映射值画面R

Question

让我们创建示例数据：

df <- data.frame(date=c("2017-01-01","2017-01-02", "2017-01-03", "2017-01-04", "2017-01-05"), X1=c("A", "B", "C", "D", "F"), 
       X2=c("B", "A", "D", "F", "C")) 
df2 <- data.frame(date=c("2017-01-01","2017-01-02", "2017-01-03", "2017-01-04", "2017-01-05"), 
        A=c("3", "4", "2", "1", "5"), 
        B=c("6", "2", "5", "1", "1"), 
        C=c("1", "4", "5", "2", "3"), 
        D=c("67", "67", "63", "61", "62"), 
        F=c("31", "33", "35", "31", "38")) 

所以，我有两个数据帧，我要匹配DF2值按日期和X1，X2为DF和创造这些新的变量。对我来说这很棘手的问题是，df2中的匹配值是在colname中。最终的结果应该是这样的：

> result 
     date X1 X2 Var1 Var2 
1 2017-01-01 A B 3 6 
2 2017-01-02 B A 2 4 
3 2017-01-03 C D 5 63 
4 2017-01-04 D F 61 31 
5 2017-01-05 F C 38 3 

result <- data.frame(date=c("2017-01-01","2017-01-02", "2017-01-03", "2017-01-04", "2017-01-05"), 
        X1=c("A", "B", "C", "D", "F"), 
        X2=c("B", "A", "D", "F", "C"), 
        Var1=c("3", "2", "5", "61", "38"), 
        Var2=c("6", "4", "63", "31", "3")) 

我想用mapvalues，但不能弄明白。第二个想法是用df2进行长格式化（融化），然后尝试，但是在那里也失败了。

好吧，这里是我最好的尝试，只是觉得可以有更高效的方式，如果你必须创建多个（> 50）新的变量数据框架。

df2.long <- melt(df2, id.vars = c("date")) 

df$Var1 <- na.omit(merge(df, df2.long, by.x = c("date", "X1"), by.y = c("date", "variable"), all.x = FALSE, all.y = TRUE))[,4] 
df$Var2 <- na.omit(merge(df, df2.long, by.x = c("date", "X2"), by.y = c("date", "variable"), all.x = FALSE, all.y = TRUE))[,5] 

Answer 1

与一种可能性：

df$Var1 <- mapply(function(day, col) df2[df2$date==day, as.character(col)], 
        day=df$date, col=df$X1) 
df$Var2 <- mapply(function(day, col) df2[df2$date==day, as.character(col)], 
        day=df$date, col=df$X2) 

df 
#  date X1 X2 Var1 Var2 
#1 2017-01-01 A B 3 6 
#2 2017-01-02 B A 2 4 
#3 2017-01-03 C D 5 63 
#4 2017-01-04 D F 61 31 
#5 2017-01-05 F C 38 3 

NB：
如果你有更多的列进行修改（不只是喜欢在你的例子），你可以使用lapply遍历列X.：

df[, paste0("Var", 1:2)] <- lapply(df[,paste0("X", 1:2)], 
            function(value) { 
             mapply(function(day, col) df2[df2$date==day, as.character(col)], 
              day=df$date, col=value)}) 

Answer 2

我们可以使用match从“X1”和“X2”栏目，cbind与行的顺序获得“DF2”的列索引，使用行/列索引中提取的价值“ DF2' ，并分配输出以创建 '瓦尔' 列

df[paste0("Var", 1:2)] <- lapply(df[2:3], function(x) 
      df2[-1][cbind(1:nrow(df2), match(x, names(df2)[-1]))]) 
df 
#  date X1 X2 Var1 Var2 
#1 2017-01-01 A B 3 6 
#2 2017-01-02 B A 2 4 
#3 2017-01-03 C D 5 63 
#4 2017-01-04 D F 61 31 
#5 2017-01-05 F C 38 3 

Answer 3

使用dplyr和tidyr：

df2_m <- group_by(df2, date) %>% 
    gather('X1', 'var', -date) 

left_join(df, df2_m) %>% 
    left_join(df2_m, by = c('date', 'X2' = 'X1')) %>% 
    rename(Var1 = var.x, Var2 = var.y) -> result 

Answer 4

使用data.table

library(data.table) # v>=1.10.0 
dcast(
    melt(setDT(df), 1L)[ # melt the first table by date 
    melt(setDT(df2), 1L), # melt the second table by date 
    on = .(date, value = variable), # join by date and the letters 
    nomatch = 0L], # remove everything that wasn't matched 
    date ~ variable, # convert back to long format 
    value.var = c("value", "i.value")) # take both values columns 

#   date value_X1 value_X2 i.value_X1 i.value_X2 
# 1: 2017-01-01  A  B   3   6 
# 2: 2017-01-02  B  A   2   4 
# 3: 2017-01-03  C  D   5   63 
# 4: 2017-01-04  D  F   61   31 
# 5: 2017-01-05  F  C   38   3 

Answer 5

利用熔融和匹配

：

df2l<-melt(df2, measure=c("A","B","C","D","F")) 
Indices <- match(paste(df$date, df$X1), paste(df2l$date,df2l$variable)) 
df$Var1 <- df2l$value[Indices] 
Indices2 <- match(paste(df$date, df$X2), paste(df2l$date,df2l$variable)) 
df$Var2 <- df2l$value[Indices2]