这是我的数据框:
date <- as.Date(c("1993-09-21", "1994-02-12", "1994-02-23", "1994-05-14", "1994-08-18", "1994-08-25", "1994-08-29", "1994-09-17", "1994-10-16", "1994-10-16", "1994-10-22", "1994-10-26", "1994-12-26", "1995-04-12", "1995-05-04", "1995-06-20", "1995-07-11", "1995-07-27", "1995-08-14", "1995-08-15", "1995-08-22", "1995-08-27", "1995-08-27", "1995-08-28", "1995-08-30", "1995-08-30", "1995-09-03", "1995-09-03", "1995-09-03", "1995-09-15"))
value <- c(2, 1, 1, 1, 2, 1, 2, 4, 2, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 1, 1)
df <- data.frame(date, value)
df$value.equals.1 <- df$value == 1
我需要两件事:(1) 值为 1 的每个连续连续值的第一个和最后一个日期 (2) 每个连续值为 1 的值的长度。
我已经用我需要的东西注释了数据框。我怎样才能在 R 中实现这一点?
我们可以用rleid
from做到这一点data.table
。使用rleid
on 'value.equals.1'创建分组变量,根据 'value.equals.1' 对 'date' 进行子集化,并提取按 'grp' 分组的第一个和最后一个 'date'
library(data.table)
setDT(df)[, date[value.equals.1], .(grp = rleid(value.equals.1))
][, .(date = c(V1[1], V1[.N]), n = .N), by = grp][, grp := NULL][]
# date n
# 1: 1994-02-12 3
# 2: 1994-05-14 3
# 3: 1994-08-25 1
# 4: 1994-08-25 1
# 5: 1994-10-22 1
# 6: 1994-10-22 1
# 7: 1994-12-26 8
# 8: 1995-08-15 8
# 9: 1995-08-27 3
#10: 1995-08-30 3
#11: 1995-09-03 2
#12: 1995-09-15 2
或者这可以用 tidyverse
library(dplyr)
df %>%
group_by(grp = rleid(value.equals.1)) %>%
filter(all(value.equals.1)) %>%
mutate(n = n()) %>%
slice(c(1, n())) %>%
ungroup %>%
select(date, n)
# A tibble: 12 x 2
# date n
# <date> <int>
# 1 1994-02-12 3
# 2 1994-05-14 3
# 3 1994-08-25 1
# 4 1994-08-25 1
# 5 1994-10-22 1
# 6 1994-10-22 1
# 7 1994-12-26 8
# 8 1995-08-15 8
# 9 1995-08-27 3
#10 1995-08-30 3
#11 1995-09-03 2
#12 1995-09-15 2
或者使用rle
frombase R
创建组
grp <- inverse.rle(within.list(rle(df$value.equals.1), values <- seq_along(values)))
do.call(c, lapply(with(df, split(date[value.equals.1],
grp[value.equals.1])), function(x) c(x[1], x[length(x)])))
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