TidyR Challenge: Data.Table Solution
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Arun Srinivasan is the man! Once he saw that his data.table solution to the TidyR Challenge had an issue, he fixed it!
His solution is below along with a quick equivalence test to my original solution, and check out this stackOverflow question for a more engaging discussion of the strengths and weaknesses of both dplyr/tidyr and data.table.
Fake Data
library(wakefield) library(tidyr) library(dplyr) d <- r_data_frame( n=100, id, r_series(date_stamp,15,name='foo_date'), r_series(level,15,name='foo_supply'), r_series(date_stamp,10,name='bar_date'), r_series(level,10,name='bar_supply'), r_series(date_stamp,3,name='baz_date'), r_series(level,3,name='baz_supply') )
Test Function for Equivalence
# Create a true ordered data frame and drop any extraneous classes for each column true_ordered_df <- function(x){ x$ID <- as.character(x$ID); class(x$ID) <- 'character' x$med_date <- as.Date(x$med_date); class(x$med_date) <- 'Date' x$med_supply <- as.integer(x$med_supply); class(x$med_supply) <- 'integer' x$med_name <- as.character(x$med_name); class(x$med_name) <- 'character' x <- data.frame( ID=x$ID, med_date=x$med_date, med_supply=x$med_supply, med_name=x$med_name, stringsAsFactors=FALSE ) x <- x[with(x,order(ID,med_date,med_supply,med_name)),] row.names(x) <- NULL x }
Data.Table Solution, thanks to Arun Srinivasan
require(data.table) # v1.9.5 dt = as.data.table(d) pattern = c("date", "supply") mcols = lapply(pattern, grep, names(dt), value=TRUE) dt.m = melt(dt, id="ID", measure=mcols, variable.name="med_name", value.name = paste("med", pattern, sep="_")) setattr(dt.m$med_name, 'levels', gsub("_.*$", "", mcols[[1L]])) scripts2 <- true_ordered_df(dt.m)
My Original Solution
# foo med_dates <- d %>% select(ID,foo_date_1:foo_date_15) %>% gather(med_seq, med_date, foo_date_1:foo_date_15) med_dates$med_seq <- as.integer(sub('^foo_date_','',med_dates$med_seq)) med_supply <- d %>% select(ID,foo_supply_1:foo_supply_15) %>% gather(med_seq, med_supply, foo_supply_1:foo_supply_15) med_supply$med_seq <- as.integer(sub('^foo_supply_','',med_supply$med_seq)) foo <- left_join(med_dates,med_supply, by=c('ID','med_seq')) %>% select(ID,med_date,med_supply) foo$med_name <- 'foo' # bar med_dates <- d %>% select(ID,bar_date_1:bar_date_10) %>% gather(med_seq, med_date, bar_date_1:bar_date_10) med_dates$med_seq <- as.integer(sub('^bar_date_','',med_dates$med_seq)) med_supply <- d %>% select(ID,bar_supply_1:bar_supply_10) %>% gather(med_seq, med_supply, bar_supply_1:bar_supply_10) med_supply$med_seq <- as.integer(sub('^bar_supply_','',med_supply$med_seq)) bar <- left_join(med_dates,med_supply, by=c('ID','med_seq')) %>% select(ID,med_date,med_supply) bar$med_name <- 'bar' # baz med_dates <- d %>% select(ID,baz_date_1:baz_date_3) %>% gather(med_seq, med_date, baz_date_1:baz_date_3) med_dates$med_seq <- as.integer(sub('^baz_date_','',med_dates$med_seq)) med_supply <- d %>% select(ID,baz_supply_1:baz_supply_3) %>% gather(med_seq, med_supply, baz_supply_1:baz_supply_3) med_supply$med_seq <- as.integer(sub('^baz_supply_','',med_supply$med_seq)) baz <- left_join(med_dates,med_supply, by=c('ID','med_seq')) %>% select(ID,med_date,med_supply) baz$med_name <- 'baz' scripts <- true_ordered_df(rbind(foo,bar,baz)) all.equal(scripts,scripts2) ## [1] TRUE
Huzzah!
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