在data.frame中按连续年份细分 [英] Subset by consecutive years in a data.frame

问题描述

我在R中有一个data.frame/data.table，如下所示:

I have a data.frame / data.table in R as follows:

df <- data.frame(
  ID = c(rep("A", 20)),
  year = c(1968, 1971, 1972, 1973, 1974, 1976, 1978, 1980, 1982, 1984, 1985, 
           1986, 1987, 1988, 1990, 1991, 1992, 1993, 1994, 1995)
)

我想对df进行子集化，以便仅保留连续至少五年的那些条目.在此示例中，这是两个时期(1984:1988和1990:1995)的情况.如何在R中执行此操作?

I'd like to subset the df in order to keep only those entries which have at least five consecutive years. In this example this is the case in two periods (1984:1988 and 1990:1995). How can I do this in R?

推荐答案

使用 diff 和 cumsum 的紧凑型解决方案:

A compact solution using diff and cumsum:

setDT(df)[, grp := cumsum(c(0, diff(year)) > 1), by = ID
          ][, if (.N > 4) .SD, by = .(ID, grp)][, grp := NULL][]

给出所需的结果:

    ID year
 1:  A 1984
 2:  A 1985
 3:  A 1986
 4:  A 1987
 5:  A 1988
 6:  A 1990
 7:  A 1991
 8:  A 1992
 9:  A 1993
10:  A 1994
11:  A 1995

说明:

• 使用 grp:= cumsum(c(0，diff(year))> 1)，通过= ID ，您可以为每个 ID连续创建一个(临时)分组变量.
• 使用 if(.N> 4).SD，通过=.(ID，grp)，您可以为每个 ID 选择仅连续5年或更长时间的组
• 使用 grp:= NULL 删除(临时)分组变量.

• With grp := cumsum(c(0, diff(year)) > 1), by = ID you create a (temporary) grouping variable for consecutive years for each ID.
• With if (.N > 4) .SD, by = .(ID, grp) you select only groups with 5 or more consecutive years for each ID.
• With grp := NULL you remove the (temporary) grouping variable.

基于R的可比较方法:

i <- with(df, ave(year, ID, FUN = function(x) { 
  r <- rle(cumsum(c(0, diff(year)) > 1));
  rep(r$lengths, r$lengths)
  } ))

df[i > 4,] # or df[which(i > 4),]

这将为您带来相同的结果.

which will get you the same result.

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