r语言 - 使用键来替换整个数据表中的值

Question

我有一个大的数据表，看起来如下:

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))

实际的数据表要大得多，但这应该作为一个示例。此外，我有一个键，我想用数字替换评级(这里是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))

我想用我在键中分配的每个评级的CreditQuality替换这些评级。这意味着有F1的单元格现在是2。有WD的单元格是6，以此类推。新表应该如下所示:

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似乎只适用于单列，mapvalues只适用于原子向量，而不适用于数据表。有些人遇到过类似的问题，但大多数人只需要替换单列中的值，而我想替换data.table中多个列的值

Answer 1

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 2

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:我假设你的源数据是df，你的评级数据是rating。我看到你的框架已经是data.table类

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包(v2.0.1)创建于2022-06-01}

Answer 4

In base R:

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

相同的逻辑在data.table:

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

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))