I'm trying to use "Do" to change the entries in a column of a dataset using two adjacent columns. When I simply assign i and execute the lin for each value of i, it works but when I enclose it in Do, the Dataset becomes null.

The code I'm trying to use is:

sample5 =Do[ReplacePart[sample4, {i, "GeoPosition"} ->
GeoPosition[{sample4[i, "Latitude"], 
sample4[i, "Longitude"]}]], {i, Length[sample4]}]

The trial Dataset is:

Dataset[{<|"Facility Name" -> "Stanford Health Care", "Latitude" -> 37.4330139, "Longitude" -> -122.1758423,"GeoPosition" -> "GeoPosition"|>, <|"Facility Name" ->  "California Pacific Medical Center - California West Campus \  Hospital", "Latitude" -> 37.7861157, "Longitude" -> -122.4560595,"GeoPosition" -> "GeoPosition"|>} ]

Mathematica 11.3 on Mac Os Sierra 10.12.6

  • 1
    $\begingroup$ There is almost never a good reason to use Do. Try Table instead. $\endgroup$ Mar 13 '19 at 20:54
  • 1
    $\begingroup$ There are plenty of good reasons to use Do (haven't checked if this is one of them) $\endgroup$
    – Jason B.
    Mar 13 '19 at 20:56
  • $\begingroup$ Can Table be used to modify a Dataset and still result in a Dataset? $\endgroup$
    – user5091
    Mar 13 '19 at 20:56

Your first problem is that

ReplacePart[ sample4, rules]

does not modify sample4, but in fact returns a new Dataset with the desired replacement.

The reason that your example5 is Null is because that is what Do returns. Do is a wonderful function, but it won't return anything (it just does something).

Minimal modification to your code:

sample5 = sample4;
    sample5 = ReplacePart[
        {i, "GeoPosition"} -> GeoPosition[{sample5[i, "Latitude"], sample5[i, "Longitude"]}]
    {i, Length @ sample4}

For this particular use case, you don't need to loop at all, you can do it all in one ReplacePart command:

    {i_, "GeoPosition"} :> GeoPosition[{sample4[i, "Latitude"], sample4[i, "Longitude"]}]

For a functional approach you may use Query directly or implicitly with Dataset. Taking sample4 as in OP then

direct use with,

Query[All, <|#, "GeoPosition" -> GeoPosition@{#["Latitude"], #["Longitude"]}|> &]@sample4

and implicit use with Slot syntax sugar

sample4[All, <|#, "GeoPosition" -> GeoPosition@{#Latitude, #Longitude}|> &]

Both of the above give

Mathematica graphics

Hope this helps.


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