Multiple filters on the same data to create multiple outputs in R

Question

I have the data below, and I'm looking to filter it into some new data frames.

I'm curious if there's a better way of doing it than the following rather than having a separate line for each city and date?

vector.sydney = dplyr::filter(vector, grepl('Sydney', Location)) %>% group_by(Location) %>% filter(Date == '2018-12-07')
vector.adelaide = dplyr::filter(vector, grepl('Adelaide', Location)) %>% group_by(Location) %>% filter(Date == '2018-12-07')
vector.melbourne = dplyr::filter(vector, grepl('Melbourne', Location)) %>% group_by(Location) %>% filter(Date == '2018-12-07')
vector.sydney = dplyr::filter(vector, grepl('Sydney', Location)) %>% group_by(Location) %>% filter(Date == '2018-12-06')
vector.adelaide = dplyr::filter(vector, grepl('Adelaide', Location)) %>% group_by(Location) %>% filter(Date == '2018-12-06')
vector.melbourne = dplyr::filter(vector, grepl('Melbourne', Location)) %>% group_by(Location) %>% filter(Date == '2018-12-06')

I assume I can probably create a list of Locations eg: c("Sydney", "Melbourne", "Adelaide") but I can't figure out how I'd name the outputs.

Can anyone please point me in the right direction?

Dateframe

vector = structure(list(Date = c("6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "6-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18", "7-Dec-18"), Delivery = c("12-Jan-19", "22-Jan-19", "31-Jan-19", "19-Jun-19", "25-Jul-19", "7-Aug-19", "8-Aug-19", "9-Aug-19", "22-Aug-19", "27-Aug-19", "28-Aug-19", "29-Aug-19", "30-Aug-19", "3-Sep-19", "2-Oct-19", "17-Jan-19", "19-Jan-19", "8-Feb-19", "9-Feb-19", "8-Jun-19", "20-Jul-19", "7-Aug-19", "1-Oct-19", "11-Oct-19", "16-Oct-19", "25-Oct-19", "31-Oct-19", "2-Nov-19", "12-Nov-19", "19-Nov-19", "12-Jan-19", "22-Jan-19", "31-Jan-19", "8-Jun-19", "19-Jun-19", "25-Jun-19", "25-Jul-19", "7-Aug-19", "27-Aug-19", "28-Aug-19", "29-Aug-19", "30-Aug-19", "3-Sep-19", "2-Oct-19", "31-Oct-19", "12-Jan-19", "22-Jan-19", "31-Jan-19", "19-Jun-19", "25-Jun-19", "25-Jul-19", "7-Aug-19", "8-Aug-19", "9-Aug-19", "22-Aug-19", "27-Aug-19", "28-Aug-19", "29-Aug-19", "30-Aug-19", "2-Oct-19", "17-Jan-19", "8-Feb-19", "9-Feb-19", "8-Jun-19", "20-Jul-19", "7-Aug-19", "1-Oct-19", "11-Oct-19", "16-Oct-19", "25-Oct-19", "26-Oct-19", "31-Oct-19", "2-Nov-19", "12-Nov-19", "19-Nov-19", "12-Jan-19", "22-Jan-19", "31-Jan-19", "5-Jun-19", "8-Jun-19", "19-Jun-19", "25-Jun-19", "25-Jul-19", "7-Aug-19", "27-Aug-19", "28-Aug-19", "29-Aug-19", "30-Aug-19", "2-Oct-19", "31-Oct-19"), Item.No = c(675L, 440L, 648L, 29L, 605L, 511L, 686L, 656L, 229L, 92L, 331L, 384L, 379L, 332L, 461L, 84L, 418L, 535L, 254L, 90L, 721L, 984L, 342L, 197L, 562L, 327L, 991L, 225L, 514L, 34L, 790L, 675L, 95L, 195L, 407L, 495L, 187L, 529L, 914L, 728L, 839L, 981L, 30L, 728L, 610L, 218L, 265L, 890L, 127L, 2L, 412L, 800L, 65L, 358L, 8L, 737L, 966L, 778L, 617L, 984L, 280L, 374L, 113L, 101L, 310L, 389L, 426L, 572L, 925L, 972L, 611L, 672L, 362L, 145L, 963L, 595L, 186L, 640L, 359L, 714L, 898L, 985L, 93L, 324L, 208L, 629L, 969L, 622L, 235L, 411L), Rank = c(3L, 8L, 7L, 6L, 9L, 5L, 13L, 14L, 11L, 12L, 4L, 2L, 10L, 15L, 1L, 14L, 10L, 11L, 7L, 9L, 13L, 2L, 1L, 6L, 15L, 5L, 4L, 12L, 3L, 8L, 4L, 8L, 6L, 15L, 7L, 12L, 9L, 3L, 10L, 5L, 2L, 13L, 14L, 1L, 11L, 3L, 7L, 8L, 6L, 15L, 9L, 5L, 13L, 14L, 11L, 12L, 4L, 2L, 10L, 1L, 13L, 12L, 6L, 10L, 14L, 1L, 2L, 7L, 11L, 5L, 15L, 3L, 9L, 4L, 8L, 4L, 7L, 8L, 15L, 14L, 6L, 11L, 9L, 3L, 10L, 5L, 2L, 12L, 1L, 13L), Location = c("Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Sydney", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Adelaide", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne", "Melbourne")), class = "data.frame", row.names = c(NA, -90L))

your vector isn't complete. It's missing some brackets maybe — Ronak Shah
What do you do next with the data? If you are doing the same "something" to all six vector, you could also get away with filter(Date %in% c("2018-12-06","2018-12-07"), grepl("Sydney|Adelaide|Melbourne", Location)) %>% group_by(Location, Date) %>% "something". (Regardless, you can reduce each vector by moving the filter(Date==...) to within the first one, as in filter(vec, grepl(...), Date==...).) — r2evans

merv merv · Accepted Answer · 2018-12-12T01:58:50

Perhaps you're looking for split? It'll give you a list back, but that should be just as functional.

dfs <- vector %>% split(list(.$Location, .$Date))

You can then access the individual data frames by "Location.Date" indexing:

> dfs$`Sydney.7-Dec-18`
       Date  Delivery Item.No Rank Location
46 7-Dec-18 12-Jan-19     218    3   Sydney
47 7-Dec-18 22-Jan-19     265    7   Sydney
48 7-Dec-18 31-Jan-19     890    8   Sydney
49 7-Dec-18 19-Jun-19     127    6   Sydney
50 7-Dec-18 25-Jun-19       2   15   Sydney
51 7-Dec-18 25-Jul-19     412    9   Sydney
52 7-Dec-18  7-Aug-19     800    5   Sydney
53 7-Dec-18  8-Aug-19      65   13   Sydney
54 7-Dec-18  9-Aug-19     358   14   Sydney
55 7-Dec-18 22-Aug-19       8   11   Sydney
56 7-Dec-18 27-Aug-19     737   12   Sydney
57 7-Dec-18 28-Aug-19     966    4   Sydney
58 7-Dec-18 29-Aug-19     778    2   Sydney
59 7-Dec-18 30-Aug-19     617   10   Sydney
60 7-Dec-18  2-Oct-19     984    1   Sydney

or you can always assign new names if you'd prefer

names(dfs) <- c("adelaide.12.6", "melbourne.12.6", "sydney.12.6",
                "adelaide.12.7", "melbourne.12.7", "sydney.12.7")

Multiple filters on the same data to create multiple outputs in R

1 Answers