I want to replace 0. value in [row_5,B] become value of [row_5,A], and the value is changing in many rows, and I don't like to type it {0.0->1.516*10^6}

One solution of mine is

1.select row_5 lines, and split the data in parts 2=part1,part2, 
2.change part1[with row_5] modifiy row B by rowA
3.combine part1 and part2

and I do this in the association, not the dataset.

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  • $\begingroup$ Like this? ReplacePart[dataset, {{2, "a"}, {5, "a"}} -> y] $\endgroup$ – Jason B. Apr 12 '16 at 12:37
  • $\begingroup$ @JasonB yes, may be values of y is y1, y2, also some ways like Query, dataset[All, something] is welcome $\endgroup$ – HyperGroups Apr 12 '16 at 12:38

To change all rows of type "row_5" to that the "B" value is replaced with the corresponding "A" value, we can write:

dataset[All, <| #, "B" -> Replace[#type, {"row_5" -> #A, _ -> #B}] |> &]

dataset screenshot

If we only want to change the cases where the "B" value is zero, we can add that as an extra condition:

dataset[All, <| #, "B" -> Replace[#type, {"row_5" :> #A /; #B == 0., _ -> #B}] |>&]

If we wanted to replace all zero values in the B column with the corresponding A value, we could do this:

dataset[All, <| #, "B" -> Replace[#B, 0. -> #A] |> &]

dataset screenshot

  • $\begingroup$ Hi, yes, but sometimes, it's not 0. but 1.15*10^100, and maybe not so easy to type it rightly, since the table is large,some rows are hidden, and If some 0.0 is not in type=row_5, I don't want it to be replaced. $\endgroup$ – HyperGroups Apr 12 '16 at 14:29
  • $\begingroup$ So you want to change all row_5 entries, even when the value is not zero? $\endgroup$ – WReach Apr 12 '16 at 14:32
  • $\begingroup$ yes, that's the real purpose, better if could do a check, if value of [row_5,B] is 0, then replaced by [row_5,A] $\endgroup$ – HyperGroups Apr 12 '16 at 14:34
  • $\begingroup$ ha, I find it , thanks. dataset[All,<|#,"B"->If[#type=="row_5",#A,#B]|>&] $\endgroup$ – HyperGroups Apr 12 '16 at 14:40

Probably more than one way to do this, but scrolling through the documentation I found this,

dataset[All,{n->f}] selectively maps $f$ onto the $n^{th}$ column

What you want to do is map a replacement rule on to the "a" column,

dataset[All, {"a" -> (# /. {2 -> y1, 5 -> y2} &)}]

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  • $\begingroup$ sorry, I changed the dataset and make the problem clearer $\endgroup$ – HyperGroups Apr 12 '16 at 14:08
  • $\begingroup$ Anyway, thanks, I find the answer, via WReach's answer. $\endgroup$ – HyperGroups Apr 12 '16 at 14:41

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