具有此数据框：
dframe1 <- structure(list(id = c(1L, 1L, 1L, 2L, 2L), name = c("Google",
"Yahoo", "Amazon", "Amazon", "Google"), date = c("2008-11-01",
"2008-11-01", "2008-11-04", "2008-11-01", "2008-11-02")), class = "data.frame", row.names = c(NA,
-5L))
第二个：
dframe2 <- structure(list(id = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
1L, 2L, 2L, 2L, 2L, 2L, 2L), date = c("2008-11-01", "2008-11-01",
"2008-11-04", "2008-10-31", "2008-10-31", "2008-11-02", "2008-11-02",
"2008-11-02", "2008-11-05", "2008-11-02", "2008-11-03", "2008-10-31",
"2008-11-01", "2008-11-01", "2008-11-02", "2008-11-02", "2008-11-03"
), name = c("Google", "Yahoo", "Amazon", "Google", "Yahoo", "Amazon",
"Google", "Yahoo", "Amazon", "Google", "Yahoo", "Amazon", "Google",
"Amazon", "Google", "Amazon", "Google"), text_sth = c("test",
"text_sth", "text here", "another text", "other", "another one",
"test", "text_sth", "text here", "another text", "other", "etc",
"test", "text_sth", "text here", "another text", "text here")), class = "data.frame", row.names = c(NA,
-17L))
使用dframe1的结果，如何从dataframe2中保留每个ID与dframe1相同名称但在dframe1记录日期之前和之后的日期的行？
这是我尝试过的
library(data.table)
library(tidyverse)
library(reshape2)
dframe1 = data.table(dframe1)
dframe1[, date := as.Date(date)]
dframe1_first = dframe1[, .(date = min(date)), .(id, name)] %>%
mutate(date_pre = date - 1,
date_after = date + 1)
req_rows = dframe2 %>%
merge(dframe1_first %>%
rename(id = id),
by = "id") %>%
filter(date >= date_pre,
date <= date_after,
date != date) %>%
mutate(period = ifelse(date<date, '1-day-pre', '1-day-after'))
预期输出：
id date name text_sth 1 2008-10-31 Google another text 1 2008-10-31 Yahoo other 1 2008-11-02 Google test 1 2008-11-02 Yahoo text_sth 1 2008-11-05 Amazon text here 1 2008-11-02 Google another text 2 2008-10-31 Amazon etc 2 2008-11-01 Google test 2 2008-11-02 Amazon another text 2 2008-11-03 Google text here