Tidyverse 解决方案通过过滤不同的变量组合获得计数

Question

 Library(tidyverse)

使用下面的代码，我想使用 table() 或 dplyr 来获取 Sat 变量（Q1Sat、Q2Sat、Q3Sat）的计数。但是，Q1Sat 与变量 Q1Used 相关，Q2Sat 与 Q2Used 相关，Q3Sat 与 Q3Used 相关。我想过滤掉每个组合的 Used 变量中的“No”，以及 House 变量中的“No”。

因此，例如，要计算 Q1Sat 的计数，我需要在 Q1used 和 House 中过滤掉“否”。对于 Q2Sat，我需要在 Q2Used 和 House 中过滤掉“No”，对于 Q3 Sat，我必须在 Q3Used 和 House 中过滤掉“No”。

使用 Tidyverse，有什么简单的方法可以做到这一点？ （最少的代码）。如有必要，我想使用最新版本的 Tidyverse 软件包，包括 dplyr 的开发版本。

Q1Sat<-c("Neutral","Neutral","VSat","Sat","Neutral","Sat","VDis","Sat","Sat","VSat")
Q2Sat<-c("Neutral","VSat","Dis","Dis","VDis","Sat","Sat","VSat","Neutral","Dis")
Q3Sat<-c("Sat","Sat","Diss","Neutral","VSat","VDis","Sat","Sat","Sat","Neutral")
Q3Used<-c("Yes","No","Yes","Yes","Yes","Yes","Yes","Yes","Yes","No")
Q2Used<-c("Yes","Yes","Yes","Yes","No","No","Yes","Yes","Yes","Yes")
Q1Used<-c("Yes","Yes","Yes","No","No","Yes","Yes","Yes","No","Yes")
House<-c("Yes","No","Unsure","Yes","Yes","No","Unsure","Unsure","Yes","Yes")

Test<-data_frame(Q1Sat,Q2Sat,Q3Sat,Q1Used,Q2Used,Q3Used,House)

Answer 1

Test %>% 
mutate(q1 = ifelse(Q1Used=="Yes", Q1Sat, NA), 
       q2 = ifelse(Q2Used=="Yes", Q2Sat, NA), 
       q3 = ifelse(Q3Used=="Yes", Q3Sat, NA)) %>% 
select(q1:q3) %>% 
sapply(., table)

$q1

Neutral     Sat    VDis    VSat 
      2       2       1       2 

$q2

    Dis Neutral     Sat    VSat 
      3       2       1       2 

$q3

   Diss Neutral     Sat    VDis    VSat 
      1       1       4       1       1 

Answer 2

这是一个使用data.table 的选项。我们将“data.frame”转换为“data.table”（setDT(Test)），通过在melt 中指定patterns 将其重塑为“long”，按“Qs”和“Sat”分组，得到计算“已使用”为“是”的位置并将其重新调整为“宽”格式

library(data.table)
dcast(melt(setDT(Test), measure = patterns("Sat", "Used"), 
   value.name = c("Sat", "Used"), variable.name = 'Qs')[
   Used == "Yes", .N , .(Qs, Sat)], Qs~Sat, fill=0)[, Qs := nm1[Qs][]
#   Qs Dis Diss Neutral Sat VDis VSat
#1: Q1   0    0       2   2    1    2
#2: Q2   3    0       2   1    0    2
#3: Q3   0    1       1   4    1    1

---

另外，我们可以用base R更简洁地做到这一点

un1 <- unique(unlist(Test[1:3]))
t(mapply(function(x,y) table(factor(x[y == "Yes"], levels = un1)), Test[1:3], Test[4:6]))

甚至更紧凑

table(col(Test[1:3]), unlist(replace(Test[1:3], Test[4:6]!= "Yes", NA)))
#    Dis Diss Neutral Sat VDis VSat
#1   0    0       2   2    1    2
#2   3    0       2   1    0    2
#3   0    1       1   4    1    1