函数和滚动应用来查找所有值等于某物的任何窗口?

Question

跟进：R中有没有办法在定义的连续行上制作ifelse？

数据：

df<-data.frame(group=c(1, 1,1, 1,1, 2, 2, 2, 4,4,4,4), 
value=c("A","B","C","B","A","A","A","B","D","A","A","B"),
want_every_c=c("yes","yes","yes","yes","yes","no","no","no","no","no","no","no"),
want_every_b=c("yes","yes","yes","yes","yes","yes","yes","yes","no","no","no","no"),
want_two_a_within_together=c("no","no","no","no","no", "yes","yes","yes", "yes","yes","yes", "yes" ))

这些解决方案非常适合在扫描三个列时获得"want_every_"列。

以前的解决方案：

f2 <- function(colNm, val){
c('no', 'yes')[(all(rollapply({{colNm}}, 3, 
FUN = function(x) val %in% x))) + 1]
}
df %>%
group_by(group) %>%
mutate(want_every_c = f2(value, "C"),
want_every_b = f2(value, "B"))

我是函数的新手，想以不同的方式使用它。我想要一个名为"want_two_a_within_together"的新列，在一个组中，任何两个扫描都是完全值='A'。这样：

group value want_two_a_within_together
1      1     A                         no
2      1     B                         no
3      1     C                         no
4      1     B                         no
5      1     A                         no
6      2     A                        yes
7      2     A                        yes
8      2     B                        yes
9      4     D                        yes
10     4     A                        yes
11     4     A                        yes
12     4     B                        yes

我想我可以在函数中的某处添加"any"并将宽度更改为 2，但不完全确定。

制作窗口 3 时的当前尝试和问题：

f2 <- function(colNm, val, w){ c('no', 'yes')[(all(rowSums(rollapply({{colNm}}, w, FUN = function(x) x %in% val)) > 0))+1]}
df<-data.frame(value=c("B","B","B","B","B","A","A","A","A","A","A"),
correct=c("yes","yes","yes", "yes","yes","yes", "yes" ,"yes","yes","yes","yes"))
df%>%  mutate(try = if(n()>2) f2(value, "A", 3) else "no")

value want_three_a_within_together try
1      B                          yes  no
2      B                          yes  no
3      B                          yes  no
4      B                          yes  no
5      B                          yes  no
6      A                          yes  no
7      A                          yes  no
8      A                          yes  no
9      A                          yes  no
10     A                          yes  no
11     A                          yes  no

Answer 1

可以修改该函数以计算rowSums

f2n <- function(colNm, val, w){ 
c('no', 'yes')[(any(rowSums(rollapply({{colNm}}, w, 
FUN = function(x) x %in% val)) == w))+1]
}
library(dplyr)
library(zoo)
df%>% 
group_by(group) %>%
mutate(new_want_two_a_within_together = if(n()>1) f2n(value, "A", 2) else "no")
# A tibble: 18 x 4
# Groups:   group [4]
#   group value want_two_a_within_together new_want_two_a_within_together
#   <dbl> <fct> <fct>                      <chr>                         
# 1     1 A     no                         no                            
# 2     1 B     no                         no                            
# 3     1 C     no                         no                            
# 4     1 B     no                         no                            
# 5     1 A     no                         no                            
# 6     2 A     yes                        yes                           
# 7     2 A     yes                        yes                           
# 8     2 B     yes                        yes                           
# 9     4 D     yes                        yes                           
#10     4 A     yes                        yes                           
#11     4 A     yes                        yes                           
#12     4 B     yes                        yes                           
#13     5 A     yes                        yes                           
#14     5 B     yes                        yes                           
#15     5 B     yes                        yes                           
#16     5 A     yes                        yes                           
#17     5 A     yes                        yes                           
#18     5 A     yes                        yes