use*_*808 6 merge r intervals dataframe data.table
虽然这个细节当然是应用程序特定的,但在SO精神中,我试图尽可能地保持这一点!基本问题是当一个data.frame具有特定日期而另一个具有日期范围时,如何按日期合并data.frames.其次,问题是如何处理给定变量的多个观察,以及如何将它们包含在最终输出data.frame中.我敢肯定其中一些是标准的,但一个非常完整的搜索几乎没有透露.
我正在尝试合并的mre对象如下.
# 'Speeches' data.frame
structure(list(Name = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("BBB",
"AAA"), class = "factor"), Date = structure(c(12543, 12404, 12404,
12404, 12373, 12362, 12345, 12320, 12207, 15450, 15449, 15449,
15449, 15449, 15449, 15449, 15449, 15448, 15448, 15448), class = "Date")), .Names = c("Name",
"Date"), row.names = c("1", "1.1", "1.2", "1.3", "1.4", "1.5",
"1.6", "1.7", "1.8", "2", "2.1", "2.2", "2.3", "2.4", "2.5",
"2.6", "2.7", "2.8", "2.9", "2.10"), class = "data.frame")
# 'History' data.frame
structure(list(Name = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 1L,
1L, 1L, 1L, 1L, 1L, 1L), .Label = c("BBB", "AAA"), class = "factor"),
Role = structure(c(1L, 2L, 3L, 3L, 3L, 4L, 1L, 2L, 3L, 3L,
3L, 3L, 4L), .Label = c("Political groups", "National parties",
"Member", "Substitute", "Vice-Chair", "Chair", "Vice-President",
"Quaestor", "President", "Co-President"), class = "factor"),
Value = structure(c(10L, 12L, 6L, 3L, 8L, 4L, 9L, 11L, 1L,
7L, 1L, 2L, 5L), .Label = c("a", "b", "c", "d", "e", "f",
"g", "h", "i", "j", "k", "l", "m", "n", "o"), class = "factor"),
Role.Start = structure(c(12149, 12149, 12150, 12150, 12152,
12150, 14439, 14439, 14441, 14503, 15358, 15411, 14441), class = "Date"),
Role.End = structure(c(12618, 12618, 12618, 12618, 12538,
12618, 15507, 15507, 15357, 15507, 15410, 15507, 15357), class = "Date")), .Names = c("Name",
"Role", "Value", "Role.Start", "Role.End"), row.names = c(NA,
13L), class = "data.frame")
我面临着许多困难.
1)尽管在演讲和历史数据中都有日期信息,但在第一个中我有每个条目的具体日期,在第二个中有一个日期范围.理想情况下,我希望能够合并,以便每个语音条目与发言者("姓名")和发言日期所属的历史记录条目相匹配.
2)所需的输出是具有data.frame或data.table,其行等于语音data.frame中的观察值,以及Name,Date和每个角色的列(将由值填充).但是,某个角色在给定日期对于给定的发言者多次出现,因此我需要能够为这些实例创建多个列.
下面的对象给出了这个输出,但是使用了非常脆弱且非常慢的for循环构造:
structure(list(Name = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("BBB",
"AAA"), class = "factor"), Date = structure(c(12543, 12404, 12404,
12404, 12373, 12362, 12345, 12320, 12207, 15450, 15449, 15449,
15449, 15449, 15449, 15449, 15449, 15448, 15448, 15448), class = "Date"),
`Political groups` = structure(c(2L, 2L, 2L, 2L, 2L, 2L,
2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c("i",
"j"), class = "factor"), `National parties` = structure(c(2L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L, 1L, 1L), .Label = c("k", "l"), class = "factor"),
Member.1 = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), .Label = c("f",
"g"), class = "factor"), Member.2 = structure(c(2L, 2L, 2L,
2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 1L), .Label = c("b", "c"), class = "factor"), Member.3 = structure(c(NA,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA), .Label = "h", class = "factor"), Substitute = structure(c(1L,
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, NA, NA, NA, NA, NA, NA, NA,
NA, NA, NA, NA), .Label = "d", class = "factor")), .Names = c("Name",
"Date", "Political groups", "National parties", "Member.1", "Member.2",
"Member.3", "Substitute"), row.names = c("1", "1.1", "1.2", "1.3",
"1.4", "1.5", "1.6", "1.7", "1.8", "2", "2.1", "2.2", "2.3",
"2.4", "2.5", "2.6", "2.7", "2.8", "2.9", "2.10"), class = "data.frame")
任何有关如何改进此问题的帮助和/或评论都将受到欢迎!
更新:在v1.9.3 +中,现在实现了重叠连接.这是一个特殊情况,其中start和end Date相同Speeches.我们可以使用foverlaps()以下方法完成此任务:
require(data.table) ## 1.9.3+
setDT(Speeches)
setDT(History)
Speeches[, `:=`(Date2 = Date, id = .I)]
setkey(History, Name, Role.Start, Role.End)
ans = foverlaps(Speeches, History, by.x=c("Name", "Date", "Date2"))[, Date2 := NULL]
ans = ans[order(id, Value)][, N := 1:.N, by=list(Name, Date, Role, id)]
ans = dcast.data.table(ans, id+Name+Date ~ Role+N, value.var="Value")
这是范围/间隔连接的情况.
这是data.table方式.它使用两个滚动连接.
require(data.table) ## 1.9.2+
dt1 = as.data.table(Speeches)
dt2 = as.data.table(History)
# first rolling join - to get end indices
setkey(dt2, Name, Role.Start)
tmp1 = dt2[dt1, roll=Inf, which=TRUE]
# second rolling join - to get start indices
setkey(dt2, Name, Role.End)
tmp2 = dt2[dt1, roll=-Inf, which=TRUE]
# generate dt1's and dt2's corresponding row indices
idx = tmp1-tmp2+1L
idx1 = rep(seq_len(nrow(dt1)), idx)
idx2 = data.table:::vecseq(tmp2, idx, sum(idx))
dt1[, id := 1:.N] ## needed for casting later
# subset using idx1 and idx2 and bind them colwise
ans = cbind(dt1[idx1], dt2[idx2, -1L, with=FALSE])
# a little reordering to get the output correctly (factors are a pain!)
ans = ans[order(id,Value)][, N := 1:.N, by=list(Name, Date, Role, id)]
# finally cast them.
f_ans = dcast.data.table(ans, id+Name+Date ~ Role+N, value.var="Value")
这是输出:
id Name Date Political groups_1 National parties_1 Member_1 Member_2 Member_3 Substitute_1
1: 1 AAA 2004-05-05 j l c f NA d
2: 2 AAA 2003-12-18 j l c f h d
3: 3 AAA 2003-12-18 j l c f h d
4: 4 AAA 2003-12-18 j l c f h d
5: 5 AAA 2003-11-17 j l c f h d
6: 6 AAA 2003-11-06 j l c f h d
7: 7 AAA 2003-10-20 j l c f h d
8: 8 AAA 2003-09-25 j l c f h d
9: 9 AAA 2003-06-04 j l c f h d
10: 10 BBB 2012-04-20 i k b g NA NA
11: 11 BBB 2012-04-19 i k b g NA NA
12: 12 BBB 2012-04-19 i k b g NA NA
13: 13 BBB 2012-04-19 i k b g NA NA
14: 14 BBB 2012-04-19 i k b g NA NA
15: 15 BBB 2012-04-19 i k b g NA NA
16: 16 BBB 2012-04-19 i k b g NA NA
17: 17 BBB 2012-04-19 i k b g NA NA
18: 18 BBB 2012-04-18 i k b g NA NA
19: 19 BBB 2012-04-18 i k b g NA NA
20: 20 BBB 2012-04-18 i k b g NA NA
或者你也可以使用GenomicRangesbioconductor中的包完成这个,它可以很好地处理Ranges,特别是当你需要一个额外的列加入by(Name)以及范围.你可以从这里安装它.
require(GenomicRanges)
require(data.table)
dt1 <- as.data.table(Speeches)
dt2 <- as.data.table(History)
gr1 = GRanges(Rle(dt1$Name), IRanges(as.numeric(dt1$Date), as.numeric(dt1$Date)))
gr2 = GRanges(Rle(dt2$Name), IRanges(as.numeric(dt2$Role.Start), as.numeric(dt2$Role.End)))
olaps = findOverlaps(gr1, gr2, type="within")
idx1 = queryHits(olaps)
idx2 = subjectHits(olaps)
# from here, you can do exactly as above
dt1[, id := 1:.N]
...
...
dcast.data.table(ans, id+Name+Date ~ Role+N, value.var="Value")
给出与上面相同的结果.