Question

我有这个数据框：

    df<-structure(list(StimulusName = c("Alpha5", "Alpha5", "Alpha5", 
"Alpha5", "Alpha5", "Alpha5", "Alpha5", "Alpha5", "Alpha5", "Alpha5", 
"Alpha5", "Alpha5", "Alpha5", "Alpha5", "Alpha5", "Alpha5", "Alpha5", 
"Alpha5", "Alpha5", "Alpha5", "Alpha5", "Alpha5", "Alpha5", "Alpha5", 
"Alpha5", "Alpha5", "Alpha5", "Alpha5", "Alpha5", "Alpha5", "Alpha5", 
"Alpha5", "Alpha5"), Label = c(NA, NA, NA, NA, NA, "Onset", NA, 
NA, NA, NA, NA, "Offset", NA, NA, NA, NA, NA, NA, NA, "Onset", 
NA, NA, NA, NA, NA, NA, NA, NA, "Offset", NA, NA, NA, NA)), row.names = c(NA, 
-33L), class = c("tbl_df", "tbl", "data.frame"))

我想在Onset和Offset的每个出现的行{em>之间填充Label列中的单词'Onset'。最终的数据帧如下所示：

    StimulusName Label
1   Alpha5       NA
2   Alpha5       NA
3   Alpha5       NA
4   Alpha5       NA
5   Alpha5       NA
6   Alpha5       Onset
7   Alpha5       Onset
8   Alpha5       Onset
9   Alpha5       Onset
10  Alpha5       Onset
11  Alpha5       Onset
12  Alpha5       Offset
13  Alpha5       NA
14  Alpha5       NA
15  Alpha5       NA
16  Alpha5       NA
17  Alpha5       NA
18  Alpha5       NA
19  Alpha5       NA
20  Alpha5       Onset
21  Alpha5       Onset
22  Alpha5       Onset
23  Alpha5       Onset
24  Alpha5       Onset
25  Alpha5       Onset
26  Alpha5       Onset
27  Alpha5       Onset
28  Alpha5       Onset
29  Alpha5       Offset
30  Alpha5       NA
31  Alpha5       NA
32  Alpha5       NA
33  Alpha5       NA

可以在dplyr中使用fill()来实现吗？

此代码显然无法正常工作，因为它会填充第一个非NA值之后的所有内容：

Test <- df%>% 
  fill(df$Label)

我愿意接受任何解决方案，但希望使用dplyr并尽可能避免循环。

Answer 1

我们可以使用基数R mapply

#Get "Onset" indices
onset_ind <- which(df$Label == "Onset")

#Get "Offset" indices
offset_ind <- which(df$Label == "Offset")

#Create a sequence of indices between onset and offset and change their value
df$Label[unlist(mapply(function(x, y) x:y, onset_ind, offset_ind - 1))] <- "Onset"

df
#   StimulusName  Label
#1        Alpha5   <NA>
#2        Alpha5   <NA>
#3        Alpha5   <NA>
#4        Alpha5   <NA>
#5        Alpha5   <NA>
#6        Alpha5  Onset
#7        Alpha5  Onset
#8        Alpha5  Onset
#9        Alpha5  Onset
#10       Alpha5  Onset
#11       Alpha5  Onset
#12       Alpha5 Offset
#13       Alpha5   <NA>
#14       Alpha5   <NA>
#.....

此处假定“开始”和“偏移”出现的次数相同。

如果要避免不必要的变量创建（onset_ind和offset_ind），可以单行执行

df$Label[unlist(mapply(function(x, y) x:y, 
      which(df$Label == "Onset"), which(df$Label == "Offset") - 1))] <- "Onset"

Answer 2

这是一种dplyr的方式，

library(tidyverse)

df %>% 
 group_by(grp = cumsum(!is.na(Label))) %>% 
 mutate(Label = replace(Label, first(Label) == 'Onset', 'Onset')) %>% 
 ungroup() %>% 
 select(-grp)

Answer 3

library(zoo) #using na.locf()

df2<-data.frame(df) # converting list to data frame

df2$Label[which((is.na(df2$Label)& !is.na(na.locf(df2)[,2]) & na.locf(df2)[,2]=="Offset")=="FALSE")]<-
  na.locf(df2)[,2][which((is.na(df2$Label)& !is.na(na.locf(df2)[,2]) & na.locf(df2)[,2]=="Offset")=="FALSE")]

df2
StimulusName  Label
1        Alpha5   <NA>
2        Alpha5   <NA>
3        Alpha5   <NA>
4        Alpha5   <NA>
5        Alpha5   <NA>
6        Alpha5  Onset
7        Alpha5  Onset
8        Alpha5  Onset
9        Alpha5  Onset
10       Alpha5  Onset
11       Alpha5  Onset
12       Alpha5 Offset
13       Alpha5   <NA>
14       Alpha5   <NA>
15       Alpha5   <NA>
16       Alpha5   <NA>
17       Alpha5   <NA>
18       Alpha5   <NA>
19       Alpha5   <NA>
20       Alpha5  Onset
21       Alpha5  Onset
22       Alpha5  Onset
23       Alpha5  Onset
24       Alpha5  Onset
25       Alpha5  Onset
26       Alpha5  Onset
27       Alpha5  Onset
28       Alpha5  Onset
29       Alpha5 Offset
30       Alpha5   <NA>
31       Alpha5   <NA>
32       Alpha5   <NA>
33       Alpha5   <NA>

Answer 4

如何创建用户定义的函数。

 for(i in 1:length(df$Label)){    
   if(df$Label[i]%in%"Onff"==TRUE)    
   {if(df$Label[i+1]%in%NA==TRUE){df$Label[i+1]<-"Onff"}}    
     else{df$Label[i]<-df$Label[i]}    
 }

填充列中两个值之间的行

4 个答案: