Replace numeric values of one vector with another value in a data frame

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1

i would like to replace (replace) the values of a given index (vector) which find corresponding in a dataframe (tbl) by another value determined by me.

I tried the following

# Criando um index aleatório
index_1 <- sample(mtcars$gear, 5)
index_2 <- sample(mtcars$hp, 5)
index_3 <- sample(mtcars$disp, 5)

# Criando um index aleatório - Todos esses valores devem ser substituído por um determinado número (escolhi 48 também aletóriamente)
index <- c(index_1, index_2, index_3)

# Tentei criar essa função
fun_return <- function(x){
  if(x==index){
    return(48)
  } else
    return(x)
}

# Os valores não são substituídos.
mt_replaced <- fun_return(mtcars)

Does anyone have any tips?

2 answers

3


The following function does what the question describes.

fun_replace <- function(x, vetor, novo = 48){
  res <- lapply(x, function(y){
    i <- y %in% vetor
    y[i] <- novo
    y
  })
  res <- do.call(cbind, res)
  if(!is.null(rownames(x))) rownames(res) <- rownames(x)
  as.data.frame(res)
}

mt_replaced <- fun_replace(mtcars, index)
  • thank you very much, Rui! I didn’t think it would be this complicated, I wouldn’t have come to that result! Super thanks, worked perfectly on the original data!

3

With dplyr you can do something like this:

library(tidyverse)
#> Warning: package 'tibble' was built under R version 3.5.2

index_1 <- sample(mtcars$gear, 5)
index_2 <- sample(mtcars$hp, 5)
index_3 <- sample(mtcars$disp, 5)

index <- c(index_1, index_2, index_3)

mtcars %>% 
  mutate_all(~ifelse(.x %in% index, 48, .x))
#>     mpg cyl  disp  hp  drat    wt  qsec vs am gear carb
#> 1  21.0   6 160.0 110  3.90 2.620 16.46  0  1   48   48
#> 2  21.0   6 160.0 110  3.90 2.875 17.02  0  1   48   48
#> 3  22.8  48 108.0  93  3.85 2.320 18.61  1  1   48    1
#> 4  21.4   6 258.0 110  3.08 3.215 19.44  1  0   48    1
#> 5  18.7   8 360.0 175  3.15 3.440 17.02  0  0   48    2
#> 6  18.1   6 225.0  48  2.76 3.460 20.22  1  0   48    1
#> 7  14.3   8 360.0 245  3.21 3.570 15.84  0  0   48   48
#> 8  24.4  48 146.7  62  3.69 3.190 20.00  1  0   48    2
#> 9  22.8  48 140.8  95  3.92 3.150 22.90  1  0   48    2
#> 10 19.2   6 167.6  48  3.92 3.440 18.30  1  0   48   48
#> 11 17.8   6 167.6  48  3.92 3.440 18.90  1  0   48   48
#> 12 16.4   8  48.0  48  3.07 4.070 17.40  0  0   48   48
#> 13 17.3   8  48.0  48  3.07 3.730 17.60  0  0   48   48
#> 14 15.2   8  48.0  48  3.07 3.780 18.00  0  0   48   48
#> 15 10.4   8 472.0 205  2.93 5.250 17.98  0  0   48   48
#> 16 10.4   8 460.0 215 48.00 5.424 17.82  0  0   48   48
#> 17 14.7   8 440.0 230  3.23 5.345 17.42  0  0   48   48
#> 18 32.4  48  78.7  66  4.08 2.200 19.47  1  1   48    1
#> 19 30.4  48  75.7  52  4.93 1.615 18.52  1  1   48    2
#> 20 33.9  48  71.1  48  4.22 1.835 19.90  1  1   48    1
#> 21 21.5  48 120.1  97  3.70 2.465 20.01  1  0   48    1
#> 22 15.5   8  48.0 150  2.76 3.520 16.87  0  0   48    2
#> 23 15.2   8 304.0 150  3.15 3.435 17.30  0  0   48    2
#> 24 13.3   8  48.0 245  3.73 3.840 15.41  0  0   48   48
#> 25 19.2   8  48.0 175  3.08 3.845 17.05  0  0   48    2
#> 26 27.3  48  79.0  66  4.08 1.935 18.90  1  1   48    1
#> 27 26.0  48 120.3  91  4.43 2.140 16.70  0  1    5    2
#> 28 30.4  48  95.1  48  3.77 1.513 16.90  1  1    5    2
#> 29 15.8   8 351.0 264  4.22 3.170 14.50  0  1    5   48
#> 30 19.7   6  48.0 175  3.62 2.770 15.50  0  1    5    6
#> 31 15.0   8 301.0 335  3.54 3.570 14.60  0  1    5    8
#> 32 21.4  48 121.0 109  4.11 2.780 18.60  1  1   48    2

Created on 2019-02-25 by the reprex package (v0.2.1)

  • Thank you, Daniel. Simple and efficient solution! I tried 100 times with dplyr and it didn’t roll! Hugs!

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