Update and select simultaneously in data.table

0
votes

I have a that looks like this: 

# Load library
library(data.table)

# Set RNG seed
set.seed(-1)

# Create data table
dt <- data.table(year = 2000:2019,
                 value = runif(20))

# Peek
dt
#>     year      value
#>  1: 2000 0.48666718
#>  2: 2001 0.19136526
#>  3: 2002 0.99327188
#>  4: 2003 0.14670268
#>  5: 2004 0.24158948
#>  6: 2005 0.53710122
#>  7: 2006 0.35821235
#>  8: 2007 0.87191898
#>  9: 2008 0.39259106
#> 10: 2009 0.21656725
#> 11: 2010 0.79346199
#> 12: 2011 0.26007283
#> 13: 2012 0.26831560
#> 14: 2013 0.53564863
#> 15: 2014 0.29142160
#> 16: 2015 0.94810504
#> 17: 2016 0.06352872
#> 18: 2017 0.09133961
#> 19: 2018 0.31097680
#> 20: 2019 0.76861987

I'd like to calculate standard scores for value and then drop the value variable. To do this, I use chaining: in my first set of square brackets I calculate my standard score (ss), then in my second set of square brackets I select the columns year and ss thus dropping value.

# Calculate standard score and drop 'value' column
dt[, ss := as.vector(scale(value))][, .(year, ss)]
#>     year         ss
#>  1: 2000  0.1656755
#>  2: 2001 -0.8473906
#>  3: 2002  1.9036392
#>  4: 2003 -1.0006105
#>  5: 2004 -0.6750908
#>  6: 2005  0.3386950
#>  7: 2006 -0.2750031
#>  8: 2007  1.4873246
#>  9: 2008 -0.1570631
#> 10: 2009 -0.7609324
#> 11: 2010  1.2181692
#> 12: 2011 -0.6116816
#> 13: 2012 -0.5834039
#> 14: 2013  0.3337118
#> 15: 2014 -0.5041362
#> 16: 2015  1.7486893
#> 17: 2016 -1.2859481
#> 18: 2017 -1.1905397
#> 19: 2018 -0.4370499
#> 20: 2019  1.1329455

Created on 2019-08-07 by the reprex package (v0.3.0)

This is my desired result.

My question: do I have to use a chain in this situation or is there a way to update (i.e., calculate ss) and select the columns I want within a single set of []?

1
rdata.table
Not sure If it can be done without chaining, but the way I 'd do it would be to define the column to drop rather the ones to keep, i.e. dt[, ss := as.vector(scale(value))][, value := NULL] - Sotos
dt[, .(year, ss = scale(value))]? - Roland
That's awesome @Roland – could you add it as an answer so that I can close the question? - Lyngbakr

1 Answers

5
votes

Apparently you are looking for the simple dt[, .(year, ss = scale(value))]. You subset the data.table by selecting columns in a list and within that list you can also create new vectors.