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) )

我要合并的两个表遵循规则：

MLID
查看位置（其N0或N1）
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（0,0.250）电荷，992 N0的4.05重量（0.251,0.500）（3.45重量） 0,0.250）992 N1等

因此最终输出应为：

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

在R中有可能吗？尤其是在dplyr软件包中？

Answer 1

可以实现使用data.table rolling连接的选项。首先，需要使用df2在long-format中转换melt，然后将df1和df2都加入。

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2：一种基于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：

为了避免不必要的警告，对OP's数据进行了略微修改，以在stringsAsFactors = FALSE中包含data.frame自变量。

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 2

我们可以对data.table使用非等号联接。使用melt将第二个数据集整形为'long'格式，并在“ MLID”，“ Position”和“ Weight”列上的非等式比较中加入第一个数据，并分配last值“费用”的字段以在“ 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 3

这是基本的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

如何基于R中的行和列合并两个表？

3 个答案: