How to subtract all columns from correponding mean?-CodePudding

I have this data frame:

        dat=structure(list(date = as.Date(c("1983-01-01", "1984-01-01", 
                        "1984-02-01", 
                        "1985-01-01", "1985-02-01", "1986-01-01")), 
               rig = c(68.1, 62.4, 
                       
                       67.5, 78.9, 81.7, 72.2), pass = c(9.57, 10.49, 11.97, 
                                                         11.43, 9.54, 
                                                         
                                                         8.98)), 
          row.names = c(NA, 6L), class = "data.frame")

I would like that for each month (i) in year, compute the mean (m) data value for that month across all years and the standard deviation (std) of the data value for that month across all years and then replace values with (i-m)/std

we can do this to compute the mean and sd

     > dat%>%     mutate(month =month(as.Date(date))) %>% group_by(month) 
    %>%summarise(across(where(is.numeric), mean, na.rm = TRUE))%>% ungroup
     # A tibble: 2 × 3
      month   rig  pass
      <dbl> <dbl> <dbl>
  1     1  70.4  10.1
   2     2  74.6  10.8
   > dat%>%     mutate(month =month(as.Date(date))) %>% group_by(month) 
   %>%summarise(across(where(is.numeric), sd, na.rm = TRUE))%>% ungroup
   # A tibble: 2 × 3
    month   rig  pass
    <dbl> <dbl> <dbl>
   1     1  6.95  1.07
   2     2 10.0   1.72

How can i subtract and divide these values to each corresponding column:expected results, example of pass:

 (9.57-10.1175)/1.073169  =   -0.5101713

         date       rig     pass
  1 1983-01-01 -0.3310809  -0.5101713
  2 1984-01-01 -1.151586 0.3471028
  3 1984-02-01 -0.7071068 0.707107
  4 1985-01-01 1.22356 1.223013
  5 1985-02-01  0.7071068  -0.707107
  6 1986-01-01 0.2591068  -1.059945

CodePudding user response：

Could you just use scale():

dat=structure(list(date = as.Date(c("1983-01-01", "1984-01-01", 
                                    "1984-02-01", 
                                    "1985-01-01", "1985-02-01", "1986-01-01")), 
                   rig = c(68.1, 62.4, 
                           
                           67.5, 78.9, 81.7, 72.2), pass = c(9.57, 10.49, 11.97, 
                                                             11.43, 9.54, 
                                                             
                                                             8.98)), 
              row.names = c(NA, 6L), class = "data.frame")

library(dplyr)
library(lubridate)
dat%>%     
  mutate(month =month(as.Date(date))) %>% 
  group_by(month) %>% 
  mutate(across(where(is.numeric), ~c(scale(.x))))
#> # A tibble: 6 × 4
#> # Groups:   month [2]
#>   date          rig   pass month
#>   <date>      <dbl>  <dbl> <dbl>
#> 1 1983-01-01 -0.331 -0.510     1
#> 2 1984-01-01 -1.15   0.347     1
#> 3 1984-02-01 -0.707  0.707     2
#> 4 1985-01-01  1.22   1.22      1
#> 5 1985-02-01  0.707 -0.707     2
#> 6 1986-01-01  0.259 -1.06      1

Or, if you wanted to be more explicit about the transformation:

dat%>%     
  mutate(month =month(as.Date(date))) %>% 
  group_by(month) %>% 
  mutate(across(where(is.numeric), ~(.x - mean(.x))/sd(.x)))
#> # A tibble: 6 × 4
#> # Groups:   month [2]
#>   date          rig   pass month
#>   <date>      <dbl>  <dbl> <dbl>
#> 1 1983-01-01 -0.331 -0.510     1
#> 2 1984-01-01 -1.15   0.347     1
#> 3 1984-02-01 -0.707  0.707     2
#> 4 1985-01-01  1.22   1.22      1
#> 5 1985-02-01  0.707 -0.707     2
#> 6 1986-01-01  0.259 -1.06      1

^{Created on 2023-01-17 by the reprex package (v2.0.1)}