基于过滤不同变量组合获取计数的Tidyverse解决方案

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 "No"。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，完成此任务的简单方法是什么？（最少的代码量）。如果需要，我想使用最新版本的 Tidyverse 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))，通过指定 melt 中的 patterns 将其重塑为 'long'，按 [= 分组28=] 和 'Sat'，获取 'Used' 为 'Yes' 的计数并将其重新整形为 'wide' 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

此外，我们可以使用 base R

更紧凑地完成此操作

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

或更紧凑

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

基于过滤不同变量组合获取计数的Tidyverse解决方案

Tidyverse Solution to get Counts Based on Filtering Different Combinations of Variables

r

dplyr

purrr

tidyverse