Closest value to a specific column in R

I would like to find the closest value to column x3 below.

data=data.frame(x1=c(24,12,76),x2=c(15,30,20),x3=c(45,27,15))

data

  x1 x2 x3

1 24 15 45

2 12 30 27

3 76 20 15

So desired output will be

Closest_Value_to_x3

   24

   30

   20

Please help. Thank you

edited Jan 23 at 13:36

markus

14.3k11236

asked Jan 23 at 13:24

melik

17012

add a comment |

I would like to find the closest value to column x3 below.

data=data.frame(x1=c(24,12,76),x2=c(15,30,20),x3=c(45,27,15))

data

  x1 x2 x3

1 24 15 45

2 12 30 27

3 76 20 15

So desired output will be

Closest_Value_to_x3

   24

   30

   20

Please help. Thank you

edited Jan 23 at 13:36

markus

14.3k11236

asked Jan 23 at 13:24

melik

17012

add a comment |

I would like to find the closest value to column x3 below.

data=data.frame(x1=c(24,12,76),x2=c(15,30,20),x3=c(45,27,15))

data

  x1 x2 x3

1 24 15 45

2 12 30 27

3 76 20 15

So desired output will be

Closest_Value_to_x3

   24

   30

   20

Please help. Thank you

edited Jan 23 at 13:36

markus

14.3k11236

asked Jan 23 at 13:24

melik

17012

I would like to find the closest value to column x3 below.

data=data.frame(x1=c(24,12,76),x2=c(15,30,20),x3=c(45,27,15))

data

  x1 x2 x3

1 24 15 45

2 12 30 27

3 76 20 15

So desired output will be

Closest_Value_to_x3

   24

   30

   20

Please help. Thank you

r dataframe closest

edited Jan 23 at 13:36

markus

14.3k11236

asked Jan 23 at 13:24

melik

17012

edited Jan 23 at 13:36

markus

14.3k11236

asked Jan 23 at 13:24

melik

17012

edited Jan 23 at 13:36

markus

14.3k11236

edited Jan 23 at 13:36

markus

14.3k11236

edited Jan 23 at 13:36

markus

14.3k11236

asked Jan 23 at 13:24

melik

17012

asked Jan 23 at 13:24

melik

17012

asked Jan 23 at 13:24

melik

17012

add a comment |

4 Answers
4

active

oldest

votes

Use max.col(-abs(data[, 3] - data[, -3])) to find the column positions of the closest values and use this result as part of a matrix to extract desired values from your data. The matrix is returned by cbind

col <- 3

data[, -col][cbind(1:nrow(data),

                   max.col(-abs(data[, col] - data[, -col])))]

#[1] 24 30 20

edited Feb 18 at 8:23

answered Jan 23 at 13:30

markus

14.3k11236

Nice answer. Although I don't think the outside [, 1:2] subset is necessary since you've already done that subset inside the abs() call.

– Rich Scriven
Jan 23 at 20:25

@RichScriven Thanks for your comment. I guess I included it in case OP changes their mind and want's to find the closest values to, say, column 1, in which case we'd need the [, 2:3] subset.

– markus
Jan 23 at 20:32

add a comment |

Here is another approach using matrixStats

x <- as.matrix(data[,-3L])

y <- abs(x - .subset2(data, 3L))

x[matrixStats::rowMins(y) == y]

# [1] 24 30 20

Or in base using vapply

x <- as.matrix(data[,-3L])

y <- abs(x - .subset2(data, 3L))

vapply(1:nrow(data), 

       function(k) x[k,][which.min(y[k,])], 

       numeric(1))

# [1] 24 30 20

edited Jan 23 at 14:12

answered Jan 23 at 13:47

nate

3,1411321

add a comment |

A tidyverse solution:

data %>%

  rowid_to_column() %>%

  gather(var, val, -c(x3, rowid)) %>%

  mutate(temp = x3 - val) %>%

  group_by(rowid) %>%

  filter(abs(temp) == min(abs(temp))) %>%

  ungroup() %>%

  select(val)



    val

  <dbl>

1    24

2    30

3    20

First, it adds a row ID. Second, it transforms the data from wide to long. Third, it calculates the difference between "x3" and the other variables. Finally, it groups by the row ID and keeps the rows where the absolute difference is the smallest.

Or:

data %>%

  rowid_to_column() %>%

  gather(var, val, -c(x3, rowid)) %>%

  mutate(temp = x3 - val) %>%

  group_by(rowid) %>%

  filter(abs(temp) == min(abs(temp))) %>%

  ungroup() %>%

  pull(val)



[1] 24 30 20

Or using an approach originally proposed by @markus (it assumes that your columns are named "x"):

data %>%

 mutate(temp = paste0("x", max.col(-abs(.[, -3] - .[, 3])))) %>%

 rowwise() %>%

 summarise(val = eval(as.symbol(temp)))



    val

  <dbl>

1   24.

2   30.

3   20.

First, it is assessing the column index of the variable where the absolute difference in regard to "x3" is the smallest and combines it with "x". Then, it evaluates the combination of x and column index as a variable and returns the appropriate value.

Also borrowing the idea from @markus (not assuming that your columns are named "x"):

data %>%

 mutate(temp = max.col(-abs(.[, -3] - .[, 3]))) %>%

 rowwise %>%

 mutate(temp = names(.)[[temp]]) %>%

 summarise(val = eval(as.symbol(temp)))

First, it is assessing the column index of the variable where the absolute difference in regard to "x3" is the smallest. Second, it returns the column name based on the column index. Finally, it evaluates it as a variable and returns the appropriate value.

Or a variant where you can reference the "x3" variable by its name and not by column index (the basic idea still from @markus):

data %>%

 mutate(temp = max.col(-abs(.[, !grepl("x3", colnames(.))] - .[, grepl("x3", colnames(.))]))) %>% 

 rowwise %>%

 mutate(temp = names(.)[[temp]]) %>%

 summarise(val = eval(as.symbol(temp)))

edited Jan 24 at 6:23

answered Jan 23 at 14:00

tmfmnk

3,0371514

I like that I can always count on you for a tidyverse approach but sometimes they look so complex and intimidating. Great all the same!

– NelsonGon
Jan 23 at 14:19

1

@NelsonGon sometimes it gets really verbose, that is true. But that is also true that the tidyverse aprroaches in general are not the ones with the shortest code. Anyway, thank you for your compliment :)

– tmfmnk
Jan 23 at 14:30

1

This is a tidyverse solution, not the only one, and not one I'd write. You can make the code a lot less verbose by following the natural logic of the other answer, no need to reshape the data.

– Konrad Rudolph
Jan 23 at 17:19

1

@tmfmnk You’re completely right but just to clarify, I think in this case you’re torturing dplyr, and it confesses to anything. You could do a simpler single mutate: data %>% mutate(d = .[, -3][cbind(row_number(), max.col(- abs(.[, 3] - .[, -3])))]) — I’d be tempted to introduce a temporary column to hold the result of max.col but otherwise that’s it.

– Konrad Rudolph
Jan 24 at 9:57

1

"I think in this case you’re torturing dplyr, and it confesses to anything", this could be on fortunes.

– RLave
Jan 24 at 10:56

|
show 3 more comments

Define a function closest_to_3 that operates on a vector and returns the value in the vector that's closest to the third member:

closest_to_3 <- function(v) v[-3][which.min(abs( v[-3]-v[3] ))]

(The idiom v[-3] deletes the 3rd member from v.) Then apply this function to each row of your data frame:

apply(data, 1, closest_to_3)

#[1] 24 30 20

answered Jan 23 at 20:06

grand_chat

24111

add a comment |

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

col <- 3

data[, -col][cbind(1:nrow(data),

                   max.col(-abs(data[, col] - data[, -col])))]

#[1] 24 30 20

edited Feb 18 at 8:23

answered Jan 23 at 13:30

markus

14.3k11236

Nice answer. Although I don't think the outside [, 1:2] subset is necessary since you've already done that subset inside the abs() call.

– Rich Scriven
Jan 23 at 20:25

@RichScriven Thanks for your comment. I guess I included it in case OP changes their mind and want's to find the closest values to, say, column 1, in which case we'd need the [, 2:3] subset.

– markus
Jan 23 at 20:32

add a comment |

col <- 3

data[, -col][cbind(1:nrow(data),

                   max.col(-abs(data[, col] - data[, -col])))]

#[1] 24 30 20

edited Feb 18 at 8:23

answered Jan 23 at 13:30

markus

14.3k11236

Nice answer. Although I don't think the outside [, 1:2] subset is necessary since you've already done that subset inside the abs() call.

– Rich Scriven
Jan 23 at 20:25

@RichScriven Thanks for your comment. I guess I included it in case OP changes their mind and want's to find the closest values to, say, column 1, in which case we'd need the [, 2:3] subset.

– markus
Jan 23 at 20:32

add a comment |

col <- 3

data[, -col][cbind(1:nrow(data),

                   max.col(-abs(data[, col] - data[, -col])))]

#[1] 24 30 20

edited Feb 18 at 8:23

answered Jan 23 at 13:30

markus

14.3k11236

col <- 3

data[, -col][cbind(1:nrow(data),

                   max.col(-abs(data[, col] - data[, -col])))]

#[1] 24 30 20

edited Feb 18 at 8:23

answered Jan 23 at 13:30

markus

14.3k11236

edited Feb 18 at 8:23

answered Jan 23 at 13:30

markus

14.3k11236

answered Jan 23 at 13:30

markus

14.3k11236

answered Jan 23 at 13:30

markus

14.3k11236

Nice answer. Although I don't think the outside [, 1:2] subset is necessary since you've already done that subset inside the abs() call.

– Rich Scriven
Jan 23 at 20:25

@RichScriven Thanks for your comment. I guess I included it in case OP changes their mind and want's to find the closest values to, say, column 1, in which case we'd need the [, 2:3] subset.

– markus
Jan 23 at 20:32

add a comment |

Nice answer. Although I don't think the outside [, 1:2] subset is necessary since you've already done that subset inside the abs() call.

– Rich Scriven
Jan 23 at 20:25

@RichScriven Thanks for your comment. I guess I included it in case OP changes their mind and want's to find the closest values to, say, column 1, in which case we'd need the [, 2:3] subset.

– markus
Jan 23 at 20:32

Nice answer. Although I don't think the outside [, 1:2] subset is necessary since you've already done that subset inside the abs() call.

– Rich Scriven
Jan 23 at 20:25

@RichScriven Thanks for your comment. I guess I included it in case OP changes their mind and want's to find the closest values to, say, column 1, in which case we'd need the [, 2:3] subset.

– markus
Jan 23 at 20:32

add a comment |

Here is another approach using matrixStats

x <- as.matrix(data[,-3L])

y <- abs(x - .subset2(data, 3L))

x[matrixStats::rowMins(y) == y]

# [1] 24 30 20

Or in base using vapply

x <- as.matrix(data[,-3L])

y <- abs(x - .subset2(data, 3L))

vapply(1:nrow(data), 

       function(k) x[k,][which.min(y[k,])], 

       numeric(1))

# [1] 24 30 20

edited Jan 23 at 14:12

answered Jan 23 at 13:47

nate

3,1411321

add a comment |

Here is another approach using matrixStats

x <- as.matrix(data[,-3L])

y <- abs(x - .subset2(data, 3L))

x[matrixStats::rowMins(y) == y]

# [1] 24 30 20

Or in base using vapply

x <- as.matrix(data[,-3L])

y <- abs(x - .subset2(data, 3L))

vapply(1:nrow(data), 

       function(k) x[k,][which.min(y[k,])], 

       numeric(1))

# [1] 24 30 20

edited Jan 23 at 14:12

answered Jan 23 at 13:47

nate

3,1411321

add a comment |

Here is another approach using matrixStats

x <- as.matrix(data[,-3L])

y <- abs(x - .subset2(data, 3L))

x[matrixStats::rowMins(y) == y]

# [1] 24 30 20

Or in base using vapply

x <- as.matrix(data[,-3L])

y <- abs(x - .subset2(data, 3L))

vapply(1:nrow(data), 

       function(k) x[k,][which.min(y[k,])], 

       numeric(1))

# [1] 24 30 20

edited Jan 23 at 14:12

answered Jan 23 at 13:47

nate

3,1411321

Here is another approach using matrixStats

x <- as.matrix(data[,-3L])

y <- abs(x - .subset2(data, 3L))

x[matrixStats::rowMins(y) == y]

# [1] 24 30 20

Or in base using vapply

x <- as.matrix(data[,-3L])

y <- abs(x - .subset2(data, 3L))

vapply(1:nrow(data), 

       function(k) x[k,][which.min(y[k,])], 

       numeric(1))

# [1] 24 30 20

edited Jan 23 at 14:12

answered Jan 23 at 13:47

nate

3,1411321

edited Jan 23 at 14:12

answered Jan 23 at 13:47

nate

3,1411321

answered Jan 23 at 13:47

nate

3,1411321

answered Jan 23 at 13:47

nate

3,1411321

add a comment |

A tidyverse solution:

data %>%

  rowid_to_column() %>%

  gather(var, val, -c(x3, rowid)) %>%

  mutate(temp = x3 - val) %>%

  group_by(rowid) %>%

  filter(abs(temp) == min(abs(temp))) %>%

  ungroup() %>%

  select(val)



    val

  <dbl>

1    24

2    30

3    20

Or:

data %>%

  rowid_to_column() %>%

  gather(var, val, -c(x3, rowid)) %>%

  mutate(temp = x3 - val) %>%

  group_by(rowid) %>%

  filter(abs(temp) == min(abs(temp))) %>%

  ungroup() %>%

  pull(val)



[1] 24 30 20

Or using an approach originally proposed by @markus (it assumes that your columns are named "x"):

data %>%

 mutate(temp = paste0("x", max.col(-abs(.[, -3] - .[, 3])))) %>%

 rowwise() %>%

 summarise(val = eval(as.symbol(temp)))



    val

  <dbl>

1   24.

2   30.

3   20.

Also borrowing the idea from @markus (not assuming that your columns are named "x"):

data %>%

 mutate(temp = max.col(-abs(.[, -3] - .[, 3]))) %>%

 rowwise %>%

 mutate(temp = names(.)[[temp]]) %>%

 summarise(val = eval(as.symbol(temp)))

Or a variant where you can reference the "x3" variable by its name and not by column index (the basic idea still from @markus):

data %>%

 mutate(temp = max.col(-abs(.[, !grepl("x3", colnames(.))] - .[, grepl("x3", colnames(.))]))) %>% 

 rowwise %>%

 mutate(temp = names(.)[[temp]]) %>%

 summarise(val = eval(as.symbol(temp)))

edited Jan 24 at 6:23

answered Jan 23 at 14:00

tmfmnk

3,0371514

I like that I can always count on you for a tidyverse approach but sometimes they look so complex and intimidating. Great all the same!

– NelsonGon
Jan 23 at 14:19

1

@NelsonGon sometimes it gets really verbose, that is true. But that is also true that the tidyverse aprroaches in general are not the ones with the shortest code. Anyway, thank you for your compliment :)

– tmfmnk
Jan 23 at 14:30

1

This is a tidyverse solution, not the only one, and not one I'd write. You can make the code a lot less verbose by following the natural logic of the other answer, no need to reshape the data.

– Konrad Rudolph
Jan 23 at 17:19

1

@tmfmnk You’re completely right but just to clarify, I think in this case you’re torturing dplyr, and it confesses to anything. You could do a simpler single mutate: data %>% mutate(d = .[, -3][cbind(row_number(), max.col(- abs(.[, 3] - .[, -3])))]) — I’d be tempted to introduce a temporary column to hold the result of max.col but otherwise that’s it.

– Konrad Rudolph
Jan 24 at 9:57

1

"I think in this case you’re torturing dplyr, and it confesses to anything", this could be on fortunes.

– RLave
Jan 24 at 10:56

|
show 3 more comments

A tidyverse solution:

data %>%

  rowid_to_column() %>%

  gather(var, val, -c(x3, rowid)) %>%

  mutate(temp = x3 - val) %>%

  group_by(rowid) %>%

  filter(abs(temp) == min(abs(temp))) %>%

  ungroup() %>%

  select(val)



    val

  <dbl>

1    24

2    30

3    20

Or:

data %>%

  rowid_to_column() %>%

  gather(var, val, -c(x3, rowid)) %>%

  mutate(temp = x3 - val) %>%

  group_by(rowid) %>%

  filter(abs(temp) == min(abs(temp))) %>%

  ungroup() %>%

  pull(val)



[1] 24 30 20

Or using an approach originally proposed by @markus (it assumes that your columns are named "x"):

data %>%

 mutate(temp = paste0("x", max.col(-abs(.[, -3] - .[, 3])))) %>%

 rowwise() %>%

 summarise(val = eval(as.symbol(temp)))



    val

  <dbl>

1   24.

2   30.

3   20.

Also borrowing the idea from @markus (not assuming that your columns are named "x"):

data %>%

 mutate(temp = max.col(-abs(.[, -3] - .[, 3]))) %>%

 rowwise %>%

 mutate(temp = names(.)[[temp]]) %>%

 summarise(val = eval(as.symbol(temp)))

Or a variant where you can reference the "x3" variable by its name and not by column index (the basic idea still from @markus):

data %>%

 mutate(temp = max.col(-abs(.[, !grepl("x3", colnames(.))] - .[, grepl("x3", colnames(.))]))) %>% 

 rowwise %>%

 mutate(temp = names(.)[[temp]]) %>%

 summarise(val = eval(as.symbol(temp)))

edited Jan 24 at 6:23

answered Jan 23 at 14:00

tmfmnk

3,0371514

I like that I can always count on you for a tidyverse approach but sometimes they look so complex and intimidating. Great all the same!

– NelsonGon
Jan 23 at 14:19

1

@NelsonGon sometimes it gets really verbose, that is true. But that is also true that the tidyverse aprroaches in general are not the ones with the shortest code. Anyway, thank you for your compliment :)

– tmfmnk
Jan 23 at 14:30

1

This is a tidyverse solution, not the only one, and not one I'd write. You can make the code a lot less verbose by following the natural logic of the other answer, no need to reshape the data.

– Konrad Rudolph
Jan 23 at 17:19

1

@tmfmnk You’re completely right but just to clarify, I think in this case you’re torturing dplyr, and it confesses to anything. You could do a simpler single mutate: data %>% mutate(d = .[, -3][cbind(row_number(), max.col(- abs(.[, 3] - .[, -3])))]) — I’d be tempted to introduce a temporary column to hold the result of max.col but otherwise that’s it.

– Konrad Rudolph
Jan 24 at 9:57

1

"I think in this case you’re torturing dplyr, and it confesses to anything", this could be on fortunes.

– RLave
Jan 24 at 10:56

|
show 3 more comments

A tidyverse solution:

data %>%

  rowid_to_column() %>%

  gather(var, val, -c(x3, rowid)) %>%

  mutate(temp = x3 - val) %>%

  group_by(rowid) %>%

  filter(abs(temp) == min(abs(temp))) %>%

  ungroup() %>%

  select(val)



    val

  <dbl>

1    24

2    30

3    20

Or:

data %>%

  rowid_to_column() %>%

  gather(var, val, -c(x3, rowid)) %>%

  mutate(temp = x3 - val) %>%

  group_by(rowid) %>%

  filter(abs(temp) == min(abs(temp))) %>%

  ungroup() %>%

  pull(val)



[1] 24 30 20

Or using an approach originally proposed by @markus (it assumes that your columns are named "x"):

data %>%

 mutate(temp = paste0("x", max.col(-abs(.[, -3] - .[, 3])))) %>%

 rowwise() %>%

 summarise(val = eval(as.symbol(temp)))



    val

  <dbl>

1   24.

2   30.

3   20.

Also borrowing the idea from @markus (not assuming that your columns are named "x"):

data %>%

 mutate(temp = max.col(-abs(.[, -3] - .[, 3]))) %>%

 rowwise %>%

 mutate(temp = names(.)[[temp]]) %>%

 summarise(val = eval(as.symbol(temp)))

Or a variant where you can reference the "x3" variable by its name and not by column index (the basic idea still from @markus):

data %>%

 mutate(temp = max.col(-abs(.[, !grepl("x3", colnames(.))] - .[, grepl("x3", colnames(.))]))) %>% 

 rowwise %>%

 mutate(temp = names(.)[[temp]]) %>%

 summarise(val = eval(as.symbol(temp)))

edited Jan 24 at 6:23

answered Jan 23 at 14:00

tmfmnk

3,0371514

A tidyverse solution:

data %>%

  rowid_to_column() %>%

  gather(var, val, -c(x3, rowid)) %>%

  mutate(temp = x3 - val) %>%

  group_by(rowid) %>%

  filter(abs(temp) == min(abs(temp))) %>%

  ungroup() %>%

  select(val)



    val

  <dbl>

1    24

2    30

3    20

Or:

data %>%

  rowid_to_column() %>%

  gather(var, val, -c(x3, rowid)) %>%

  mutate(temp = x3 - val) %>%

  group_by(rowid) %>%

  filter(abs(temp) == min(abs(temp))) %>%

  ungroup() %>%

  pull(val)



[1] 24 30 20

Or using an approach originally proposed by @markus (it assumes that your columns are named "x"):

data %>%

 mutate(temp = paste0("x", max.col(-abs(.[, -3] - .[, 3])))) %>%

 rowwise() %>%

 summarise(val = eval(as.symbol(temp)))



    val

  <dbl>

1   24.

2   30.

3   20.

Also borrowing the idea from @markus (not assuming that your columns are named "x"):

data %>%

 mutate(temp = max.col(-abs(.[, -3] - .[, 3]))) %>%

 rowwise %>%

 mutate(temp = names(.)[[temp]]) %>%

 summarise(val = eval(as.symbol(temp)))

Or a variant where you can reference the "x3" variable by its name and not by column index (the basic idea still from @markus):

data %>%

 mutate(temp = max.col(-abs(.[, !grepl("x3", colnames(.))] - .[, grepl("x3", colnames(.))]))) %>% 

 rowwise %>%

 mutate(temp = names(.)[[temp]]) %>%

 summarise(val = eval(as.symbol(temp)))

edited Jan 24 at 6:23

answered Jan 23 at 14:00

tmfmnk

3,0371514

edited Jan 24 at 6:23

answered Jan 23 at 14:00

tmfmnk

3,0371514

answered Jan 23 at 14:00

tmfmnk

3,0371514

answered Jan 23 at 14:00

tmfmnk

3,0371514

I like that I can always count on you for a tidyverse approach but sometimes they look so complex and intimidating. Great all the same!

– NelsonGon
Jan 23 at 14:19

1

@NelsonGon sometimes it gets really verbose, that is true. But that is also true that the tidyverse aprroaches in general are not the ones with the shortest code. Anyway, thank you for your compliment :)

– tmfmnk
Jan 23 at 14:30

1

This is a tidyverse solution, not the only one, and not one I'd write. You can make the code a lot less verbose by following the natural logic of the other answer, no need to reshape the data.

– Konrad Rudolph
Jan 23 at 17:19

1

@tmfmnk You’re completely right but just to clarify, I think in this case you’re torturing dplyr, and it confesses to anything. You could do a simpler single mutate: data %>% mutate(d = .[, -3][cbind(row_number(), max.col(- abs(.[, 3] - .[, -3])))]) — I’d be tempted to introduce a temporary column to hold the result of max.col but otherwise that’s it.

– Konrad Rudolph
Jan 24 at 9:57

1

"I think in this case you’re torturing dplyr, and it confesses to anything", this could be on fortunes.

– RLave
Jan 24 at 10:56

|
show 3 more comments

I like that I can always count on you for a tidyverse approach but sometimes they look so complex and intimidating. Great all the same!

– NelsonGon
Jan 23 at 14:19

1

@NelsonGon sometimes it gets really verbose, that is true. But that is also true that the tidyverse aprroaches in general are not the ones with the shortest code. Anyway, thank you for your compliment :)

– tmfmnk
Jan 23 at 14:30

1

This is a tidyverse solution, not the only one, and not one I'd write. You can make the code a lot less verbose by following the natural logic of the other answer, no need to reshape the data.

– Konrad Rudolph
Jan 23 at 17:19

1

@tmfmnk You’re completely right but just to clarify, I think in this case you’re torturing dplyr, and it confesses to anything. You could do a simpler single mutate: data %>% mutate(d = .[, -3][cbind(row_number(), max.col(- abs(.[, 3] - .[, -3])))]) — I’d be tempted to introduce a temporary column to hold the result of max.col but otherwise that’s it.

– Konrad Rudolph
Jan 24 at 9:57

1

"I think in this case you’re torturing dplyr, and it confesses to anything", this could be on fortunes.

– RLave
Jan 24 at 10:56

I like that I can always count on you for a tidyverse approach but sometimes they look so complex and intimidating. Great all the same!

– NelsonGon
Jan 23 at 14:19

@NelsonGon sometimes it gets really verbose, that is true. But that is also true that the tidyverse aprroaches in general are not the ones with the shortest code. Anyway, thank you for your compliment :)

– tmfmnk
Jan 23 at 14:30

This is a tidyverse solution, not the only one, and not one I'd write. You can make the code a lot less verbose by following the natural logic of the other answer, no need to reshape the data.

– Konrad Rudolph
Jan 23 at 17:19

@tmfmnk You’re completely right but just to clarify, I think in this case you’re torturing dplyr, and it confesses to anything. You could do a simpler single mutate: data %>% mutate(d = .[, -3][cbind(row_number(), max.col(- abs(.[, 3] - .[, -3])))]) — I’d be tempted to introduce a temporary column to hold the result of max.col but otherwise that’s it.

– Konrad Rudolph
Jan 24 at 9:57

"I think in this case you’re torturing dplyr, and it confesses to anything", this could be on fortunes.

– RLave
Jan 24 at 10:56

|
show 3 more comments

Define a function closest_to_3 that operates on a vector and returns the value in the vector that's closest to the third member:

closest_to_3 <- function(v) v[-3][which.min(abs( v[-3]-v[3] ))]

(The idiom v[-3] deletes the 3rd member from v.) Then apply this function to each row of your data frame:

apply(data, 1, closest_to_3)

#[1] 24 30 20

answered Jan 23 at 20:06

grand_chat

24111

add a comment |

Define a function closest_to_3 that operates on a vector and returns the value in the vector that's closest to the third member:

closest_to_3 <- function(v) v[-3][which.min(abs( v[-3]-v[3] ))]

(The idiom v[-3] deletes the 3rd member from v.) Then apply this function to each row of your data frame:

apply(data, 1, closest_to_3)

#[1] 24 30 20

answered Jan 23 at 20:06

grand_chat

24111

add a comment |

Define a function closest_to_3 that operates on a vector and returns the value in the vector that's closest to the third member:

closest_to_3 <- function(v) v[-3][which.min(abs( v[-3]-v[3] ))]

(The idiom v[-3] deletes the 3rd member from v.) Then apply this function to each row of your data frame:

apply(data, 1, closest_to_3)

#[1] 24 30 20

answered Jan 23 at 20:06

grand_chat

24111

Define a function closest_to_3 that operates on a vector and returns the value in the vector that's closest to the third member:

closest_to_3 <- function(v) v[-3][which.min(abs( v[-3]-v[3] ))]

(The idiom v[-3] deletes the 3rd member from v.) Then apply this function to each row of your data frame:

apply(data, 1, closest_to_3)

#[1] 24 30 20

answered Jan 23 at 20:06

grand_chat

24111

answered Jan 23 at 20:06

grand_chat

24111

answered Jan 23 at 20:06

grand_chat

24111

answered Jan 23 at 20:06

grand_chat

24111

add a comment |

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