, n}}
list /. x : ...
1
I think the easy way is still use Gather and reorder the index. Here we handle the general situation.
SeedRandom[400];
list = Table[RandomSample[Alphabet[], 3], 40];
newlist = Thread[Range[Length@list] -> list];
result = Gather[newlist,
Length[Intersection[Last@#1, Last@#2]] >= 2 &];
keys = Keys /@ result;
keyc = Thread[keys -> RandomColor[...
1
A variant of OP's solution that avoids nesting of Highlighted:
list //. {a___, x : ({_, _, _} | Highlighted[{_, _, _}, ___]), b___,
y : ({_, _, _}), c___} /; Length[Intersection[x /. Highlighted -> (# &), y]] >= 2 :>
{a, Highlighted[x /. Highlighted -> (# &)], b, Highlighted@y, c}
Same approach using Style:
list //. {a___, x : (...
3
rg = RelationGraph[UnsameQ @ ## && Length@Intersection[##] >= 2 &, list]
hl = VertexList @ EdgeList @ rg
{{a, b, c}, {b, c, d}, {c, a, m}, {c, d, n}}
list /. x : Alternatives @@ hl :> Style[x, Gray]
list /. x : Alternatives @@ hl :> Highlighted[x, BaseStyle -> Red]
HighlightGraph[rg, hl]
You can also use ConnectedComponents and ...
2
ClearAll[formatList]
formatList[list_] := Module[{rules},
rules =
AssociationThread[
list -> (If[Max[#] >= 2, Gray, Black] & /@
Function[{element},
Length@Intersection[element, #] & /@
Complement[list, {element}]] /@ list)
];
Style[#, rules[#]] & /@ list
]
formatList[list]
3
You can use SequenceCases:
SequenceCases[
data,
{textHeader, s:PatternSequence[moveTo, text]..} -> {textHeader, s}
]
{{textHeader, moveTo, text, moveTo, text}, {textHeader, moveTo, text}}
2
Let us introduce a function factor that will be able to factorize the expression taking out a desired multiplicand:
factor[expr_, fact_, fun1_ : Expand, fun2_ : Identity] :=
Module[{a = fact, b = expr/fact},fun2[Evaluate[a]]*fun1[Evaluate[b]]]
This is the expression:
expr = -t^-b + (t - t^(1/2 + 1/(8 k)))^-b;
By applying the function factor as follows
...
1
Based on Mathematica documentation, and alternative solution could be implement a simple replacement rule as follows in the next function:
NoUnits[f_] := f /. q_Quantity :> QuantityMagnitude[q]
Then, implement the Plot as a pure function to increase drastically the execution time:
AbsoluteTiming[Plot[
#,
{l, .1, 15}, Frame -> True,
...
0
If already in the input of the actual notebook: KnownUnitQ["Microflicks"] gives True. So there is need for someaction to get this referentially connceted to other known unit entities.
Mathematica V 12.0.0 has FormularLookup and FormulaData for lot of the questions purposes.
FormulaLookup["planck's law"]
The unit system in Physics is in ...
2
Instead of using UnitConvert, you can just divide the output by the relevant output unit. Here's one implementation of the idea, based on the OP's version:
ClearAll @ NoUnits;
SetAttributes[NoUnits, HoldAll];
NoUnits[fun_, quants : {__Quantity}, output : Quantity[1, out_]] :=
Expand[fun/output /. Thread[quants[[All, 1]] -> quants]] /. x_ y_Piecewise ...
4
data = {{"S", "T", "C", "R1", "R2"}, {a, 1, x, 2.9}, {a, 1, y,
2.6}, {a, 1, z, 8.7}, {a, 2, x, 9.4, 0.372}, {a, 2, y, 8.1,
0.208}, {a, 2, z, 7.6, 0.154}, {b, 1, x, 7.5}, {b, 1, y, 7.3}, {b,
1, z, 1.7}, {b, 2, x, 3.9, 0.213}, {b, 2, y, 7.9, 0.435}, {b, 2,
z, 2.5, 0.294}, {c, 1, x, 6.2}, {c, ...
4
I am not sure wether I understood correctly which results should be recombined but this code does the following: First the data is gathered by condition, then by test, then subsets of length 2 are generated and in the final step the numerical computation on the results is applied:
test1[{{a_,1,c_,R1a_},{b_,1,c_,R1b_}}]:={a,b,1,c,R1a/R1b}
test2[{{a_,2,c_,R1a_,...
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