merge two data frames on non overlapping intervals

问题:

I want to merge two data frames. Both of them have a begin date and an end date.

If the given intervals are overlapping, I want to split the resulting rows in non overlapping intevals.

Please see this example:

a
 id      beg_a      end_a prop_a
  1 2000-01-01 2002-12-31      A
  2 2000-01-01 2000-02-15      B
  2 2000-04-01 2000-04-15      A
  2 2002-01-01 2002-12-31      B
  3 2000-01-01 2000-06-15      A

b
 id      beg_b      end_b prop_b
  1 1999-06-01 2000-05-15      D
  1 2003-01-15 2003-01-31      D
  2 1999-01-01 2003-01-15      D
  3 2000-07-01 2001-08-01      E

merged
  id      beg_a      end_a prop_a      beg_b      end_b prop_b overallBeg overallEnd
   1       <NA>       <NA>   <NA> 1999-06-01 2000-05-15      D 1999-06-01 1999-12-31
   1 2000-01-01 2002-12-31      A 1999-06-01 2000-05-15      D 2000-01-01 2000-05-15
   1 2000-01-01 2002-12-31      A       <NA>       <NA>   <NA> 2000-05-16 2002-12-31
   1       <NA>       <NA>   <NA> 2003-01-15 2003-01-31      D 2003-01-15 2003-01-31
   2       <NA>       <NA>   <NA> 1999-01-01 2003-01-15      D 1999-01-01 1999-12-31
   2 2000-01-01 2000-02-15      B 1999-01-01 2003-01-15      D 2000-01-01 2000-02-15
   2       <NA>       <NA>   <NA> 1999-01-01 2003-01-15      D 2000-02-16 2000-03-31
   2 2000-04-01 2000-04-15      A 1999-01-01 2003-01-15      D 2000-04-01 2000-04-15
   2       <NA>       <NA>   <NA> 1999-01-01 2003-01-15      D 2000-04-16 2001-12-31
   2 2002-01-01 2002-12-31      B 1999-01-01 2003-01-15      D 2002-01-01 2002-12-31
   2       <NA>       <NA>   <NA> 1999-01-01 2003-01-15      D 2003-01-01 2003-01-15
   3 2000-01-01 2000-06-15      A       <NA>       <NA>   <NA> 2000-01-01 2000-06-15
   3       <NA>       <NA>   <NA> 2000-07-01 2001-08-01      E 2000-07-01 2001-08-01

(or simply use these commands in R)

a <- structure(list(id = c(1, 2, 2, 2, 3), beg_a = structure(c(10957, 
  10957, 11048, 11688, 10957), class = "Date"), end_a = structure(c(12052, 
  11002, 11062, 12052, 11123), class = "Date"), prop_a = structure(c(1L, 
  2L, 1L, 2L, 1L), .Label = c("A", "B"), class = "factor")), .Names = c("id", 
  "beg_a", "end_a", "prop_a"), row.names = c(NA, -5L), class = "data.frame")

b <- structure(list(id = c(1, 1, 2, 3), beg_b = structure(c(10743, 
  12067, 10592, 11139), class = "Date"), end_b = structure(c(11092, 
  12083, 12067, 11535), class = "Date"), prop_b = structure(c(1L, 
  1L, 1L, 2L), .Label = c("D", "E"), class = "factor")), .Names = c("id", 
  "beg_b", "end_b", "prop_b"), row.names = c(NA, -4L), class = "data.frame")

merged <- structure(list(id = c(1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3), 
      beg_a = structure(c(NA, 10957, 10957, NA, NA, 10957, NA, 
      11048, NA, 11688, NA, 10957, NA), class = "Date"), end_a = structure(c(NA, 
      12052, 12052, NA, NA, 11002, NA, 11062, NA, 12052, NA, 11123, 
      NA), class = "Date"), prop_a = structure(c(NA, 1L, 1L, NA, 
      NA, 2L, NA, 1L, NA, 2L, NA, 1L, NA), .Label = c("A", "B"), class = "factor"), 
      beg_b = structure(c(10743, 10743, NA, 12067, 10592, 10592, 
      10592, 10592, 10592, 10592, 10592, NA, 11139), class = "Date"), 
      end_b = structure(c(11092, 11092, NA, 12083, 12067, 12067, 
      12067, 12067, 12067, 12067, 12067, NA, 11535), class = "Date"), 
      prop_b = structure(c(1L, 1L, NA, 1L, 1L, 1L, 1L, 1L, 1L, 
      1L, 1L, NA, 2L), .Label = c("D", "E"), class = "factor"), 
      overallBeg = structure(c(10743, 10957, 11093, 12067, 10592, 
      10957, 11003, 11048, 11063, 11688, 12053, 10957, 11139), class = "Date"), 
      overallEnd = structure(c(10956, 11092, 12052, 12083, 10956, 
      11002, 11047, 11062, 11687, 12052, 12067, 11123, 11535), class = "Date")), .Names = c("id", 
  "beg_a", "end_a", "prop_a", "beg_b", "end_b", "prop_b", "overallBeg", 
  "overallEnd"), row.names = c(NA, -13L), class = "data.frame")

I think there are some similarities with another question of mine: "smoothing" time data - can it be done more efficient?

But also slightly different.

Thank you in advance for your help!

回答1:

sqldf will work, but I tried a 'pure' R solution. It works, but it is a little sloppy. I haven't figured out how to 'vectorize' the solution (remove the two for loops in the split.interval, and remove the need to lapply over id.split).

First I create two functions that can take one id, and merge 'a' and 'b' together:

split.interval = function(sub.a, sub.b) {
    begs = c(sub.a$beg_a,sub.b$beg_b)  
    ends = c(sub.a$end_a,sub.b$end_b)
    dates=c(begs,ends)
    dates = dates[order(dates)]
    d = data.frame(overallBeg = dates[-length(dates)], overallEnd = dates[-1])
    date.match = function(x,y) {
            s = match(x, d$overallBeg )
            e = match(y, d$overallEnd )
            join=as.Date(rep(NA,length(d$overallBeg)))
            for (i in 1:length(x)) join [s[i]:e[i]]= x[i]
            join
    }

    d$a_join = date.match(sub.a$beg_a,sub.a$end_a)
    d$b_join = date.match(sub.b$beg_b,sub.b$end_b)

    d = merge(sub.a,d,by.x='beg_a',by.y='a_join',all.y=T)
    d = merge(sub.b,d,by.x='beg_b',by.y='b_join',all.y=T)

    d$id=pmax(d$id.x,d$id.y,na.rm=T)
    d = d [order(d$overallBeg),c('id','beg_a','end_a','prop_a','beg_b','end_b','prop_b','overallBeg','overallEnd')]
    # This next line will lead to a bug if overallBeg == overallEnd
    d$overallEnd [d$overallEnd == c(d$overallBeg[-1],F)] = d$overallEnd [d$overallEnd == c(d$overallBeg[-1],F)] - 1  
    d

}

id.split = function (ids) {
    sub.a=a[a$id==ids,]
    sub.b=b[b$id==ids,]

    split.interval ( sub.a , sub.b )
}

Then I run the function for every ID, and bind them all together.

l=lapply(unique(c(a$id,b$id)), id.split) 
res = do.call(rbind,l)
row.names(res) = NULL
res

回答2:

You can do that in two steps: first, compute all the desired intervals and put them in an intermediary table, then join this table with the two initial data.frames.

# First build all the desired intervals
names(a) <- c( "id", "valid_from", "valid_until", "prop_a" )
names(b) <- c( "id", "valid_from", "valid_until", "prop_b" )

intervals <- rbind( 
  data.frame( id = a$id, date = a$valid_from ),
  data.frame( id = a$id, date = a$valid_until ),
  data.frame( id = b$id, date = b$valid_from ),
  data.frame( id = b$id, date = b$valid_until )
)
intervals <- unique( intervals )
intervals <- intervals[ order(intervals$id, intervals$date), ]
n <- dim(intervals)[1]
intervals <- data.frame(
  id = intervals$id[-n],
  id2 = intervals$id[-1],
  valid_from = intervals$date[-n],
  valid_until = intervals$date[-1]
)
intervals <- intervals[ 
  intervals$id == intervals$id2, 
  c("id", "valid_from", "valid_until") 
]

Since the condition on which we join the data is not a simple equality, let us use sqldf.

library(sqldf)
d <- sqldf( "
  SELECT intervals.id,
         intervals.valid_from, intervals.valid_until, 
         a.prop_a, b.prop_b
  FROM intervals
  LEFT JOIN a
  ON          a.valid_from  <= intervals.valid_from 
  AND intervals.valid_until <=         a.valid_until
  AND intervals.id = a.id
  LEFT JOIN b
  ON          b.valid_from  <= intervals.valid_from 
  AND intervals.valid_until <=         b.valid_until
  AND intervals.id = b.id
" )