从逻辑 DF 和布尔 DF 构造 DF

Question

我有两个 DF 以下是布尔值

> Portfolio 1 

Name | X1992 | X1993 | X1994 
XHHD   False   False   True
Jdqd   False   False   False
Jhds   False   True    False
lkaz   False   False   True                      
nqb    True    False   False
jbqs   False   False   False
jbq    False   True    False
knd    True    False   True
njvd   False   True    False
kjiz   True    False   True
khza   False   False   False
akhd   False   False   True
jkaze  True    True    False
bzae   True    False   False

第二个 DF 有一个月的频率

返回

Name | 1992/01 | 1992/02 ..... 1994/12 
XHHD   0.23       0.564
Jdqd   0.3        0.654
Jhds   0.234      0.456
lkaz   0.54       0.472               
nqb    0.99       0.761
jbqs   0.01       1.765
jbq    0.23       0.002
knd    0.59       2.32
njvd   0.123      0.43
kjiz   0.987      -0.12
khza   1.34       0.12
akhd   0.76       0.23
jkaze  0.654      0.98
bzae   0.43       0.73

我想要一个 DF，它可以根据布尔 DF 计算公司的平均每月回报，如果 Companies TRUE 然后将她的回报包括在计算中。 结果将是

Date     Portfolio 1
1992/01   mean
1992/02   mean
1992/03   mean
1992/04   mean

Answer 1

我们可以通过转换为“长”格式来做到这一点，进行连接，然后通过“日期”获取mean

library(data.table)
dM1 <- melt(setDT(df1), id.var = "Name")
na.omit(melt(setDT(df2), id.var = "Name", variable.name = "Date")[,
     variable := paste0("X", substring(Date, 1, 4))
      ][dM1, on = .(Name, variable)
       ][i.value == "True", .(Portfolio = mean(value, na.rm = TRUE)) , Date])
#     Date  Portfolio  
#1: 1992/01    0.7302
#2: 1992/02    0.9342

数据

 df1 <- structure(list(Name = c("XHHD", "Jdqd", "Jhds", "lkaz", "nqb", 
"jbqs", "jbq", "knd", "njvd", "kjiz", "khza", "akhd", "jkaze", 
"bzae"), X1992 = c("False", "False", "False", "False", "True", 
"False", "False", "True", "False", "True", "False", "False", 
"True", "True"), X1993 = c("False", "False", "True", "False", 
"False", "False", "True", "False", "True", "False", "False", 
"False", "True", "False"), X1994 = c("True", "False", "False", 
"True", "False", "False", "False", "True", "False", "True", "False", 
"True", "False", "False")), .Names = c("Name", "X1992", "X1993", 
"X1994"), class = "data.frame", row.names = c(NA, -14L))

df2 <-  structure(list(Name = c("XHHD", "Jdqd", "Jhds", "lkaz", "nqb", 
"jbqs", "jbq", "knd", "njvd", "kjiz", "khza", "akhd", "jkaze", 
"bzae"), `1992/01` = c(0.23, 0.3, 0.234, 0.54, 0.99, 0.01, 0.23, 
0.59, 0.123, 0.987, 1.34, 0.76, 0.654, 0.43), `1992/02` = c(0.564, 
0.654, 0.456, 0.472, 0.761, 1.765, 0.002, 2.32, 0.43, -0.12, 
0.12, 0.23, 0.98, 0.73)), .Names = c("Name", "1992/01", "1992/02"
), class = "data.frame", row.names = c(NA, -14L))