2
$\begingroup$

Given an Association list w/ a common Key but also some distinct ones:

data3 = {<|"commonKey" -> 1, "b" -> 3|>, <|"commonKey" -> 2, 
    "c" -> 4|>} // Dataset

enter image description here

How to implement logic similar to Switch and Which using named slots?

data3[All, Switch[#commonKey, 1, #b] &] // Normal

{3, Missing["Failed"]}

data3[All, Switch[#commonKey, 1, #b, 2, #c] &] //Normal

{Missing["Failed"], Missing["Failed"]}

Similarly, using Which:

data3[All, Which[#commonKey == 1, #b, #commonKey == 2, #c] &]//Normal

{Missing["Failed"], Missing["Failed"]}

The cause is Function::slota Named Slot can't be filled from the association with the Key that is missing.

Is there an elegant workaround that can be mapped on existing Association list, as opposed to aggregating via GroupBy[#commonKey&]? (Reason: the actual data set that is simplified here is chronologically sorted, and grouping would not preserve the item sequence).

Improve this question
$\endgroup$
2
  • $\begingroup$ You really are trying to save time by abbreviating with w/ w/?? $\endgroup$ – halirutan♦ May 12 '15 at 0:27
  • $\begingroup$ Programmable: w/, w/i, w/o ... $\endgroup$ – alancalvitti May 12 '15 at 18:03
1
$\begingroup$

You could use the whole association and access it inside your function:

data3[All, Switch[#commonKey, 1, #["b"], _, 0] &]

Mathematica graphics

Is this an alternative?

Improve this answer
$\endgroup$
3
  • $\begingroup$ Not sure where the 0 comes from but did you mean to write this? data3[All, Switch[#commonKey, 1, #["b"], 2, #["c"]] &]? That works while Switch[#commonKey, 1, #b, 2, #c] . $\endgroup$ – alancalvitti May 12 '15 at 18:09
  • $\begingroup$ @alancalvitti I meant to write it because I thought you might want a fallback rule if there isn't any #b in the association. You can just remove the part and adapt if like you want. $\endgroup$ – halirutan♦ May 12 '15 at 19:18
  • $\begingroup$ You may also be interested in this related q: mathematica.stackexchange.com/questions/83414/… $\endgroup$ – alancalvitti May 14 '15 at 18:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.