使用键替换整个值 data.table

Question

我有一个大数据 table 如下所示：

    V1              V2           V3  V4  V5  V6  V7  V8  V9
1: XS0285400197 TR.IssuerRating  F1  F1  F1  F1  F1  F1  F1
2: XS0041971275 TR.IssuerRating AAA AAA AAA AAA  F1  F1  AAA
3: XS0043098127 TR.IssuerRating  WD  WD  WD  WD  WD  WD  WD

structure(list(V1 = c("XS0285400197", "XS0041971275", "XS0043098127"
), V2 = c("TR.IssuerRating", "TR.IssuerRating", "TR.IssuerRating"
), V3 = c("F1", "AAA", "WD"), V4 = c("F1", "AAA", "WD"), V5 = c("F1", 
"AAA", "WD"), V6 = c("F1", "AAA", "WD"), V7 = c("F1", "F1", "WD"
), V8 = c("F1", "F1", "WD"), V9 = c("F1", "AAA", "WD")), class = "data.frame", row.names = c(NA, 
-3L))

实际数据table要大得多，但这只是一个例子。此外，我有一把钥匙，我想用数字替换评级（此处为 F1、AAA 和 WD）。

    Rating CreditQuality
1:   F1               2
2:  AAA               1
3:  WD                6
4:  (P)B2             6
5: (P)Ba1             4
6: (P)Ba2             5

structure(list(Rating = c("F1", "AAA", "WD", "(P)B2", "(P)Ba1", 
"(P)Ba2"), CreditQuality = c(2L, 1L, 6L, 6L, 4L, 5L)), class = "data.frame", row.names = c(NA, 
-6L))

我想用我在密钥中分配给每个评级的 CreditQuality 替换这些评级。这意味着带有 F1 的单元格现在是 2。带有 WD 的单元格将是 6，依此类推。新的 table 应该如下所示：

    V1              V2           V3  V4  V5  V6  V7  V8  V9
1: XS0285400197 TR.IssuerRating   2   2   2   2  2    2  2
2: XS0041971275 TR.IssuerRating   1   1   1   1  2    2  1
3: XS0043098127 TR.IssuerRating   6   6   6   6  6    6  6

我试过使用 match 和 mapvalues 但是 match 似乎只适用于单个列而 mapvalues 只适用于原子向量而不适用于 data.table。有些人遇到过类似的问题，但是他们中的大多数只需要替换单个列中的值，而我想替换 data.table

中多个列的值

Answer 1

您可以使用 melt 和 dcast:

dcast(
  rating[melt(df, id=c("V1", "V2"),value.name = "Rating"), on="Rating"],
  V1+V2~variable, value.var = "CreditQuality"
)

输出：

             V1              V2 V3 V4 V5 V6 V7 V8 V9
1: XS0041971275 TR.IssuerRating  1  1  1  1  2  2  1
2: XS0043098127 TR.IssuerRating  6  6  6  6  6  6  6
3: XS0285400197 TR.IssuerRating  2  2  2  2  2  2  2

注意：我假设您的源数据是 df，您的评分数据是 rating。我看到你的相框已经是 class data.table

Answer 2

您可以使用 dplyr 和 across。

library(dplyr)

# Define input data
df <- data.frame(
  V1 = c("XS0285400197", "XS0041971275", "XS0043098127"),
  V2 = c("TR.IssuerRating", "TR.IssuerRating", "TR.IssuerRating"),
  V3 = c("F1", "AAA", "WD"),
  V4 = c("F1", "AAA", "WD"),
  V5 = c("F1", "AAA", "WD"),
  V6 = c("F1", "AAA", "WD"),
  V7 = c("F1", "F1", "WD"),
  V8 = c("F1", "F1", "WD"),
  V9 = c("F1", "AAA", "WD"),
  stringsAsFactors = FALSE
)

lookup <- data.frame(
  Rating = c("F1", "AAA", "WD", "(P)B2", "(P)Ba1", "(P)Ba2"),
  CreditQuality = c(2, 1, 6, 6, 4, 5)
)

# Make a look up vector
lookup_vec <- lookup$CreditQuality
names(lookup_vec) <- lookup$Rating

# Use dplyr across to apply look up
df_mod <- df %>%
  mutate(across(seq(3, dim(df)[2]), ~ lookup_vec[.x]))

# View
df_mod

#             V1              V2 V3 V4 V5 V6 V7 V8 V9
# 1 XS0285400197 TR.IssuerRating  2  2  2  2  2  2  2
# 2 XS0041971275 TR.IssuerRating  1  1  1  1  2  2  1
# 3 XS0043098127 TR.IssuerRating  6  6  6  6  6  6  6

Answer 3

df <-
  structure(
    list(
      V1 = c("XS0285400197", "XS0041971275", "XS0043098127"),
      V2 = c("TR.IssuerRating", "TR.IssuerRating", "TR.IssuerRating"),
      V3 = c("F1", "AAA", "WD"),
      V4 = c("F1", "AAA", "WD"),
      V5 = c("F1", "AAA", "WD"),
      V6 = c("F1", "AAA", "WD"),
      V7 = c("F1", "F1", "WD"),
      V8 = c("F1", "F1", "WD"),
      V9 = c("F1", "AAA", "WD")),
    class = "data.frame",
    row.names = c(NA,-3L)
  )

rating <-
  structure(list(
    Rating = c("F1", "AAA", "WD", "(P)B2", "(P)Ba1", "(P)Ba2"),
    CreditQuality = c(2L, 1L, 6L, 6L, 4L, 5L)),
  class = "data.frame",
  row.names = c(NA,-6L))

df
#>             V1              V2  V3  V4  V5  V6 V7 V8  V9
#> 1 XS0285400197 TR.IssuerRating  F1  F1  F1  F1 F1 F1  F1
#> 2 XS0041971275 TR.IssuerRating AAA AAA AAA AAA F1 F1 AAA
#> 3 XS0043098127 TR.IssuerRating  WD  WD  WD  WD WD WD  WD

#tidyverse
library(tidyverse)
df %>% 
  mutate(across(V3:V9, ~with(rating, CreditQuality[match(.x, table = Rating)])))
#>             V1              V2 V3 V4 V5 V6 V7 V8 V9
#> 1 XS0285400197 TR.IssuerRating  2  2  2  2  2  2  2
#> 2 XS0041971275 TR.IssuerRating  1  1  1  1  2  2  1
#> 3 XS0043098127 TR.IssuerRating  6  6  6  6  6  6  6

# base
df[, 3:9] <- sapply(df[ ,3:9], function(x) with(rating, CreditQuality[match(x, table = Rating)]))
df
#>             V1              V2 V3 V4 V5 V6 V7 V8 V9
#> 1 XS0285400197 TR.IssuerRating  2  2  2  2  2  2  2
#> 2 XS0041971275 TR.IssuerRating  1  1  1  1  2  2  1
#> 3 XS0043098127 TR.IssuerRating  6  6  6  6  6  6  6

^{由 reprex package (v2.0.1)}

创建于 2022-06-01

Answer 4

在基数 R 中：

lut     = with(B, setNames(CreditQuality, Rating))
vars    = paste0("V", 3:9)
A[vars] = lapply(A[vars], \(x) lut[x])

#             V1              V2 V3 V4 V5 V6 V7 V8 V9
# 1 XS0285400197 TR.IssuerRating  2  2  2  2  2  2  2
# 2 XS0041971275 TR.IssuerRating  1  1  1  1  2  2  1
# 3 XS0043098127 TR.IssuerRating  6  6  6  6  6  6  6

data.table中的相同逻辑：

setDT(A)
A[, (vars) := lapply(.SD, \(x) lut[x]), .SDcols = vars]

数据

A = structure(list(V1 = c("XS0285400197", "XS0041971275", "XS0043098127"
), V2 = c("TR.IssuerRating", "TR.IssuerRating", "TR.IssuerRating"
), V3 = c("F1", "AAA", "WD"), V4 = c("F1", "AAA", "WD"), V5 = c("F1", 
"AAA", "WD"), V6 = c("F1", "AAA", "WD"), V7 = c("F1", "F1", "WD"
), V8 = c("F1", "F1", "WD"), V9 = c("F1", "AAA", "WD")), class = "data.frame", row.names = c(NA, 
-3L))

B = structure(list(Rating = c("F1", "AAA", "WD", "(P)B2", "(P)Ba1", 
"(P)Ba2"), CreditQuality = c(2L, 1L, 6L, 6L, 4L, 5L)), class = "data.frame", row.names = c(NA, 
-6L))

使用键替换整个值 data.table

Using a key to replace values across a whole data.table

r

plyr

dataframe

data.table