Combine Multiple Columns Into Tidy Data [duplicate

问题:

This question already has an answer here:

• Reshaping multiple sets of measurement columns (wide format) into single columns (long format) 6 answers

My dataset looks like this:

unique.id abx.1    start.1     stop.1 abx.2    start.2     stop.2 abx.3    start.3     stop.3 abx.4    start.4
1         1  Moxi 2014-01-01 2014-01-07  PenG 2014-01-01 2014-01-07 Vanco 2014-01-01 2014-01-07  Moxi 2014-01-01
2         2  Moxi 2014-01-01 2014-01-02 Cipro 2014-01-01 2014-01-02  PenG 2014-01-01 2014-01-02 Vanco 2014-01-01
3         3 Cipro 2014-01-01 2014-01-05 Vanco 2014-01-01 2014-01-05 Cipro 2014-01-01 2014-01-05 Vanco 2014-01-01
4         4 Vanco 2014-01-02 2014-01-03 Cipro 2014-01-02 2014-01-03 Cipro 2014-01-02 2014-01-03  PenG 2014-01-02
5         5 Vanco 2014-01-01 2014-01-02  PenG 2014-01-01 2014-01-02  PenG 2014-01-01 2014-01-02 Cipro 2014-01-01
      stop.4    intervention
1 2014-01-07       0
2 2014-01-02       0
3 2014-01-05       1
4 2014-01-03       1
5 2014-01-02       0

With some code to create this:

 abxoptions <- c(\"Cipro\", \"Moxi\", \"PenG\", \"Vanco\")
      df3 <- data.frame(
      unique.id = 1:5,
      abx.1 = sample(abxoptions,5, replace=TRUE),
      start.1 = as.Date(c(\'2014-01-01\', \'2014-01-01\', \'2014-01-01\', \'2014-01-02\', \'2014-01-01\')),
      stop.1  = as.Date(c(\'2014-01-07\', \'2014-01-02\', \'2014-01-05\', \'2014-01-03\', \'2014-01-02\')),
      abx.2 = sample(abxoptions,5, replace=TRUE),         
      start.2 = as.Date(c(\'2014-01-01\', \'2014-01-01\', \'2014-01-01\', \'2014-01-02\', \'2014-01-01\')),
      stop.2  = as.Date(c(\'2014-01-07\', \'2014-01-02\', \'2014-01-05\', \'2014-01-03\', \'2014-01-02\')),
      abx.3 = sample(abxoptions,5, replace=TRUE),         
      start.3 = as.Date(c(\'2014-01-01\', \'2014-01-01\', \'2014-01-01\', \'2014-01-02\', \'2014-01-01\')),
      stop.3  = as.Date(c(\'2014-01-07\', \'2014-01-02\', \'2014-01-05\', \'2014-01-03\', \'2014-01-02\')),
      abx.4 = sample(abxoptions,5, replace=TRUE),         
      start.4 = as.Date(c(\'2014-01-01\', \'2014-01-01\', \'2014-01-01\', \'2014-01-02\', \'2014-01-01\')),
      stop.4  = as.Date(c(\'2014-01-07\', \'2014-01-02\', \'2014-01-05\', \'2014-01-03\', \'2014-01-02\')),
      intervention = c(0,0,1,1,0)

)

I would like to tidy this data to look like this:

unique.id    abx     start    stop           intervention
1            Moxi    2014-01-10 2014-01-07      0
1            Pen G   2014-01-01 2014-01-07      0
1            Vanco   2014-01-01 2014-01-07      0
1            Moxi    2014-01-01 2014-01-07      0  etc etc

The following solutions didn\'t get me where I needed: Gather multiple sets of columns and Combining multiple columns into one

I suspect that Hadley\'s amazing tidyr pakcage is the way to go...just can\'t figure this out. Any help would be greatly appreciated.

回答1:

Almost every data tidying problem can be solved in three steps:

1. Gather all non-variable columns
2. Separate \"colname\" column into multiple variables
3. Re-spread the data

(often you\'ll only need one or two of these, but I think they\'re almost always in this order).

For your data:

1. The only column that\'s already a variable is unique.id
2. You need to split current column names into variable and number
3. Then you need to put the \"variable\" variable back into columns

This looks like:

library(tidyr)
library(dplyr)

df3 %>%
  gather(col, value, -unique.id, -intervention) %>%
  separate(col, c(\"variable\", \"number\")) %>%
  spread(variable, value, convert = TRUE) %>%
  mutate(start = as.Date(start, \"1970-01-01\"), stop = as.Date(stop, \"1970-01-01\"))

Your case is a bit more complicated because you have two types of variables, so you need to restore the types at the end.

回答2:

You could try reshape from base R

reshape(df3, direction=\'long\', varying=2:ncol(df3), sep=\".\")

Or use merged.stack from splitstackshape

 library(splitstackshape)
 merged.stack(df3, var.stubs=c(\'abx\', \'start\', \'stop\'), sep=\'.\')[,
    c(\'start\', \'stop\') := lapply(.SD, as.Date,
                   origin=\'1970-01-01\'), .SDcols=4:5][]

回答3:

Recently, a new feature has been added to melt.data.table, which allows melting into multiple columns painless. All you\'ve to do is provide the columns you\'d want to melt separately in a list in measure.vars argument.

You can grab the development version by following these instructions.

require(data.table) ## v1.9.5
setDT(dat) # dat is now a data.table
melt(dat, id = 1L, measure = patterns(\"^abx\", \"^start\", \"^stop\"), 
          value.name = c(\"abx\", \"start\", \"stop\"))

#     unique.id variable   abx      start       stop
#  1:         1        1  Moxi 2014-01-01 2014-01-07
#  2:         2        1  Moxi 2014-01-01 2014-01-02
#  3:         3        1 Cipro 2014-01-01 2014-01-05
#  4:         4        1 Vanco 2014-01-02 2014-01-03
#  5:         5        1 Vanco 2014-01-01 2014-01-02
#  6:         1        2  PenG 2014-01-01 2014-01-07
#  7:         2        2 Cipro 2014-01-01 2014-01-02
#  8:         3        2 Vanco 2014-01-01 2014-01-05
#  9:         4        2 Cipro 2014-01-02 2014-01-03
# 10:         5        2  PenG 2014-01-01 2014-01-02
# 11:         1        3 Vanco 2014-01-01 2014-01-07
# 12:         2        3  PenG 2014-01-01 2014-01-02
# 13:         3        3 Cipro 2014-01-01 2014-01-05
# 14:         4        3 Cipro 2014-01-02 2014-01-03
# 15:         5        3  PenG 2014-01-01 2014-01-02
# 16:         1        4  Moxi 2014-01-01 2014-01-07
# 17:         2        4 Vanco 2014-01-01 2014-01-02
# 18:         3        4 Vanco 2014-01-01 2014-01-05
# 19:         4        4  PenG 2014-01-02 2014-01-03
# 20:         5        4 Cipro 2014-01-01 2014-01-02

I\'ve used column numbers here, but you can provide column names as well.