R 折叠具有重叠范围的行
我有一个带有开始和结束时间的data.frame:R 折叠具有重叠范围的行,r,subset,rows,overlap,overlapping,R,Subset,Rows,Overlap,Overlapping,我有一个带有开始和结束时间的data.frame: ranges<- data.frame(start = c(65.72000,65.72187, 65.94312,73.75625,89.61625),stop = c(79.72187,79.72375,79.94312,87.75625,104.94062)) > ranges start stop 1 65.72000 79.72187 2 65.72187 79.72375 3 65.94312
ranges<- data.frame(start = c(65.72000,65.72187, 65.94312,73.75625,89.61625),stop = c(79.72187,79.72375,79.94312,87.75625,104.94062))
> ranges
start stop
1 65.72000 79.72187
2 65.72187 79.72375
3 65.94312 79.94312
4 73.75625 87.75625
5 89.61625 104.94062
我试过这个:
mdat <- outer(ranges$start, ranges$stop, function(x,y) y > x)
mdat[upper.tri(mdat)|col(mdat)==row(mdat)] <- NA
mdat
mdat x)
mdat[upper.tri(mdat)| col(mdat)=row(mdat)]您可以尝试以下方法:
library(dplyr)
ranges %>%
arrange(start) %>%
group_by(g = cumsum(cummax(lag(stop, default = first(stop))) < start)) %>%
summarise(start = first(start), stop = max(stop))
# A tibble: 2 × 3
# g start stop
# <int> <dbl> <dbl>
#1 0 65.72000 87.75625
#2 1 89.61625 104.94062
库(dplyr)
范围%>%
安排(开始)%%>%
分组依据(g=cumsum(cummax(滞后(停止,默认值=first(停止)))%
总结(开始=第一(开始),停止=最大(停止))
#一个tibble:2×3
#g起止点
#
#1 0 65.72000 87.75625
#2 1 89.61625 104.94062
这是一个数据表
解决方案
library(data.table)
setDT(ranges)
ranges[, .(start=min(start), stop=max(stop)),
by=.(group=cumsum(c(1, tail(start, -1) > head(stop, -1))))]
group start stop
1: 1 65.72000 87.75625
2: 2 89.61625 104.94062
在这里,通过检查上一次启动是否大于停止,然后使用cumsum
来构建组。在每个组中,计算最小开始时间和最大停止时间。对于基本R
和熔化/取消堆叠
,让我们再添加一些日期,使问题更有趣和通用:
ranges<- data.frame(start = c(65.72000,65.72187, 65.94312,73.75625,89.61625,105.1,104.99),stop = c(79.72187,79.72375,79.94312,87.75625,104.94062,110.22,108.01))
ranges
# start stop
#1 65.72000 79.72187
#2 65.72187 79.72375
#3 65.94312 79.94312
#4 73.75625 87.75625
#5 89.61625 104.94062
#6 105.10000 110.22000
#7 104.99000 108.01000
library(reshape2)
ranges <- melt(ranges)
ranges <- ranges[order(ranges$value),]
ranges
# variable value
#1 start 65.72000
#2 start 65.72187
#3 start 65.94312
#4 start 73.75625
#8 stop 79.72187
#9 stop 79.72375
#10 stop 79.94312
#11 stop 87.75625
#5 start 89.61625
#12 stop 104.94062
#7 start 104.99000
#6 start 105.10000
#14 stop 108.01000
#13 stop 110.22000
看起来不太容易。你已经看过dplyr了吗?这将是我第一次尝试解决。我尝试了以下方法:mdat x)mdat[upper.tri(mdat)| col(mdat)=row(mdat)]您将什么定义为重叠?相关:,。我猜如果下一行的开始在前一行的范围内,那么我希望保持下一行的“停止”。因此,如果下一行的起点是73.75625(终点是87.75625),而上一行的范围是65.94312到79.94312,那么我想有一个范围,作为两者的组合,基本上包括这两个范围:65.94312到87.75625。这很好地解决了我的问题,但我还不太明白如何解决。感谢你的小提琴技巧,同时我再次检查代码,试图辨别它的魔力。@Jemus42代码首先按start
列对行进行排序lag(stop)
添加一个虚拟列,其中包含先前的stop
值cummax
将保留以前的stop
值中的最高值,以便您可以与start
列进行比较。如果start
值大于新组中前面stop
值的最大值cumsum
将累计TRUE
s,以便您拥有每组的标识符(g
)<代码>总结
将按组id进行聚合,并获取乞讨和间隔结束时间。非常智能的解决方案@Psidom:)
ranges<- data.frame(start = c(65.72000,65.72187, 65.94312,73.75625,89.61625,105.1,104.99),stop = c(79.72187,79.72375,79.94312,87.75625,104.94062,110.22,108.01))
ranges
# start stop
#1 65.72000 79.72187
#2 65.72187 79.72375
#3 65.94312 79.94312
#4 73.75625 87.75625
#5 89.61625 104.94062
#6 105.10000 110.22000
#7 104.99000 108.01000
library(reshape2)
ranges <- melt(ranges)
ranges <- ranges[order(ranges$value),]
ranges
# variable value
#1 start 65.72000
#2 start 65.72187
#3 start 65.94312
#4 start 73.75625
#8 stop 79.72187
#9 stop 79.72375
#10 stop 79.94312
#11 stop 87.75625
#5 start 89.61625
#12 stop 104.94062
#7 start 104.99000
#6 start 105.10000
#14 stop 108.01000
#13 stop 110.22000
indices <- intersect(which(ranges$variable=='start')-1, which(ranges$variable=='stop'))
unstack(ranges[c(1, sort(c(indices, indices+1)), nrow(ranges)),], value~variable)
# start stop
#1 65.72000 87.75625
#2 89.61625 104.94062
#3 104.99000 110.22000