[英]Repeating variable in group by category in R
我有要為其創建新變量的數據:標志
數據以縱向格式設置,重復 id 並具有相關日期。
另外兩個重要的變量是category 和 company 。
類別:對於每個 id,至少會有一個類別"a"和"b" ,但大多數時候會有多個 "a" 和 "b"。 公司:同一ID可能有多個公司。 有時,類別“b”與特定 ID 的類別“a”具有相同的公司。 為了方便起見,我只包括了三個公司,分別是 x、y、z。
我想創建一個標志。 這樣當 group_by id
下面是帶有標志變量的數據框(預期輸出)
id<- c(1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,5,5,5)
date<- as.Date(c("2001-01-04", "2007-09-23", "2008-11-14",
"2009-11-13", "2012-07-21", "2014-09-15",
"2000-04-01", "2008-07-14", "2008-07-14",
"2001-03-21", "2019-05-23", "2019-05-08",
"2004-07-06", "2007-08-12", "2011-09-20",
"2011-09-20", "2014-08-15", "2014-08-15"))
category<- c("a", "b", "b", "a", "b", "b", "a", "b", "b",
"a", "b", "b", "a", "a", "b", "b", "b", "b")
company<-c("x", "x", "x", "x", "y", "y", "x", "x", "x",
"x", "y", "z", "x", "x", "x", "x", "x", "y")
flag<-c ("rp","p1", "p2", "nr", "p0", "p0", "rp", "p1",
"p1", "nr", "p0", "p0", "rp", "rp", "p1", "p1",
"p2", "p0")
dfx <- data.frame(id, date, category, company, flag)
如果我正確理解邏輯,一種可能的方法是tidyverse
。 同時按id
和company
分組后,可以看到“a”和“b”這兩個類別是否都存在; 如果是這樣,用“rp”標記類別為“a”的那些行。
一個更復雜的case_when
可以考慮您的不同規則,但在您需要“p”和一系列數字的情況下留下缺失的NA
情況。 可以根據這些缺失值制作一個包含計數器的臨時列,為您提供“p1”、“p2”等。
library(tidyverse)
dfx %>%
group_by(id, company) %>%
mutate(new_flag = case_when(
all(c("a", "b") %in% category) & category == "a" ~ "rp",
category == "a" ~ "nr",
TRUE ~ NA_character_)) %>%
group_by(id) %>%
mutate(new_flag = case_when(
category == "b" & new_flag[category == "a"][1] == "nr" ~ "p0",
category == "b" & new_flag[category == "a"][1] == "rp" &
company == company[category == "a"][1] ~ NA_character_,
category == "b" & new_flag[category == "a"][1] == "rp" &
company != company[category == "a"][1] ~ "p0",
TRUE ~ new_flag)) %>%
group_by(id, company) %>%
mutate(ctr = cumsum(is.na(new_flag) & date != lag(date, default = first(date[is.na(new_flag)])))) %>%
mutate(new_flag = ifelse(is.na(new_flag), paste0("p", ctr), new_flag)) %>%
select(-ctr)
Output
id date category company flag new_flag
<dbl> <date> <chr> <chr> <chr> <chr>
1 1 2001-01-04 a x rp rp
2 1 2007-09-23 b x p1 p1
3 1 2008-11-14 b x p2 p2
4 2 2009-11-13 a x nr nr
5 2 2012-07-21 b y p0 p0
6 2 2014-09-15 b y p0 p0
7 3 2000-04-01 a x rp rp
8 3 2008-07-14 b x p1 p1
9 3 2008-07-14 b x p1 p1
10 4 2001-03-21 a x nr nr
11 4 2019-05-23 b y p0 p0
12 4 2019-05-08 b z p0 p0
13 5 2004-07-06 a x rp rp
14 5 2007-08-12 a x rp rp
15 5 2011-09-20 b x p1 p1
16 5 2011-09-20 b x p1 p1
17 5 2014-08-15 b x p2 p2
18 5 2014-08-15 b y p0 p0
關鍵是編寫一個 function 以根據您的條件正確標記類別。 對於每組id
和company
,您的條件簡化為三個互斥的條件:
因此,考慮以下 function
flag_category <- function(x, date) {
out <- character(length(x))
a <- which(x == "a")
b <- which(x == "b")
if (length(a) > 0L && length(b) > 0L) {
out[a] <- "rp"
dateb <- date[b] # get the date where category is "b"
udateb <- unique(dateb) # get the unique dates
out[b] <- paste0("p", rank(udateb)[match(dateb, udateb)]) # `rank` finds the order for each unique date; use `match` to get the positions in `dateb` to which those ranks belong
return(out)
}
if (length(a) > 0L) {
out[] <- "nr"
return(out)
}
out[] <- "p0"
out
}
然后你可以將它應用到每組id
和company
。
dfx %>% group_by(id, company) %>% mutate(flag2 = flag_category(category, date))
Output
# A tibble: 18 x 6
# Groups: id, company [9]
id date category company flag flag2
<dbl> <date> <chr> <chr> <chr> <chr>
1 1 2001-01-04 a x rp rp
2 1 2007-09-23 b x p1 p1
3 1 2008-11-14 b x p2 p2
4 2 2009-11-13 a x nr nr
5 2 2012-07-21 b y p0 p0
6 2 2014-09-15 b y p0 p0
7 3 2000-04-01 a x rp rp
8 3 2008-07-14 b x p1 p1
9 3 2008-07-14 b x p1 p1
10 4 2001-03-21 a x nr nr
11 4 2019-05-23 b y p0 p0
12 4 2019-05-08 b z p0 p0
13 5 2004-07-06 a x rp rp
14 5 2007-08-12 a x rp rp
15 5 2011-09-20 b x p1 p1
16 5 2011-09-20 b x p1 p1
17 5 2014-08-15 b x p2 p2
18 5 2014-08-15 b y p0 p0
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