Mixing
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!
r list apply lapply
asked Jan 2 at 0:16
JeffJeff
135
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))
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!
r list apply lapply
asked Jan 2 at 0:16
JeffJeff
135
I think the 4th value ofx.varshould 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))
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!
r list apply lapply
asked Jan 2 at 0:16
JeffJeff
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))
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!
r list apply lapply
r list apply lapply
asked Jan 2 at 0:16
JeffJeff
135
asked Jan 2 at 0:16
JeffJeff
135
edited Jan 2 at 3:59
Jeff
asked Jan 2 at 0:16
JeffJeff
135
asked Jan 2 at 0:16
JeffJeff
135
asked Jan 2 at 0:16
JeffJeff
135
135
I think the 4th value ofx.varshould 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 ofx.varshould 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
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
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$cluster, FUN = min) - x)
df
# cluster x y x.var y.var
#1 A 3 4 -1 -3
#2 B 4 5 -3 -2
#3 B 1 3 0 0
#4 A 5 1 -3 0
#5 A 2 2 0 -1
#6 B 6 6 -5 -3
answered Jan 2 at 6:12
akrunakrun
415k13204278
1
Thank you very much for the recommended approach. This is extremely helpful.
– Jeff
Jan 2 at 11:37
add a comment |
Here's an approach using dplyr.
library(dplyr)
df <- data.frame(
cluster = c("A","B","B","A","A","B"),
x = c(3,4,1,5,2,6),
y = c(4,5,3,1,2,6))
df %>%
group_by(cluster) %>%
mutate(x.var = min(x) - x,
y.var = min(y) - y)
#> # A tibble: 6 x 5
#> # Groups: cluster [2]
#> cluster x y x.var y.var
#> <fct> <dbl> <dbl> <dbl> <dbl>
#> 1 A 3 4 -1 -3
#> 2 B 4 5 -3 -2
#> 3 B 1 3 0 0
#> 4 A 5 1 -3 0
#> 5 A 2 2 0 -1
#> 6 B 6 6 -5 -3
Created on 2019-01-01 by the reprex package (v0.2.1)
answered Jan 2 at 0:39
Jake KauppJake Kaupp
5,70721428
Hi Jake. Thanks a ton for the recommendation. This approach completely solves my inquiry.
– Jeff
Jan 2 at 4:02
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Here is a base R method that uses ave with lapply. Loop through the columns of dataset excluding the 'cluster', then with ave get the min grouped by 'cluster', subtract from the column and assign the list of vectors to new columns
df[paste0(names(df)[-1], ".var")] <- lapply(df[-1], function(x)
ave(x, df$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
akrunakrun
415k13204278
1
Thank you very much for the recommended approach. This is extremely helpful.
– Jeff
Jan 2 at 11:37
add a comment |
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$cluster, FUN = min) - x)
df
# cluster x y x.var y.var
#1 A 3 4 -1 -3
#2 B 4 5 -3 -2
#3 B 1 3 0 0
#4 A 5 1 -3 0
#5 A 2 2 0 -1
#6 B 6 6 -5 -3
answered Jan 2 at 6:12
akrunakrun
415k13204278
1
Thank you very much for the recommended approach. This is extremely helpful.
– Jeff
Jan 2 at 11:37
add a comment |
Here is a base R method that uses ave with lapply. Loop through the columns of dataset excluding the 'cluster', then with ave get the min grouped by 'cluster', subtract from the column and assign the list of vectors to new columns
df[paste0(names(df)[-1], ".var")] <- lapply(df[-1], function(x)
ave(x, df$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
akrunakrun
415k13204278
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
akrunakrun
415k13204278
answered Jan 2 at 6:12
akrunakrun
415k13204278
answered Jan 2 at 6:12
akrunakrun
415k13204278
answered Jan 2 at 6:12
akrunakrun
415k13204278
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
1
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 KauppJake 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 KauppJake 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 KauppJake 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 KauppJake Kaupp
5,70721428
answered Jan 2 at 0:39
Jake KauppJake Kaupp
5,70721428
answered Jan 2 at 0:39
Jake KauppJake Kaupp
5,70721428
answered Jan 2 at 0:39
Jake KauppJake Kaupp
5,70721428
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
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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I think the 4th value of
x.varshould 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