Timeline for How can I add a column into a existing Dataset?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 31, 2018 at 11:34 | history | edited | bobthechemist | CC BY-SA 3.0 | Added important (IMO) point from comments about valid symbol names. |
| Jul 30, 2017 at 22:11 | history | edited | WReach | CC BY-SA 3.0 | deleted 4 characters in body |
| Jul 30, 2017 at 22:03 | history | edited | WReach | CC BY-SA 3.0 | added the _2017 Update_ which discussses the absence of the "shorter form" from the documentation |
| Apr 10, 2016 at 19:08 | history | edited | WReach | CC BY-SA 3.0 | for completeness, added an example that uses nested association syntax |
| Apr 10, 2016 at 18:56 | comment | added | WReach | @HyperGroups When a key name is not a valid symbol we can write, for example, #["col_name"]. | |
| Apr 10, 2016 at 4:52 | comment | added | HyperGroups | Hhi, Sometimes, one cannot use #col1, for example, col1_name, Is there simpler method to do such a thing? like pandas dataframe?col3=dataSetFinal["col1"]/dataSetFinal["col2"] | |
| Jul 4, 2015 at 19:17 | comment | added | Murta | @TaliesinBeynon I´m studying R data.table, and now I understand why they are so popular. These kind of operations are much simpler to be performed. In data.table syntax, add a new column would be as simpler as: ds[, col3 := col1+col2], and the values would be changed by reference. No need to do ds = ds[, col3 := col1+col2]. Here is a data.table sheet if nice ideas I miss in Dataset | |
| May 21, 2015 at 15:12 | comment | added | Taliesin Beynon | @Szabolcs this requires kernel hooks that I haven't had enough time to actually implement. It is still planned, though not for 10.2. | |
| May 19, 2015 at 12:03 | comment | added | Szabolcs | @TaliesinBeynon This does not yet work in v10.1. Is it still planned? | |
| Jul 12, 2014 at 7:59 | comment | added | Taliesin Beynon | @WReach, that's a great answer. Eventually, though, you should be able to write dataset[[All, "foo"]] = {...} and have that just work (as long as the list is the right length). | |
| Jul 11, 2014 at 20:29 | vote | accept | Murta | ||
| Jun 25, 2014 at 2:46 | comment | added | Murta | Cool! Tks. I'll not accept now just to respect the protocol +1 | |
| Jun 25, 2014 at 2:36 | history | answered | WReach | CC BY-SA 3.0 |