R 使用自定义函数按列处理数据的最佳方式

Question

library(data.table)

df <- structure(list(
continuousNumericOne = c(3.82495116149284, 0.915662542284416, 0.751001771620762, NA, NA, 8.07583989184169, 4.57303752008246, 4.02747047825306, 2.79953011697721, 4.28614794390785), 
catagoricalFactorOne = structure(c(3L, 3L, 3L, NA, 3L, NA, 2L, 2L, 2L, NA), .Label = c("blue", "green", "red"), class = "factor"), 
continuousNumericTwo = c(NA, NA, 2.58285715825289, -2.71316582700148, 3.95645652249594, 1.96862094118233, 4.96960533647993, 6.15199683070215, 3.98091405116921, NA), 
catagoricalFactorTwo = structure(c(3L, 3L, 3L, NA, 3L, 3L, 2L, 2L, 2L, 1L), .Label = c("blue", "orange", "red"), class = "factor"), 
continuousNumericThree = c(3.43332616062442, 2.21448227693603, 2.31889349781533, NA, NA, 3.57539465909581, 3.28076535012702, NA, 3.15063300766727, 2.9556632429251), 
continuousNumericFour = c(7.77131807052585, NA, 6.5830522592014, NA, 7.36003333388333, 8.25217350122047, 7.18282902739316, 8.60641407074177, 4.87689328481095, NA)), 
.Names = c("continuousNumericOne", "catagoricalFactorOne", "continuousFactorTwo", "catagoricalFactorTwo", "continuousNumericThree", "continuousNumericFour"), 
row.names = c(NA, -10L), 
class = c("data.table", "data.frame"))

> df
    continuousNumericOne catagoricalFactorOne continuousFactorTwo catagoricalFactorTwo continuousNumericThree continuousNumericFour
 1:            3.8249512                  red                  NA                  red               3.433326              7.771318
 2:            0.9156625                  red                  NA                  red               2.214482                    NA
 3:            0.7510018                  red            2.582857                  red               2.318893              6.583052
 4:                   NA                   NA           -2.713166                   NA                     NA                    NA
 5:                   NA                  red            3.956457                  red                     NA              7.360033
 6:            8.0758399                   NA            1.968621                  red               3.575395              8.252174
 7:            4.5730375                green            4.969605               orange               3.280765              7.182829
 8:            4.0274705                green            6.151997               orange                     NA              8.606414
 9:            2.7995301                green            3.980914               orange               3.150633              4.876893
10:            4.2861479                   NA                  NA                 blue               2.955663                    NA

如何制作自定义函数来处理数据如下...

1. 如果列是分类（因子），请将所有 NA 替换为“空白”

2. 如果该列是连续的（数字），则可以更加灵活地进一步处理数据，例如，首先将数据从 0 缩放到 1，然后根据需要替换 NA，可能使用 -1.1。

我花了很多时间制作清单， 尝试跟踪列名以及给定的列名是否为因子， 尝试通过 apply 方法应用不同的功能，仍然没有运气。

如果有更好的方法，我会全力以赴。

Answer 1

我们可以创建一个函数

f1 <- function(dat){
 iCat <- which(sapply(dat, is.factor))
 iNum <- which(sapply(dat, is.numeric))
 dat[, (iCat) := lapply(.SD, function(x) {
              levels(x) <- c(levels(x), "")
              x[is.na(x)] <- ""
              x}), .SDcols = iCat]  
  dat[, (iNum) := lapply(.SD, function(x) {
                 x1 <- as.vector(scale(x))
                 x1[is.na(x1)] <- -1.1
                 x1}), .SDcols = iNum][]
 }

f1(df)
#continuousNumericOne catagoricalFactorOne continuousFactorTwo
# 1:           0.07257304                  red          -1.1000000
# 2:          -1.18235090                  red          -1.1000000
# 3:          -1.25337745                  red          -0.1400258
# 4:          -1.10000000                               -1.9826003
# 5:          -1.10000000                  red           0.3378723
# 6:           1.90619723                               -0.3537288
# 7:           0.39526068                green           0.6903636
# 8:           0.15992990                green           1.1017373
# 9:          -0.36974314                green           0.3463815
#10:           0.27151063                               -1.1000000
#    catagoricalFactorTwo continuousNumericThree continuousNumericFour
# 1:                  red             0.83246346            0.43436598
# 2:                  red            -1.45562130           -1.10000000
# 3:                  red            -1.25961447           -0.52487557
# 4:                                 -1.10000000           -1.10000000
# 5:                  red            -1.10000000            0.10235154
# 6:                  red             1.09916272            0.82254218
# 7:               orange             0.54606741           -0.04069872
# 8:               orange            -1.10000000            1.10850704
# 9:               orange             0.30177540           -1.90219245
#10:                 blue            -0.06423321           -1.10000000

Answer 2

首先，您的示例数据的数字列名称为 "Factor"。所以我提供了我自己的示例数据。

library(data.table)

set.seed(1)
df <- data.table(
  num1 = runif(10),
  fac1 = factor(sample(letters, 10)),
  num2 = runif(10),
  fac2 = factor(sample(letters, 10)),
  char = sample(letters, 10)
)

至于任务，带有通用函数的方法分派符合要求。

process.factor <- function(x) {
  # Replace with actual logic
  rep_len("f", length(x))
}

process.numeric <- function(x) {
  # Replace with actual logic
  rep_len("n", length(x))
}

process.default <- function(x) {
  # Replace with actual logic for "other" classes not specifically handled
  rep_len("d", length(x))
}

process <- function(x) {
  UseMethod("process")
}

然后我们只是lapplydata.table 内的列，这导致另一个data.table。

df[, lapply(.SD, process)]
#     num1 fac1 num2 fac2 char
#  1:    n    f    n    f    d
#  2:    n    f    n    f    d
#  3:    n    f    n    f    d
#  4:    n    f    n    f    d
#  5:    n    f    n    f    d
#  6:    n    f    n    f    d
#  7:    n    f    n    f    d
#  8:    n    f    n    f    d
#  9:    n    f    n    f    d
# 10:    n    f    n    f    d