我想合并两个数据帧。 它们都有开始日期和结束日期。
如果给定的间隔重叠，我想将结果行拆分为非重叠的间隔。
请看这个例子：
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
（或简单地在R中使用这些命令）
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")
我认为与我的另一个问题有一些相似之处： "smoothing" time data - can it be done more efficient?
但也略有不同。
提前感谢您的帮助！
答案 0 :(得分：1)
您可以分两步完成： 首先，计算所有需要的间隔并将它们放在中间表中，然后将该表与两个初始data.frames连接。
# 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")
]
由于我们加入数据的条件不是简单的相等，所以让我们使用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
" )
答案 1 :(得分：1)
sqldf会起作用，但我试过'纯'R解决方案。它有效，但它有点草率。我还没弄明白如何“向量化”解决方案（在split.interval中删除两个for循环，并删除了对id.split进行lapply的需要）。
首先，我创建两个可以占用一个id的函数，并将'a'和'b'合并在一起：
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$id==ids,]
split.interval ( sub.a , sub.b )
}
然后我为每个ID运行该函数，并将它们全部绑定在一起。
l=lapply(unique(c(a$id,b$id)), id.split)
res = do.call(rbind,l)
row.names(res) = NULL
res