我有一个数据框架，其中第一列表示工作(经理、员工或工人(，第二列表示此人是否在晚上工作，最后一列是家庭代码(如果两个人共享相同的代码，则意味着他们共享同一栋房子(。
#Here is the reproductible data :
PCS <- c("worker", "manager","employee","employee","worker","worker","manager","employee","manager","employee")
work_night <- c("Yes","Yes","No", "No","No","Yes","No","Yes","No","Yes")
HHnum <- c(1,1,2,2,3,3,4,4,5,5)
df <- data.frame(PCS,work_night,HHnum)
我的问题是，我希望有一个新的家庭数据框架，而不是个人。我想根据HHnum对个人进行分组，然后合并他们的答案。
对于变量"；PCS"；我有了基于答案组合的新类别：经理+工作="；我"；经理+员工="；II"；，雇员+雇员=VI，工人+工人=III等
对于变量"0"；"work_night"；，我想申请一个分数(如果两个都回答是，那么分数=2，如果一个回答是，则分数=1，如果两个回答否，那么分数=0(。
为了清楚起见，我希望我的数据帧看起来像这样：
HHnum PCS work_night
1 "I" 2
2 "VI" 0
3 "III" 1
4 "II" 1
5 "II" 1
如何使用dplyr在R上执行此操作？我知道我需要group_by((，但我不知道该用什么。
最佳，Victor
这里有一种方法(尽管我承认这很冗长(。我创建了一个参考数据帧(即combos(，以防类别超过3，然后将其与主数据帧(如df_new(连接，以引入PCS罗马数字。
library(dplyr)
library(tidyr)
# Create a dataframe with all of the combinations of PCS.
combos <- expand.grid(unique(df$PCS), unique(df$PCS))
combos <- unique(t(apply(combos, 1, sort))) %>%
as.data.frame() %>%
dplyr::mutate(PCS = as.roman(row_number()))
# Create another dataframe with the columns reversed (will make it easier to join to the main dataframe).
combos2 <- data.frame(V1 = c(combos$V2), V2 = c(combos$V1), PCS = c(combos$PCS)) %>%
dplyr::mutate(PCS = as.roman(PCS))
combos <- rbind(combos, combos2)
# Get the count of "Yes" for each HHnum group.
# Then, put the PCS into 2 columns to join together with "combos" df.
df_new <- df %>%
dplyr::group_by(HHnum) %>%
dplyr::mutate(work_night = sum(work_night == "Yes")) %>%
dplyr::group_by(grp = rep(1:2, length.out = n())) %>%
dplyr::ungroup() %>%
tidyr::pivot_wider(names_from = grp, values_from = PCS) %>%
dplyr::rename("V1" = 3, "V2" = 4) %>%
dplyr::left_join(combos, by = c("V1", "V2")) %>%
unique() %>%
dplyr::select(HHnum, PCS, work_night)