Question

我想知道如何对此代码进行矢量化。

dates = list(as.Date(c("2000-02-08", "2000-02-11")),
     as.Date(c("2000-03-02", "2000-03-07")),
     as.Date(c("2000-03-02", "2000-03-07")),
     as.Date(c("2000-03-03", "2000-03-07")),
     as.Date(c("2000-03-16", "2000-03-30")),
     as.Date(c("2000-03-16")))

i = 2
while(i <= length(dates))
{
     if(dates[[i]][1] < dates[[i-1]][2])
     {
        dates[[i]] = NULL
        i = i-1
    }

  i = i+1
}

我想只得到几个不相交的日期。

Date1 = as.Date(c("2000-03-02", "2000-03-07"))
Date2 = as.Date(c("2000-03-03", "2000-03-07"))

例如，如果Date2包含在Date1的范围内，那么我们删除Date2。

Answer 1

来自包foverlaps的{{1}}：

data.table

输出（我添加的情况略有不同）：

dates = list(as.Date(c("2000-02-08", "2000-02-11")),
            as.Date(c("2000-03-02", "2000-03-07")),
            as.Date(c("2000-03-02", "2000-03-05")),
            as.Date(c("2000-03-09", "2000-03-15")),
            as.Date(c("2000-03-16", "2000-03-30")),
            as.Date(c("2000-03-16")))

dt<-as.data.table(do.call(rbind,dates))
setkey(dt)
# Get id of the ranges within others
tmp <- foverlaps(dt,dt,which=T,type="within")[,xid]
# summarize this
t<-table(tmp)

# Filter for ranges appearing only once, hence not included in another one.
res <- dt[ as.integer(names(t[t==1])) , ]
# not aboslutely necessary, but it's to retrieve date objects which were converted by the rbind call.
res[, `:=`( V1=as.Date(V1,origin="1970-01-01"), V2=as.Date(V2, origin="1970-01-01"))][]

如果您希望排除任何交叉点，请在foverlaps调用中设置V1 V2 1: 2000-02-08 2000-02-11 2: 2000-03-02 2000-03-07 3: 2000-03-09 2000-03-15 4: 2000-03-16 2000-03-30以获取此输出：

type="any"

Answer 2

取决于您正在寻找的方向。在我的示例中，您查看以下任何数据行是否重叠（我只是查看开始日期，但您可以扩展它）。

dates = list(as.Date(c("2000-02-08", "2000-02-11")),
             as.Date(c("2000-03-02", "2000-03-07")),
             as.Date(c("2000-03-02", "2000-03-07")),
             as.Date(c("2000-03-03", "2000-03-07")),
             as.Date(c("2000-03-16", "2000-03-30")),
             as.Date(c("2000-03-16")))

m <- do.call(rbind,dates)

rem <- sapply(seq_along(m[,1]),function(x){any(which(
  m[x,1]<m[,2] & m[x,1]>=m[,1])>x)})

m[!rem,]

如何获得不重叠的日期？

2 个答案: