如何根据R中的行和列合并两个表？

Question

df1 <- data.frame(MLID=c('992','992','BJR'),
              Position=c('N0','N1','N1'),
              Weight=c(0.125,0.58,0.69))


df2 <- data.frame(MLID=c('992','992','992','992',
                     'BJR','BJR','BJR','BJR'),
              Weight=c(0,0.251,0.501,1.001,
                       0,0.251,0.501,1.001),
              N0=c(2.80,4.05,4.05,4.05,
                   4.05,4.05,4.05,4.05),
              N1=c(3.47,4.73,4.95,5.15,
                   4.73,7.73,4.95,5.15) )

我想要的是合并这两个表遵循规则：

1. MLID
2. 查看位置（N0 或 N1）
3. 找到重量所在的范围（如 excel 中的近似 vlookup）（2.8 表示 992 N0 的重量收费 (0,0.250)，重量收费 4.05 (0.251,0.500) 992 N0，3.47 重量（ 0,0.250) 992 N1 等。

所以最终的输出应该是：

MILD  Position  Weight  Charge
992      N0     0.125    2.8
992      N1     0.580    4.95
BJR      N1     0.690    4.95

可以在 R 中实现吗？特别是在 dplyr 包中？

Answer 1

我们可以对data.table 使用非等连接。使用melt 将第二个数据集重塑为“长”格式，并与“MLID”、“位置”的第一个数据和“重量”列上的非等比较连接，并分配“电荷”的last 值在“df1”中创建列

library(data.table)
setDT(df1)[setnames(melt(setDT(df2), measure = c("NO", "N1"), 
       variable.name = "Position", value.name = "Charge"), "Weight", "wt"), 
      Charge := Charge, on = .(MLID, Position, Weight > wt), mult = "last"] 

df1
#   MLID Position Weight Charge
#1:  992       NO  0.125   2.80
#2:  992       N1  0.580   4.95
#3:  BJR       N1  0.690   4.95

Answer 2

可以实现使用data.tablerolling连接的选项。首先，df2 需要使用melt 转换为long-format，然后加入df1 和df2。

library(data.table)

setDT(df1, key = c("MLID", "Position","Weight") )

df2 <- melt(df2, id.vars = c("MLID","Weight"), variable.name = "Position", 
                                                      value.name = "Charge")

setDT(df2, key = c("MLID", "Position","Weight"))

df2[df1, roll = "nearest"]
#    MLID Weight Position Charge
# 1:  992  0.580       N1   4.95
# 2:  992  0.125       NO   2.80
# 3:  BJR  0.690       N1   4.95

选项#2：基于tidyverse 的方法可以是：

library(tidyverse)
df2 %>% gather(Position, Charge, -MLID, -Weight) %>%
  right_join(df1, by=c("MLID", "Position")) %>%
  filter(Weight.x <= Weight.y) %>%
  group_by(MLID, Position) %>%
  arrange(Weight.y-Weight.x) %>% 
  slice(1) %>%
  select(MLID, Weight = Weight.y, Position, Charge)

# # A tibble: 3 x 4
# # Groups: MLID, Position [3]
#   MLID  Weight Position Charge
#   <chr>  <dbl> <chr>     <dbl>
# 1 992    0.580 N1         4.95
# 2 992    0.125 NO         2.80
# 3 BJR    0.690 N1         4.95

数据：

OP's 数据稍作修改，以在 data.frame 中包含 stringsAsFactors = FALSE 参数以避免不必要的警告。

df1 <- data.frame(MLID=c('992','992','BJR'),
                  Position=c('NO','N1','N1'),
                  Weight=c(0.125,0.58,0.69), stringsAsFactors = FALSE)


df2 <- data.frame(MLID=c('992','992','992','992',
                         'BJR','BJR','BJR','BJR'),
                  Weight=c(0,0.251,0.501,1.001,
                           0,0.251,0.501,1.001),
                  NO=c(2.80,4.05,4.05,4.05,
                       4.05,4.05,4.05,4.05),
                  N1=c(3.47,4.73,4.95,5.15,
                       4.73,7.73,4.95,5.15), stringsAsFactors = FALSE )

Answer 3

这是一个基本的 R 版本：

outdf <- merge(df1, df2, by = "MLID")
outdf$dist <- abs(outdf$Weight.x - outdf$Weight.y)
ting <- aggregate(dist ~ MLID + Position, FUN = function(x) min(x), data = outdf)
outdf2 <- merge(outdf, ting, by.x = c("MLID", "Position", "dist"))
outdf2$charge <- ifelse(outdf2$Position == "N1", outdf2$N1, outdf2$NO)
outdf2 <- outdf2[,c("MLID", "Position", "Weight.x", "charge")]
outdf2
# MLID Position Weight.x charge
# 1  992       N1    0.580   4.95
# 2  992       NO    0.125   2.80
# 3  BJR       N1    0.690   4.95