Using lapply to create new variables based on multiple conditions and subsets

I am trying to create a list of new variables that represent the deviations from the minimum values of variables based on subsets of another variable.

Consider the following:

df <- data.frame(

      cluster = c("A","B","B","A","A","B"),

      x = c(3,4,1,5,2,6),

      y = c(4,5,3,1,2,6))

I would like to create two new variables, call them x.var and y.var, that take on the deviation from the minimum value of the respective underlying variables contingent on cluster. Thus, x.var and y.var would hopefully be:

x.var y.var

-1    -3

-3    -2

 0     0

-3     0

 0    -1

-5    -3

I have tried unsuccessfully to use lapply with an anonymous function to accomplish this:

vars <- lapply(df[,c(2:3)],function(x) 

    ifelse(df$cluster=="A",

    min(df[df$cluster=="A",x])-x,

    min(df[df$cluster=="B",x])-x))

I receive the following error:

Error in `[.data.frame`(df, df$cluster == "A", x) : undefined columns selected

Any help would be much appreciated!

edited Jan 2 at 3:59

asked Jan 2 at 0:16

Jeff

135

I think the 4th value of x.var should be -3 (2-5) ?

– thelatemail
Jan 2 at 0:26

You are right. Thanks for catching that. I adjusted the value.

– Jeff
Jan 2 at 4:00

add a comment |

I am trying to create a list of new variables that represent the deviations from the minimum values of variables based on subsets of another variable.

Consider the following:

df <- data.frame(

      cluster = c("A","B","B","A","A","B"),

      x = c(3,4,1,5,2,6),

      y = c(4,5,3,1,2,6))

x.var y.var

-1    -3

-3    -2

 0     0

-3     0

 0    -1

-5    -3

I have tried unsuccessfully to use lapply with an anonymous function to accomplish this:

vars <- lapply(df[,c(2:3)],function(x) 

    ifelse(df$cluster=="A",

    min(df[df$cluster=="A",x])-x,

    min(df[df$cluster=="B",x])-x))

I receive the following error:

Error in `[.data.frame`(df, df$cluster == "A", x) : undefined columns selected

Any help would be much appreciated!

edited Jan 2 at 3:59

asked Jan 2 at 0:16

Jeff

135

I think the 4th value of x.var should be -3 (2-5) ?

– thelatemail
Jan 2 at 0:26

You are right. Thanks for catching that. I adjusted the value.

– Jeff
Jan 2 at 4:00

add a comment |

I am trying to create a list of new variables that represent the deviations from the minimum values of variables based on subsets of another variable.

Consider the following:

df <- data.frame(

      cluster = c("A","B","B","A","A","B"),

      x = c(3,4,1,5,2,6),

      y = c(4,5,3,1,2,6))

x.var y.var

-1    -3

-3    -2

 0     0

-3     0

 0    -1

-5    -3

I have tried unsuccessfully to use lapply with an anonymous function to accomplish this:

vars <- lapply(df[,c(2:3)],function(x) 

    ifelse(df$cluster=="A",

    min(df[df$cluster=="A",x])-x,

    min(df[df$cluster=="B",x])-x))

I receive the following error:

Error in `[.data.frame`(df, df$cluster == "A", x) : undefined columns selected

Any help would be much appreciated!

edited Jan 2 at 3:59

asked Jan 2 at 0:16

Jeff

135

I am trying to create a list of new variables that represent the deviations from the minimum values of variables based on subsets of another variable.

Consider the following:

df <- data.frame(

      cluster = c("A","B","B","A","A","B"),

      x = c(3,4,1,5,2,6),

      y = c(4,5,3,1,2,6))

x.var y.var

-1    -3

-3    -2

 0     0

-3     0

 0    -1

-5    -3

I have tried unsuccessfully to use lapply with an anonymous function to accomplish this:

vars <- lapply(df[,c(2:3)],function(x) 

    ifelse(df$cluster=="A",

    min(df[df$cluster=="A",x])-x,

    min(df[df$cluster=="B",x])-x))

I receive the following error:

Error in `[.data.frame`(df, df$cluster == "A", x) : undefined columns selected

Any help would be much appreciated!

r list apply lapply

edited Jan 2 at 3:59

asked Jan 2 at 0:16

Jeff

135

edited Jan 2 at 3:59

asked Jan 2 at 0:16

Jeff

135

edited Jan 2 at 3:59

asked Jan 2 at 0:16

Jeff

135

asked Jan 2 at 0:16

Jeff

135

asked Jan 2 at 0:16

Jeff

135

I think the 4th value of x.var should be -3 (2-5) ?

– thelatemail
Jan 2 at 0:26

You are right. Thanks for catching that. I adjusted the value.

– Jeff
Jan 2 at 4:00

add a comment |

I think the 4th value of x.var should be -3 (2-5) ?

– thelatemail
Jan 2 at 0:26

You are right. Thanks for catching that. I adjusted the value.

– Jeff
Jan 2 at 4:00

I think the 4th value of x.var should be -3 (2-5) ?

– thelatemail
Jan 2 at 0:26

You are right. Thanks for catching that. I adjusted the value.

– Jeff
Jan 2 at 4:00

add a comment |

2 Answers
2

active

oldest

votes

Here is a base R method that uses ave with lapply. Loop through the columns of dataset excluding the 'cluster', then with ave get the min grouped by 'cluster', subtract from the column and assign the list of vectors to new columns

df[paste0(names(df)[-1], ".var")] <- lapply(df[-1], function(x)

                                         ave(x, df$cluster, FUN = min) - x)

df

#  cluster x y x.var y.var

#1       A 3 4    -1    -3

#2       B 4 5    -3    -2

#3       B 1 3     0     0

#4       A 5 1    -3     0

#5       A 2 2     0    -1

#6       B 6 6    -5    -3

answered Jan 2 at 6:12

akrun

415k13204278

1

Thank you very much for the recommended approach. This is extremely helpful.

– Jeff
Jan 2 at 11:37

add a comment |

Here's an approach using dplyr.

library(dplyr)



df <- data.frame(

  cluster = c("A","B","B","A","A","B"),

  x = c(3,4,1,5,2,6),

  y = c(4,5,3,1,2,6))



df %>% 

  group_by(cluster) %>% 

  mutate(x.var = min(x) - x,

         y.var = min(y) - y)



#> # A tibble: 6 x 5

#> # Groups:   cluster [2]

#>   cluster     x     y x.var y.var

#>   <fct>   <dbl> <dbl> <dbl> <dbl>

#> 1 A           3     4    -1    -3

#> 2 B           4     5    -3    -2

#> 3 B           1     3     0     0

#> 4 A           5     1    -3     0

#> 5 A           2     2     0    -1

#> 6 B           6     6    -5    -3

^{Created on 2019-01-01 by the reprex package (v0.2.1)}

answered Jan 2 at 0:39

Jake Kaupp

5,70721428

Hi Jake. Thanks a ton for the recommendation. This approach completely solves my inquiry.

– Jeff
Jan 2 at 4:02

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

df[paste0(names(df)[-1], ".var")] <- lapply(df[-1], function(x)

                                         ave(x, df$cluster, FUN = min) - x)

df

#  cluster x y x.var y.var

#1       A 3 4    -1    -3

#2       B 4 5    -3    -2

#3       B 1 3     0     0

#4       A 5 1    -3     0

#5       A 2 2     0    -1

#6       B 6 6    -5    -3

answered Jan 2 at 6:12

akrun

415k13204278

1

Thank you very much for the recommended approach. This is extremely helpful.

– Jeff
Jan 2 at 11:37

add a comment |

df[paste0(names(df)[-1], ".var")] <- lapply(df[-1], function(x)

                                         ave(x, df$cluster, FUN = min) - x)

df

#  cluster x y x.var y.var

#1       A 3 4    -1    -3

#2       B 4 5    -3    -2

#3       B 1 3     0     0

#4       A 5 1    -3     0

#5       A 2 2     0    -1

#6       B 6 6    -5    -3

answered Jan 2 at 6:12

akrun

415k13204278

1

Thank you very much for the recommended approach. This is extremely helpful.

– Jeff
Jan 2 at 11:37

add a comment |

df[paste0(names(df)[-1], ".var")] <- lapply(df[-1], function(x)

                                         ave(x, df$cluster, FUN = min) - x)

df

#  cluster x y x.var y.var

#1       A 3 4    -1    -3

#2       B 4 5    -3    -2

#3       B 1 3     0     0

#4       A 5 1    -3     0

#5       A 2 2     0    -1

#6       B 6 6    -5    -3

answered Jan 2 at 6:12

akrun

415k13204278

df[paste0(names(df)[-1], ".var")] <- lapply(df[-1], function(x)

                                         ave(x, df$cluster, FUN = min) - x)

df

#  cluster x y x.var y.var

#1       A 3 4    -1    -3

#2       B 4 5    -3    -2

#3       B 1 3     0     0

#4       A 5 1    -3     0

#5       A 2 2     0    -1

#6       B 6 6    -5    -3

answered Jan 2 at 6:12

akrun

415k13204278

answered Jan 2 at 6:12

akrun

415k13204278

answered Jan 2 at 6:12

akrun

415k13204278

answered Jan 2 at 6:12

akrun

415k13204278

1

Thank you very much for the recommended approach. This is extremely helpful.

– Jeff
Jan 2 at 11:37

add a comment |

1

Thank you very much for the recommended approach. This is extremely helpful.

– Jeff
Jan 2 at 11:37

Thank you very much for the recommended approach. This is extremely helpful.

– Jeff
Jan 2 at 11:37

add a comment |

Here's an approach using dplyr.

library(dplyr)



df <- data.frame(

  cluster = c("A","B","B","A","A","B"),

  x = c(3,4,1,5,2,6),

  y = c(4,5,3,1,2,6))



df %>% 

  group_by(cluster) %>% 

  mutate(x.var = min(x) - x,

         y.var = min(y) - y)



#> # A tibble: 6 x 5

#> # Groups:   cluster [2]

#>   cluster     x     y x.var y.var

#>   <fct>   <dbl> <dbl> <dbl> <dbl>

#> 1 A           3     4    -1    -3

#> 2 B           4     5    -3    -2

#> 3 B           1     3     0     0

#> 4 A           5     1    -3     0

#> 5 A           2     2     0    -1

#> 6 B           6     6    -5    -3

^{Created on 2019-01-01 by the reprex package (v0.2.1)}

answered Jan 2 at 0:39

Jake Kaupp

5,70721428

Hi Jake. Thanks a ton for the recommendation. This approach completely solves my inquiry.

– Jeff
Jan 2 at 4:02

add a comment |

Here's an approach using dplyr.

library(dplyr)



df <- data.frame(

  cluster = c("A","B","B","A","A","B"),

  x = c(3,4,1,5,2,6),

  y = c(4,5,3,1,2,6))



df %>% 

  group_by(cluster) %>% 

  mutate(x.var = min(x) - x,

         y.var = min(y) - y)



#> # A tibble: 6 x 5

#> # Groups:   cluster [2]

#>   cluster     x     y x.var y.var

#>   <fct>   <dbl> <dbl> <dbl> <dbl>

#> 1 A           3     4    -1    -3

#> 2 B           4     5    -3    -2

#> 3 B           1     3     0     0

#> 4 A           5     1    -3     0

#> 5 A           2     2     0    -1

#> 6 B           6     6    -5    -3

^{Created on 2019-01-01 by the reprex package (v0.2.1)}

answered Jan 2 at 0:39

Jake Kaupp

5,70721428

Hi Jake. Thanks a ton for the recommendation. This approach completely solves my inquiry.

– Jeff
Jan 2 at 4:02

add a comment |

Here's an approach using dplyr.

library(dplyr)



df <- data.frame(

  cluster = c("A","B","B","A","A","B"),

  x = c(3,4,1,5,2,6),

  y = c(4,5,3,1,2,6))



df %>% 

  group_by(cluster) %>% 

  mutate(x.var = min(x) - x,

         y.var = min(y) - y)



#> # A tibble: 6 x 5

#> # Groups:   cluster [2]

#>   cluster     x     y x.var y.var

#>   <fct>   <dbl> <dbl> <dbl> <dbl>

#> 1 A           3     4    -1    -3

#> 2 B           4     5    -3    -2

#> 3 B           1     3     0     0

#> 4 A           5     1    -3     0

#> 5 A           2     2     0    -1

#> 6 B           6     6    -5    -3

^{Created on 2019-01-01 by the reprex package (v0.2.1)}

answered Jan 2 at 0:39

Jake Kaupp

5,70721428

Here's an approach using dplyr.

library(dplyr)



df <- data.frame(

  cluster = c("A","B","B","A","A","B"),

  x = c(3,4,1,5,2,6),

  y = c(4,5,3,1,2,6))



df %>% 

  group_by(cluster) %>% 

  mutate(x.var = min(x) - x,

         y.var = min(y) - y)



#> # A tibble: 6 x 5

#> # Groups:   cluster [2]

#>   cluster     x     y x.var y.var

#>   <fct>   <dbl> <dbl> <dbl> <dbl>

#> 1 A           3     4    -1    -3

#> 2 B           4     5    -3    -2

#> 3 B           1     3     0     0

#> 4 A           5     1    -3     0

#> 5 A           2     2     0    -1

#> 6 B           6     6    -5    -3

^{Created on 2019-01-01 by the reprex package (v0.2.1)}

answered Jan 2 at 0:39

Jake Kaupp

5,70721428

answered Jan 2 at 0:39

Jake Kaupp

5,70721428

answered Jan 2 at 0:39

Jake Kaupp

5,70721428

answered Jan 2 at 0:39

Jake Kaupp

5,70721428

Hi Jake. Thanks a ton for the recommendation. This approach completely solves my inquiry.

– Jeff
Jan 2 at 4:02

add a comment |

Hi Jake. Thanks a ton for the recommendation. This approach completely solves my inquiry.

– Jeff
Jan 2 at 4:02

Hi Jake. Thanks a ton for the recommendation. This approach completely solves my inquiry.

– Jeff
Jan 2 at 4:02

add a comment |

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