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"))

所以我有两个数据框，我希望按日期和X1和X2匹配df2到df的值，并为这些创建新的变量。这对我来说很棘手的是df2中的匹配值是在colnames中。最终结果应如下所示：

> 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"))

我想使用mapvalues，但无法理解。第二个想法是用df2进行长时间格式化（熔化），然后尝试，但也失败了。

好的，这是我最好的尝试，只是觉得可以有更有效的方法，如果你必须为数据框创建多个（> 50）新变量。

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

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 2

双熔＆gt;加入＆gt;使用data.table

的dcast选项

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 3

我们可以使用${COOKIE_FOO}来获取＆＃39; df2＆＃39;的列索引来自＆＃39; X1＆＃39;和＆＃39; X2＆＃39;带有行序列的列match，使用行/列索引提取＆＃39; df2＆＃39;中的值，并指定输出以创建＆＃39; Var＆＃39;列

cbind

Answer 4

mapply的可能性：

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

<强> NB：
如果您要修改更多列（不仅仅是示例中的2个），则可以使用lapply循环遍历列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 5

使用融合和匹配：

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]

映射数据帧之间的值R.

5 个答案: