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Add a column to some Dataset

Claude Mante, Retired
Posted 4 years ago

Hello everybody!

It's likely a silly question, but I can't find the way to add a column to some dataset. Consider:

dataset = Dataset[{
   <|"a" -> 1, "b" -> 3, "c" -> {1}|>,
   <|"a" -> 2, "b" -> 8, "c" -> {2, 3}|>,
   <|"a" -> 3, "b" -> 0, "c" -> {3}|>,
   <|"a" -> 4, "b" -> 1, "c" -> {4, 5}|>,
   <|"a" -> 5, "b" -> 12, "c" -> {5, 6, 7}|>,
   <|"a" -> 6, "b" -> -3, "c" -> {}|>}]

and compute

Prov = dataset[All, {#a - #b &}]

How could I build another dataset with Prov as a supplementary column, named "Prov" ,for instance?

Thanks
POSTED BY: Claude Mante
3 Replies
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Try:

dataset[All, <|#, "Prov" -> #a - #b|> &]

For working with Dataset I found this nice post inspiring.
POSTED BY: Henrik Schachner

Thank, Claudio!

It works on this small data set, but not on the real one, unfortunately... It's a data set about fishing (joined).

We need first routines to estimate the second parameter (probability of success) of a negative binomial (or Pascal, in my case) distribution:

ClearAll["Global`*"];

Place = NotebookDirectory[];
SetDirectory[Place <> "/Données"];
\[Epsilon] = 10^-6;
Brut = Import["Peches2005.xlsx", {"Dataset", 1}, HeaderLines -> 1];
pEstPonct[temps_, fish_] := 
  If[N@fish <= 1 + \[Epsilon], 
   0, (fish - 1)/(temps + fish - 1)];  (* Haldane; 1945, Sahai; 1979 *)


pEstGroup[temps_, fish_ /; fish >= 1] := 
 Block[{freq, \[Theta], x, fx, p},
     freq = Tally[temps];
     x = First@Transpose@freq;
     fx = Last@Transpose@freq;
     \[Theta] = (x.fx)/((x.fx) + fish Total@fx - 1);
     p = (1 - \[Theta]);
  N[p]  
  ]

Then, the sub-dataset associated with trouts (coded "TRF") is extracted:

Cherche = "TRF";
Lepoisson = Brut[Select[#"nom_sp" == Cherche &]];

Now, estimation of the probability (2 methods); Diff denotes the time spent, nb (the variable "Total" , an unfortunate coding) the number of fishes; Pas is the time unit.

Pas = 4;     (* Unité de temps, en minutes : temps entre deux actes \
de pêche (lancers,...)  *)
Diff = Flatten@
   Normal@Lepoisson[
     All, {N[(AbsoluteTime@#"hcont" - 
           AbsoluteTime@#"hdeb") (1/(60 Pas))] &}];
nb = Lepoisson[All, {"Total"}];
Paire = Dataset@
   Join[{Join[Keys@nb[[1]], {"Diff"}]}, 
    Table[Join[(Keys@nb[[1]] /. nb)[[y]], {Diff[[y]]}], {y, 
      Length@Diff}]];
Head@Paire

"Paire" denotes the pair (total, Diff), built with your method. It seems correct but, for some reason, the following instructions don't work:

Paire = Select[Paire, #Diff > 0 &];
Length@Paire
Paire = Select[Paire, #Total =!= "" &];
Length@Paire

Any idea?

Claude

Attachments:
Peches2005.xlsx
POSTED BY: Claude Mante

Hi Claude,

See if that's what you want to do:

dataset2 = {<|"a" -> 1, "b" -> 3, "c" -> {1}|>, <|"a" -> 2, "b" -> 8, 
    "c" -> {2, 3}|>, <|"a" -> 3, "b" -> 0, "c" -> {3}|>, <|"a" -> 4, 
    "b" -> 1, "c" -> {4, 5}|>, <|"a" -> 5, "b" -> 12, 
    "c" -> {5, 6, 7}|>, <|"a" -> 6, "b" -> -3, "c" -> {}|>};
Prov2 = Table[
   dataset2[[x, 1]] - dataset2[[x, 2]], {x, 1, Length@dataset2}];

Dataset@Join[{Join[Keys@dataset2[[1]], {"Prov"}]}, 
  Table[Join[(Keys@dataset2[[1]] /. dataset2)[[y]], {Prov2[[y]]}], {y,
     1, Length@Prov2}]]

im1

Maybe there is another way to do it, but you can do it this way...
POSTED BY: Claudio Chaib
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