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Data Science Wolfram Language

Convert the following list into a Dataset?

Tsai Ming-Chou

Tsai Ming-Chou, National Defense Medical Center

Posted 6 years ago

The first row should be treated as a field name instead of data. temp = {{a, b, c}, {1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; Dataset[temp]

POSTED BY: Tsai Ming-Chou

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Girish Arabale

Posted 6 years ago

One way of doing I can think of quickly is: p = aqi[All /* Catenate /* (Partition[#, 2] &) /* GeoPosition, {"Latitude", "Longitude"}] which will create GeoPosition arrays. You can generate map: GeoListPlot[p]

POSTED BY: Girish Arabale

Tsai Ming-Chou

Tsai Ming-Chou, National Defense Medical Center

Posted 6 years ago

If I need to use GeoSmoothHistogram[locs,espec], I want to use GeoPosition and AQI indicator (aqi [[;;, 3]]). What should I do?Is it possible to consider GeoElevationData at the same time?

POSTED BY: Tsai Ming-Chou

Girish Arabale

Posted 6 years ago

pos = Values[Normal[aqi[All, {"Latitude", "Longitude", "AQI"}]]] /. {lat_, long_, aqi_} :> Thread[GeoPosition[{lat, long}] -> aqi]; GeoSmoothHistogram[Association[pos], ColorFunction -> "Rainbow", Mesh -> 20, MeshStyle -> Directive[Opacity[0.085], Blue], GeoBackground -> "StreetMap"] You need to generate GeoElevationData : ged= GeoElevationData /@(Values[Normal[aqi[All, {"Latitude", "Longitude"}]]] /. {lat_, long_} :> GeoPosition[{lat, long}]) Check with GeoBackground -> "ReliefMap" option in the above plot. Play with GeoElevationData too. Have been just throwing quick solutions while commuting...

pos = Values[Normal[aqi[All, {"Latitude", "Longitude", "AQI"}]]] /. {lat_, long_, aqi_} :> Thread[GeoPosition[{lat, long}] -> aqi];

GeoSmoothHistogram[Association[pos], ColorFunction -> "Rainbow", 
Mesh -> 20, MeshStyle -> Directive[Opacity[0.085], Blue], 
GeoBackground -> "StreetMap"]

enter image description here

You need to generate GeoElevationData :

   ged= GeoElevationData /@(Values[Normal[aqi[All, {"Latitude", "Longitude"}]]] /. {lat_, long_} :> GeoPosition[{lat, long}])

Check with GeoBackground -> "ReliefMap" option in the above plot. Play with GeoElevationData too. Have been just throwing quick solutions while commuting...

POSTED BY: Girish Arabale

Tsai Ming-Chou

Tsai Ming-Chou, National Defense Medical Center

Posted 6 years ago

Thank you for your help, the code seems to be really short and effective. But for a beginner, the logic between the instructions, I still need to learn more. Especially the use of Association, Map, Replace and Rule.It is really a headache.

POSTED BY: Tsai Ming-Chou

Tsai Ming-Chou

Tsai Ming-Chou, National Defense Medical Center

Posted 6 years ago

It's amazing! Different instructions produce different data structures! I am an environmental epidemiology student and mathematica beginner. I would like to use public data to observe the distribution of air pollution concentrations but plagued by a large number of monitoring station coordinates and observations (CSV file). The coordinates and observations of these coordinates and observations are converted to GeoPosition or GeoShothistogram every day. Now that the CSV file has converted to Dataset, and establish a long-term database and the workload reduced. The next step is to convert the station coordinates to GeoPosition. The coordinate data is already known as aqi[[;;,{-2,-1}]]. How to put coordinates into GeoPosition? aqi = Import[ "http://opendata.epa.gov.tw/ws/Data/AQI/?$format=csv", {"CSV", "Dataset"}, HeaderLines -> 1]; I have tried the following instructions but it does not work p={#Latitude, #Longitude} & /@ aqi;GeoListPlot[p]

POSTED BY: Tsai Ming-Chou

Gustavo Delfino

Posted 6 years ago

Given: temp = {{a, b, c}, {1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; Here are some additional styles of solving it. Here we see that we can use `<\| \|>` In place of `Association[]` for a little shorter notation: <\|ToString /@ First@temp -> Transpose@Rest@temp // Thread\|> (* <\|"a" -> {1, 4, 7}, "b" -> {2, 5, 8}, "c" -> {3, 6, 9}\|> ) Also if you don't like having too many # and & in your code, you can use the operator form of `Map` and also `Curry` which was introduced in version 11.3: Transpose@Rest@temp // Map[Curry[AssociationThread, 2][ToString /@ First@temp]] ({<\|"a" -> 1, "b" -> 4, "c" -> 7\|>, <\|"a" -> 2, "b" -> 5, "c" -> 8\|>, <\|"a" -> 3, "b" -> 6, "c" -> 9\|>}*)

Given:

temp = {{a, b, c}, {1, 2, 3}, {4, 5, 6}, {7, 8, 9}};

Here are some additional styles of solving it. Here we see that we can use <| |> In place of Association[] for a little shorter notation:

<|ToString /@ First@temp -> Transpose@Rest@temp // Thread|>
(* <|"a" -> {1, 4, 7}, "b" -> {2, 5, 8}, "c" -> {3, 6, 9}|> *)

Also if you don't like having too many # and & in your code, you can use the operator form of Map and also Curry which was introduced in version 11.3:

Transpose@Rest@temp //
   Map[Curry[AssociationThread, 2][ToString /@ First@temp]]
(*{<|"a" -> 1, "b" -> 4, "c" -> 7|>,
   <|"a" -> 2, "b" -> 5, "c" -> 8|>,
   <|"a" -> 3, "b" -> 6, "c" -> 9|>}*)

POSTED BY: Gustavo Delfino

Christopher French

Christopher French, instant ramen noodles

Posted 6 years ago

These two approaches give the same result: AssociationThread[keys -> #] & /@ values ({<\|"a" -> 1, "b" -> 2, "c" -> 3\|>, <\|"a" -> 4, "b" -> 5, "c" -> 6\|>, <\|"a" -> 7, "b" -> 8, "c" -> 9\|>}) Association /@ (MapThread[Rule, {keys, #}] & /@ values) ({<\|"a" -> 1, "b" -> 2, "c" -> 3\|>, <\|"a" -> 4, "b" -> 5, "c" -> 6\|>, <\|"a" -> 7, "b" -> 8, "c" -> 9\|>}) Three approaches: SemanticImportString[ExportString[temp, "CSV"]]

These two approaches give the same result:

AssociationThread[keys -> #] & /@ values
(*{<|"a" -> 1, "b" -> 2, "c" -> 3|>, <|"a" -> 4, "b" -> 5, "c" -> 6|>, <|"a" -> 7, "b" -> 8, "c" -> 9|>}*)

Association /@ (MapThread[Rule, {keys, #}] & /@ values)
(*{<|"a" -> 1, "b" -> 2, "c" -> 3|>, <|"a" -> 4, "b" -> 5, "c" -> 6|>, <|"a" -> 7, "b" -> 8, "c" -> 9|>}*)

Three approaches:

SemanticImportString[ExportString[temp, "CSV"]]

POSTED BY: Christopher French

Tsai Ming-Chou

Tsai Ming-Chou, National Defense Medical Center

Posted 6 years ago

First of all, thank for your generous help! From everyone's suggestion, I try various methods and observe what is different! temp = {{a, b, c}, {1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; keys = ToString /@ First@temp; values = Rest@temp; Dataset[Association /@ MapThread[Rule, {keys, #}] & /@ values] Dataset[Association /@ (MapThread[Rule, {keys, #}] & /@ values)] The second dataset can match my imagination of structural data. But if keys = First@temp; keys = First@temp; Dataset[Association /@ MapThread[Rule, {keys, #}] & /@ values] Dataset[Association /@ (MapThread[Rule, {keys, #}] & /@ values)] Although there is no difference in the appearance of the dataset, the actual structure is not the same. {{<\|a -> 1\|>, <\|b -> 2\|>, <\|c -> 3\|>}, {<\|a -> 4\|>, <\|b -> 5\|>, <\|c -> 6\|>}, {<\|a -> 7\|>, <\|b -> 8\|>, <\|c -> 9\|>}} {<\|a -> 1, b -> 2, c -> 3\|>, <\|a -> 4, b -> 5, c -> 6\|>, <\|a -> 7, b -> 8, c -> 9\|>}

POSTED BY: Tsai Ming-Chou

Neil Singer

Neil Singer, AC Kinetics, Inc.

Posted 6 years ago

If your data is starting in a file (ie CSV or TSV file), you can go direct from file to dataset with SemanticImport[]. Regards

POSTED BY: Neil Singer

Henrik Schachner

Henrik Schachner, Radiation Therapy Center, Weilheim, Germany

Posted 6 years ago

Most likely there is a simpler way, but here is a first attempt : temp = {{a, b, c}, {1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; keys = ToString /@ First[temp]; values = Rest[temp]; dataset = Dataset[Association /@ (MapThread[Rule, {keys, #}] & /@ values)] which gives Regards -- Henrik