Проверьте, являются ли все значения числовыми в нескольких столбцах, и преобразуйте их в числовые.

Question

У меня есть датафрейм, в котором все столбцы являются символами.

ID <- c("A","A","A","A","A","A","A","A","B","B","B","B","B","B","B","B")
ToolID <- c("CCP_A","CCP_A","CCQ_A","CCQ_A","IOT_B","CCP_B","CCQ_B","IOT_B",
            "CCP_A","CCP_A","CCQ_A","CCQ_A","IOT_B","CCP_B","CCQ_B","IOT_B")
Step <- c("Step_A","Step_A","Step_B","Step_C","Step_D","Step_D","Step_E","Step_F",
          "Step_A","Step_A","Step_B","Step_C","Step_D","Step_D","Step_E","Step_F")
Measurement <- c("Length","Breadth","Width","Height",NA,NA,NA,NA,
                 "Length","Breadth","Width","Height",NA,NA,NA,NA)
Passfail <- c("Pass","Pass","Fail","Fail","Pass","Pass","Pass","Pass",
              "Pass","Pass","Fail","Fail","Pass","Pass","Pass","Pass")
Points <- as.character(c(7,5,3,4,0,0,0,0,17,15,13,14,0,0,0,0))
Average <- as.character(c(7.5,6.5,7.1,6.6,NA,NA,NA,NA,17.5,16.5,17.1,16.6,NA,NA,NA,NA))
Sigma <- as.character(c(2.5,2.5,2.1,2.6,NA,NA,NA,NA,12.5,12.5,12.1,12.6,NA,NA,NA,NA))
Tool <- c("ABC_1","ABC_2","ABD_1","ABD_2","COB_1","COB_2","COB_1","COB_2",
          "ABC_1","ABC_2","ABD_1","ABD_2","COB_1","COB_2","COB_1","COB_2")
Dose <- as.character(c(NA,NA,NA,NA,17.1,NA,NA,17.3,NA,NA,NA,NA,117.1,NA,NA,117.3))
Machine <- c("CO2","CO6","CO3","CO6","CO2,CO6","CO2,CO3,CO4","CO2,CO3","CO2",
             "CO2","CO6","CO3","CO6","CO2,CO6","CO2,CO3,CO4","CO2,CO3","CO2")

df <- data.frame(ID,ToolID,Step,Measurement,Passfail,Points,Average,Sigma,Tool,Dose,Machine)

Я пытаюсь проверить эти символьные векторы на числовые значения, а затем преобразовать их с числовыми значениями в числовые.Я использую пакет "varhandle" в R, чтобы сделать это

library(varhandle)

if(all(check.numeric(df$Machine, na.rm=TRUE))){
  # convert the vector to numeric
  df$Machine <- as.numeric(df$Machine)
}

Это работает, но неэффективно, потому что мне нужно вручную вводить имена столбцов, как указано выше.Как я могу сделать это более эффективно в цикле или использовать векторизацию для нескольких столбцов?Мой фактический набор данных имеет около 350 столбцов.Может ли кто-нибудь указать мне правильное направление?

Answer 1

Другое решение - это перепечатка из пакета hablar:

library(hablar)

df %>% retype()

, что дает:

# A tibble: 16 x 11
   ID    ToolID Step   Measurement Passfail Points Average Sigma Tool   Dose Machine    
   <chr> <chr>  <chr>  <chr>       <chr>     <int>   <dbl> <dbl> <chr> <dbl> <chr>      
 1 A     CCP_A  Step_A Length      Pass          7    7.50  2.50 ABC_1  NA   CO2        
 2 A     CCP_A  Step_A Breadth     Pass          5    6.50  2.50 ABC_2  NA   CO6        
 3 A     CCQ_A  Step_B Width       Fail          3    7.10  2.10 ABD_1  NA   CO3        
 4 A     CCQ_A  Step_C Height      Fail          4    6.60  2.60 ABD_2  NA   CO6        
 5 A     IOT_B  Step_D NA          Pass          0   NA    NA    COB_1  17.1 CO2,CO6    
 6 A     CCP_B  Step_D NA          Pass          0   NA    NA    COB_2  NA   CO2,CO3,CO4
 7 A     CCQ_B  Step_E NA          Pass          0   NA    NA    COB_1  NA   CO2,CO3    

Answer 2

Мы можем сделать это за base R

df[] <- lapply(df, function(x) type.convert(as.character(x), as.is = TRUE))
str(df)
#'data.frame':  16 obs. of  11 variables:
# $ ID         : chr  "A" "A" "A" "A" ...
# $ ToolID     : chr  "CCP_A" "CCP_A" "CCQ_A" "CCQ_A" ...
# $ Step       : chr  "Step_A" "Step_A" "Step_B" "Step_C" ...
# $ Measurement: chr  "Length" "Breadth" "Width" "Height" ...
# $ Passfail   : chr  "Pass" "Pass" "Fail" "Fail" ...
# $ Points     : int  7 5 3 4 0 0 0 0 17 15 ...
# $ Average    : num  7.5 6.5 7.1 6.6 NA NA NA NA 17.5 16.5 ...
# $ Sigma      : num  2.5 2.5 2.1 2.6 NA NA NA NA 12.5 12.5 ...
# $ Tool       : chr  "ABC_1" "ABC_2" "ABD_1" "ABD_2" ...
# $ Dose       : num  NA NA NA NA 17.1 NA NA 17.3 NA NA ...
# $ Machine    : chr  "CO2" "CO6" "CO3" "CO6" ...

Answer 3

С варочной ручкой и прицелом:

df %>% mutate_if(purrr::compose(all,check.numeric),as.numeric)

Answer 4

Я думаю, что самое простое решение - использовать all.is.numeric из Hmisc.Вот простой пример:

Hmisc::all.is.numeric(c("A", "B", "1"), what = "vector", extras = NA)
## [1] "A" "B" "1"
Hmisc::all.is.numeric(c("3", "2", "1", NA), what = "vector", extras = NA)
## [1]  3  2  1 NA

Затем вы можете использовать mutate_all из dplyr, чтобы выполнить всю работу для data.frame:

library(dplyr)
ID <- c("A","A","A","A","A","A","A","A","B","B","B","B","B","B","B","B")
ToolID <- c("CCP_A","CCP_A","CCQ_A","CCQ_A","IOT_B","CCP_B","CCQ_B","IOT_B",
            "CCP_A","CCP_A","CCQ_A","CCQ_A","IOT_B","CCP_B","CCQ_B","IOT_B")
Step <- c("Step_A","Step_A","Step_B","Step_C","Step_D","Step_D","Step_E","Step_F",
          "Step_A","Step_A","Step_B","Step_C","Step_D","Step_D","Step_E","Step_F")
Measurement <- c("Length","Breadth","Width","Height",NA,NA,NA,NA,
                 "Length","Breadth","Width","Height",NA,NA,NA,NA)
Passfail <- c("Pass","Pass","Fail","Fail","Pass","Pass","Pass","Pass",
              "Pass","Pass","Fail","Fail","Pass","Pass","Pass","Pass")
Points <- as.character(c(7,5,3,4,0,0,0,0,17,15,13,14,0,0,0,0))
Average <- as.character(c(7.5,6.5,7.1,6.6,NA,NA,NA,NA,17.5,16.5,17.1,16.6,NA,NA,NA,NA))
Sigma <- as.character(c(2.5,2.5,2.1,2.6,NA,NA,NA,NA,12.5,12.5,12.1,12.6,NA,NA,NA,NA))
Tool <- c("ABC_1","ABC_2","ABD_1","ABD_2","COB_1","COB_2","COB_1","COB_2",
          "ABC_1","ABC_2","ABD_1","ABD_2","COB_1","COB_2","COB_1","COB_2")
Dose <- as.character(c(NA,NA,NA,NA,17.1,NA,NA,17.3,NA,NA,NA,NA,117.1,NA,NA,117.3))
Machine <- c("CO2","CO6","CO3","CO6","CO2,CO6","CO2,CO3,CO4","CO2,CO3","CO2",
             "CO2","CO6","CO3","CO6","CO2,CO6","CO2,CO3,CO4","CO2,CO3","CO2")

df <- data.frame(ID,ToolID,Step,Measurement,Passfail,Points,Average,Sigma,Tool,Dose,Machine)

dt2 <- df %>% mutate_all(function(x) Hmisc::all.is.numeric(x, what = "vector", extras = NA))

## check classes
sapply(dt2, class)

##         ID      ToolID        Step Measurement    Passfail      Points 
## "character" "character" "character" "character" "character"   "numeric" 
##    Average       Sigma        Tool        Dose     Machine 
##  "numeric"   "numeric" "character"   "numeric" "character" 

Answer 5

Мы можем использовать функцию parse_guess из пакета readr, которая в основном пытается угадать тип столбцов.

library(readr)
library(dplyr)

df1 <- df %>% mutate_all(parse_guess)


str(df1)
#'data.frame':  16 obs. of  11 variables:
# $ ID         : chr  "A" "A" "A" "A" ...
# $ ToolID     : chr  "CCP_A" "CCP_A" "CCQ_A" "CCQ_A" ...
# $ Step       : chr  "Step_A" "Step_A" "Step_B" "Step_C" ...
# $ Measurement: chr  "Length" "Breadth" "Width" "Height" ...
# $ Passfail   : chr  "Pass" "Pass" "Fail" "Fail" ...
# $ Points     : int  7 5 3 4 0 0 0 0 17 15 ...
# $ Average    : num  7.5 6.5 7.1 6.6 NA NA NA NA 17.5 16.5 ...
# $ Sigma      : num  2.5 2.5 2.1 2.6 NA NA NA NA 12.5 12.5 ...
# $ Tool       : chr  "ABC_1" "ABC_2" "ABD_1" "ABD_2" ...
# $ Dose       : num  NA NA NA NA 17.1 NA NA 17.3 NA NA ...
# $ Machine    : chr  "CO2" "CO6" "CO3" "CO6" ...