Question

假设我有以下data.frame：

df <- data.frame(color = c("G","G","G","R","R","R","R","R","R","R","G","G"),
           trial = c(1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4))

如果我想从之前的color中提取trial，我该怎么做？最终目标是最终得到一个data.frame：

  color trial prevcolor
1      G     1      <NA>
2      G     1      <NA>
3      G     1      <NA>
4      R     2         G
5      R     2         G
6      R     2         G
7      R     3         R
8      R     3         R
9      R     3         R
10     R     3         R
11     G     4         R
12     G     4         R

Answer 1

我们可以使用lag（假设'试用'是有序的）

df$prevcolor <- with(df, lag(color, n=sum(trial==trial[1L])))
df
#   color trial prevcolor
#1      G     1      <NA>
#2      G     1      <NA>
#3      G     1      <NA>
#4      R     2         G
#5      R     2         G
#6      R     2         G
#7      R     3         R
#8      R     3         R
#9      R     3         R
#10     R     3         R
#11     G     4         R
#12     G     4         R

评论中@ rawr解决方案的变体（如果'试用'不是数字列）

Un1 <- unique(df$trial)
with(df, color[match(factor(trial, levels= Un1, labels = c(NA, head(Un1,-1))), trial)])

使用dplyr，我们可以使用group_indices来获取群组的索引

library(dplyr)
df %>% 
     mutate(prev_color = color[match(group_indices_(.,.dots = 'trial')-1, trial)])
#   color trial prev_color
#1      G     1       <NA>
#2      G     1       <NA>
#3      G     1       <NA>
#4      R     2          G
#5      R     2          G
#6      R     2          G
#7      R     3          R
#8      R     3          R
#9      R     3          R
#10     R     3          R
#11     G     4          R
#12     G     4          R

Answer 2

以下是使用for循环的解决方案：

df <- data.frame(color = c("G","G","G","R","R","R","R","R","R","R","G","G"),
                 trial = c(1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4))

# iterate through trial numbers
for (trial in unique(df$trial)) {
  # select color of previous trial number
  prev_color <-  as.character(df$color[df$trial == trial - 1])[1] 

  # assign previous color to current trial number
  df$prevcolor[df$trial == trial] <- prev_color
}
df

##   color trial prevcolor
##1      G     1      <NA>
##2      G     1      <NA>
##3      G     1      <NA>
##4      R     2         G
##5      R     2         G
##6      R     2         G
##7      R     3         R
##8      R     3         R
##9      R     3         R
##10     R     3         R
##11     G     4         R
##12     G     4         R

Answer 3

这是使用R中的简单merge函数的另一种解决方案。您的数据框：

df <- data.frame(color = c("G","G","G","R","R","R","R","R","R","R","G","G"),
                 trial = c(1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4))

现在使用merge功能。它仅用于合并数据帧。因此：

df2<-merge(data.frame(prevtrial=c(df$trial-1)),unique(df), by.x="prevtrial",by.y="trial",all.x=T)

现在为您的输出创建一个新的数据框：

newdf<-data.frame(color=df$color,trial=df$trial,prevtrial=df2$prevtrial,prevcolor=df2$color)

将给出：

> newdf
   color trial prevtrial prevcolor
1      G     1         0      <NA>
2      G     1         0      <NA>
3      G     1         0      <NA>
4      R     2         1         G
5      R     2         1         G
6      R     2         1         G
7      R     3         2         R
8      R     3         2         R
9      R     3         2         R
10     R     3         2         R
11     G     4         3         R
12     G     4         3         R
>

从先前交互R中拉取值

3 个答案: