Question

希望通过cutoff df将df.listing框架拆分为嵌套index_cutoff列表：

数据：

df <- data.frame(m=c("A","T","W","Z","B","A","A","W","T","K","G","B","T","B"))
index_cutoff <- c("A","B")

尝试代码：

df.listing <- split(df, df$m %in% keyword_cutoff) #failed, not working

当前输出：

$`FALSE`
   m
2  T
3  W
4  Z
8  W
9  T
10 K
11 G
13 T

$`TRUE`
   m
1 A
5 B
6 A
7 A
12 B
14 B

所需的输出第1阶段：

df.listing[[1]]
A
T
W
Z

df.listing[[2]]
B

df.listing[[3]]
A

df.listing[[4]]
A
W
T
K
G

df.listing[[5]]
B
T

df.listing[[6]]
B

期望的输出结果：

df.listing[[1]]
A
T
W
Z

df.listing[[2]]
B

df.listing[[3]]
A #since at stage 1 they are the same cutoff, hence self merge into next list
A
W
T
K
G

df.listing[[4]]
B #since at stage 1 they begin the same with "B" cutoff
T
B

感谢您并且不能通过R数据集提供可重现的示例。

Answer 1

我们需要将逻辑索引的累积和作为拆分组

split(df, cumsum(df$m %in% index_cutoff))

在OP的代码中，df$m %in% index_cutoff只有两个组，即TRUE和FALSE。通过执行cumsum，可以通过在每个TRUE值

处添加1来更改它

Answer 2

您可以尝试类似

的内容

library(dplyr)
library(zoo)

df1 <- df %>%
  mutate_if(is.factor, as.character) %>%
  mutate(grp = ifelse(m %in% index_cutoff, row_number(), NA))

df2 <- df1 %>%
  filter(!is.na(grp)) %>%
  mutate(new_grp = na.locf(ifelse(m != lag(m, default='0'), grp, NA))) %>%
  right_join(df1, by = c("m", "grp")) %>%
  select(-grp) %>%
  mutate(new_grp = na.locf(new_grp))

将最终所需的分组作为

df2
#   m new_grp
#1  A       1
#2  T       1
#3  W       1
#4  Z       1
#5  B       5
#6  A       6
#7  A       6
#8  W       6
#9  T       6
#10 K       6
#11 G       6
#12 B      12
#13 T      12
#14 B      12

现在运行

split(df2$m, df2$new_grp)

你会得到

$`1`
[1] "A" "T" "W" "Z"

$`5`
[1] "B"

$`6`
[1] "A" "A" "W" "T" "K" "G"

$`12`
[1] "B" "T" "B"

按用户定义的截止值将帧拆分为列表

2 个答案: