使用 dplyr 按组获取累积计数 [英] Using dplyr to get cumulative count by group

问题描述

提前致谢.我有以下数据:

Thanks in advance. I have the following data:

df <- data.frame(person=c(1,1,1,1,2,2,2,2,3,3,3,3), 
             neighborhood=c("A","A","A","A","B","B","C","C","D","D","E","F"))

我想生成一个新列，该列提供每个人随着面板的进行移动经过的社区的累积计数.像这样:

I would like to generate a new column that gives the cumulative count of neighborhoods that each person moves through as the panel progresses. Like such:

df2 <- data.frame(person=c(1,1,1,1,2,2,2,2,3,3,3,3), 
             neighborhood=c("A","A","A","A","B","B","C","C","D","D","E","F"),
             moved=c(0,0,0,0,0,0,1,1,0,0,1,2)
             )

再次感谢.

推荐答案

我们可以使用 group by 'person'，然后通过 matching 'neighborhood' 和它的 unique 值以获取索引并减去 1.

We can use group by 'person', then create the 'moved' by matching the 'neighborhood' with its unique values to get the index and subtract 1.

df %>%
   group_by(person) %>% 
   mutate(moved = match(neighborhood, unique(neighborhood))-1)
#   person neighborhood moved
#    <dbl>       <fctr> <dbl>
#1       1            A     0
#2       1            A     0
#3       1            A     0
#4       1            A     0
#5       2            B     0
#6       2            B     0
#7       2            C     1
#8       2            C     1
#9       3            D     0
#10      3            D     0
#11      3            E     1
#12      3            F     2

或使用 factor 和 levels 指定为 'neighborhood' 中的 unique 值，强制为 'integer' 并减去 1.

or use factor with levels specified as the unique values in 'neighborhood', coerce to 'integer' and subtract 1.

df %>%
   group_by(person) %>% 
   mutate(moved = as.integer(factor(neighborhood, levels = unique(neighborhood)))-1)
#   person neighborhood moved
#    <dbl>       <fctr> <dbl>
#1       1            A     0
#2       1            A     0
#3       1            A     0
#4       1            A     0
#5       2            B     0
#6       2            B     0
#7       2            C     1
#8       2            C     1
#9       3            D     0
#10      3            D     0
#11      3            E     1
#12      3            F     2

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