Use lapply on condition in R-CodePudding

it's easier to explain what I want to do if you look at the code first but essentially I think I want to use lapply on a condition but I wasn't able to do it.

library("tidyverse")
names <- rep(c("City A", "City B"), each = 11)
year <- rep(c(2010:2020), times = 2)
col_1 <- c(1, 17, 34, 788, 3, 4, 78, 98, 650, 45, 20,
           23, 45, 56, 877, 54, 12, 109, 167, 12, 19, 908)
col_2 <- c(3, 4, 23, 433, 2, 45, 34, 123, 98, 76, 342,
           760, 123, 145, 892, 23, 5, 90, 40, 12, 67, 98)

df <- as.data.frame(cbind(names, year, col_1, col_2))

df <- df %>%
  mutate(col_1 = as.numeric(col_1),
         col_2 = as.numeric(col_2))

I want every numeric column in the year 2018 and later to be rounded with round_any to a value which is a multiple of three (plyr::round_any, 3) What I tried is this:

df_2018 <- df %>%
  filter(year >= 2018)

df <- df %>%
  filter(!(year >= 2018))
  
df_2018[, c(3:4)] <- lapply(df_2018[, c(3:4)], plyr::round_any, 3)
  
df <- rbind(df, df_2018)

In reality, there's about 50 numeric columns and tons of rows. What I tried works in theory but I would like to achieve it with less code and cleaner. I am new to using lapply and I failed trying to combine it with an ifelse because I don't want it to change my year column.

Thank you for everyone who takes the time out of their day to look at this :)

CodePudding user response：

Using dplyr::across and if_else you could do:

library(dplyr)

df |> 
  mutate(across(-c(names, year), ~ if_else(year >= 2018, plyr::round_any(.x, 3), .x)))
#>     names year col_1 col_2
#> 1  City A 2010     1     3
#> 2  City A 2011    17     4
#> 3  City A 2012    34    23
#> 4  City A 2013   788   433
#> 5  City A 2014     3     2
#> 6  City A 2015     4    45
#> 7  City A 2016    78    34
#> 8  City A 2017    98   123
#> 9  City A 2018   651    99
#> 10 City A 2019    45    75
#> 11 City A 2020    21   342
#> 12 City B 2010    23   760
#> 13 City B 2011    45   123
#> 14 City B 2012    56   145
#> 15 City B 2013   877   892
#> 16 City B 2014    54    23
#> 17 City B 2015    12     5
#> 18 City B 2016   109    90
#> 19 City B 2017   167    40
#> 20 City B 2018    12    12
#> 21 City B 2019    18    66
#> 22 City B 2020   909    99

CodePudding user response：

Using data.table:

cols <- grep("^col_[0-9] $", names(df), value = TRUE)
setDT(df)[year >= 2018, (cols) := round(.SD / 3) * 3, .SDcols = cols]