Defining a custom Error measure (sMAPE) function

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I am defining a custom sMAPE (Symmetric Mean Absolute Percent Error) function in R. I want to add an exception in the code where both actual and predicted are zero.

Here is the sample data:

Actual Predicted

0          0

2          1

1          0

2          4

2          1

1          3

The code is shown below:

sMAAPE <- function(actual, predicted){

if (actual == 0 & predicted == 0){

    return(0)

       }

else {

  output<- mean(abs(actual-predicted)/(abs(actual)+abs(predicted)))

  return(output)

    }

  }

As I mentioned, i want to add an exception, that is wherever both actual and predicted are 0, my error would be zero. The problem with my code is that, when it is encountering both actual and predicted to be 0 in first observation, it is giving the mean error to be zero, but in reality I want mean of all errors.

Can somebody help me with this?

asked Jan 3 at 9:26

Sourabh

2

Instead of if/else use ifelse with index created using rowSums ie. i1 <- !rowSums(df1 != 0); ifelse(i1, 0, othervalue) Wrap it in a function sMAAPE <- function(dat) { i1 <- !rowSums(dat !=0); val <- with(dat, mean(abs(actual-predicted)/(abs(actual)+abs(predicted)))); val[i1] <- 0; val}

– akrun
Jan 3 at 9:28

add a comment |

I am defining a custom sMAPE (Symmetric Mean Absolute Percent Error) function in R. I want to add an exception in the code where both actual and predicted are zero.

Here is the sample data:

Actual Predicted

0          0

2          1

1          0

2          4

2          1

1          3

The code is shown below:

sMAAPE <- function(actual, predicted){

if (actual == 0 & predicted == 0){

    return(0)

       }

else {

  output<- mean(abs(actual-predicted)/(abs(actual)+abs(predicted)))

  return(output)

    }

  }

Can somebody help me with this?

asked Jan 3 at 9:26

Sourabh

2

Instead of if/else use ifelse with index created using rowSums ie. i1 <- !rowSums(df1 != 0); ifelse(i1, 0, othervalue) Wrap it in a function sMAAPE <- function(dat) { i1 <- !rowSums(dat !=0); val <- with(dat, mean(abs(actual-predicted)/(abs(actual)+abs(predicted)))); val[i1] <- 0; val}

– akrun
Jan 3 at 9:28

add a comment |

I am defining a custom sMAPE (Symmetric Mean Absolute Percent Error) function in R. I want to add an exception in the code where both actual and predicted are zero.

Here is the sample data:

Actual Predicted

0          0

2          1

1          0

2          4

2          1

1          3

The code is shown below:

sMAAPE <- function(actual, predicted){

if (actual == 0 & predicted == 0){

    return(0)

       }

else {

  output<- mean(abs(actual-predicted)/(abs(actual)+abs(predicted)))

  return(output)

    }

  }

Can somebody help me with this?

asked Jan 3 at 9:26

Sourabh

I am defining a custom sMAPE (Symmetric Mean Absolute Percent Error) function in R. I want to add an exception in the code where both actual and predicted are zero.

Here is the sample data:

Actual Predicted

0          0

2          1

1          0

2          4

2          1

1          3

The code is shown below:

sMAAPE <- function(actual, predicted){

if (actual == 0 & predicted == 0){

    return(0)

       }

else {

  output<- mean(abs(actual-predicted)/(abs(actual)+abs(predicted)))

  return(output)

    }

  }

Can somebody help me with this?

asked Jan 3 at 9:26

Sourabh

asked Jan 3 at 9:26

Sourabh

asked Jan 3 at 9:26

Sourabh

asked Jan 3 at 9:26

Sourabh

asked Jan 3 at 9:26

Sourabh

2

Instead of if/else use ifelse with index created using rowSums ie. i1 <- !rowSums(df1 != 0); ifelse(i1, 0, othervalue) Wrap it in a function sMAAPE <- function(dat) { i1 <- !rowSums(dat !=0); val <- with(dat, mean(abs(actual-predicted)/(abs(actual)+abs(predicted)))); val[i1] <- 0; val}

– akrun
Jan 3 at 9:28

add a comment |

2

Instead of if/else use ifelse with index created using rowSums ie. i1 <- !rowSums(df1 != 0); ifelse(i1, 0, othervalue) Wrap it in a function sMAAPE <- function(dat) { i1 <- !rowSums(dat !=0); val <- with(dat, mean(abs(actual-predicted)/(abs(actual)+abs(predicted)))); val[i1] <- 0; val}

– akrun
Jan 3 at 9:28

Instead of if/else use ifelse with index created using rowSums ie. i1 <- !rowSums(df1 != 0); ifelse(i1, 0, othervalue) Wrap it in a function

sMAAPE <- function(dat) { i1 <- !rowSums(dat !=0); val <- with(dat, mean(abs(actual-predicted)/(abs(actual)+abs(predicted)))); val[i1] <- 0; val}

– akrun
Jan 3 at 9:28

Instead of if/else use ifelse with index created using rowSums ie. i1 <- !rowSums(df1 != 0); ifelse(i1, 0, othervalue) Wrap it in a function

sMAAPE <- function(dat) { i1 <- !rowSums(dat !=0); val <- with(dat, mean(abs(actual-predicted)/(abs(actual)+abs(predicted)))); val[i1] <- 0; val}

– akrun
Jan 3 at 9:28

add a comment |

2 Answers
2

active

oldest

votes

Applying @akrun's suggestion in the comments to your problem. I think this is what you need.

# Create a dataset    

Actual <- c(0,2,1,2,2,1)

Predicted <- c(0,1,0,4,1,3)

df <- data.frame(cbind(Actual, Predicted))

df





sMAAPE <- function(actual, predicted){

     output<- ifelse(actual == 0 & predicted == 0,0, # If actual and predicted are 0, error is 0 

     abs(actual- predicted)/(abs(actual)+abs(predicted))) # otherwise your function

     return(mean(output)) # Return mean of all errors

}

The output:

> sMAAPE(df$Actual, df$Predicted)

[1] 0.4166667

Which is the mean of the following vector [1] 0 0.3333333 1.0000000 0.3333333 0.3333333 0.5000000

You could add na.rm = T to the mean function so it handles NA's as well.

edited Jan 3 at 10:29

answered Jan 3 at 10:22

Niek

974518

This is also working. Also, if we add na.rm = T in mean function, this will not consider those NA's at all, therefore the mean would be wrong as the length of df will change after excluding NA's. To deal with this, I used na.rm = T is sum function and then diving this sum with length of dataframe.

– Sourabh
Jan 4 at 11:48

Adding to my previous comment, Here is the code: output_sum<- sum(abs(actual-predicted)/(abs(actual)+abs(predicted)), na.rm = T) output <- output_sum/NROW(actual) return(output)

– Sourabh
Jan 4 at 11:57

add a comment |

The problem is that actual == 0 & predicted == 0 is a logical vector, while if you put a logical vector as the condition of if sentence, only the first element will be judged.

Use sum(abs(actual))+sum(abs(predicted)) == 0 instead in the condition.

answered Jan 3 at 9:43

Xinz

543

This worked. Using sum function allows R to go the else condition. Thanks

– Sourabh
Jan 4 at 11:47

add a comment |

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

Applying @akrun's suggestion in the comments to your problem. I think this is what you need.

# Create a dataset    

Actual <- c(0,2,1,2,2,1)

Predicted <- c(0,1,0,4,1,3)

df <- data.frame(cbind(Actual, Predicted))

df





sMAAPE <- function(actual, predicted){

     output<- ifelse(actual == 0 & predicted == 0,0, # If actual and predicted are 0, error is 0 

     abs(actual- predicted)/(abs(actual)+abs(predicted))) # otherwise your function

     return(mean(output)) # Return mean of all errors

}

The output:

> sMAAPE(df$Actual, df$Predicted)

[1] 0.4166667

Which is the mean of the following vector [1] 0 0.3333333 1.0000000 0.3333333 0.3333333 0.5000000

You could add na.rm = T to the mean function so it handles NA's as well.

edited Jan 3 at 10:29

answered Jan 3 at 10:22

Niek

974518

This is also working. Also, if we add na.rm = T in mean function, this will not consider those NA's at all, therefore the mean would be wrong as the length of df will change after excluding NA's. To deal with this, I used na.rm = T is sum function and then diving this sum with length of dataframe.

– Sourabh
Jan 4 at 11:48

Adding to my previous comment, Here is the code: output_sum<- sum(abs(actual-predicted)/(abs(actual)+abs(predicted)), na.rm = T) output <- output_sum/NROW(actual) return(output)

– Sourabh
Jan 4 at 11:57

add a comment |

Applying @akrun's suggestion in the comments to your problem. I think this is what you need.

# Create a dataset    

Actual <- c(0,2,1,2,2,1)

Predicted <- c(0,1,0,4,1,3)

df <- data.frame(cbind(Actual, Predicted))

df





sMAAPE <- function(actual, predicted){

     output<- ifelse(actual == 0 & predicted == 0,0, # If actual and predicted are 0, error is 0 

     abs(actual- predicted)/(abs(actual)+abs(predicted))) # otherwise your function

     return(mean(output)) # Return mean of all errors

}

The output:

> sMAAPE(df$Actual, df$Predicted)

[1] 0.4166667

Which is the mean of the following vector [1] 0 0.3333333 1.0000000 0.3333333 0.3333333 0.5000000

You could add na.rm = T to the mean function so it handles NA's as well.

edited Jan 3 at 10:29

answered Jan 3 at 10:22

Niek

974518

This is also working. Also, if we add na.rm = T in mean function, this will not consider those NA's at all, therefore the mean would be wrong as the length of df will change after excluding NA's. To deal with this, I used na.rm = T is sum function and then diving this sum with length of dataframe.

– Sourabh
Jan 4 at 11:48

Adding to my previous comment, Here is the code: output_sum<- sum(abs(actual-predicted)/(abs(actual)+abs(predicted)), na.rm = T) output <- output_sum/NROW(actual) return(output)

– Sourabh
Jan 4 at 11:57

add a comment |

Applying @akrun's suggestion in the comments to your problem. I think this is what you need.

# Create a dataset    

Actual <- c(0,2,1,2,2,1)

Predicted <- c(0,1,0,4,1,3)

df <- data.frame(cbind(Actual, Predicted))

df





sMAAPE <- function(actual, predicted){

     output<- ifelse(actual == 0 & predicted == 0,0, # If actual and predicted are 0, error is 0 

     abs(actual- predicted)/(abs(actual)+abs(predicted))) # otherwise your function

     return(mean(output)) # Return mean of all errors

}

The output:

> sMAAPE(df$Actual, df$Predicted)

[1] 0.4166667

Which is the mean of the following vector [1] 0 0.3333333 1.0000000 0.3333333 0.3333333 0.5000000

You could add na.rm = T to the mean function so it handles NA's as well.

edited Jan 3 at 10:29

answered Jan 3 at 10:22

Niek

974518

Applying @akrun's suggestion in the comments to your problem. I think this is what you need.

# Create a dataset    

Actual <- c(0,2,1,2,2,1)

Predicted <- c(0,1,0,4,1,3)

df <- data.frame(cbind(Actual, Predicted))

df





sMAAPE <- function(actual, predicted){

     output<- ifelse(actual == 0 & predicted == 0,0, # If actual and predicted are 0, error is 0 

     abs(actual- predicted)/(abs(actual)+abs(predicted))) # otherwise your function

     return(mean(output)) # Return mean of all errors

}

The output:

> sMAAPE(df$Actual, df$Predicted)

[1] 0.4166667

Which is the mean of the following vector [1] 0 0.3333333 1.0000000 0.3333333 0.3333333 0.5000000

You could add na.rm = T to the mean function so it handles NA's as well.

edited Jan 3 at 10:29

answered Jan 3 at 10:22

Niek

974518

edited Jan 3 at 10:29

answered Jan 3 at 10:22

Niek

974518

answered Jan 3 at 10:22

Niek

974518

answered Jan 3 at 10:22

Niek

974518

This is also working. Also, if we add na.rm = T in mean function, this will not consider those NA's at all, therefore the mean would be wrong as the length of df will change after excluding NA's. To deal with this, I used na.rm = T is sum function and then diving this sum with length of dataframe.

– Sourabh
Jan 4 at 11:48

Adding to my previous comment, Here is the code: output_sum<- sum(abs(actual-predicted)/(abs(actual)+abs(predicted)), na.rm = T) output <- output_sum/NROW(actual) return(output)

– Sourabh
Jan 4 at 11:57

add a comment |

This is also working. Also, if we add na.rm = T in mean function, this will not consider those NA's at all, therefore the mean would be wrong as the length of df will change after excluding NA's. To deal with this, I used na.rm = T is sum function and then diving this sum with length of dataframe.

– Sourabh
Jan 4 at 11:48

Adding to my previous comment, Here is the code: output_sum<- sum(abs(actual-predicted)/(abs(actual)+abs(predicted)), na.rm = T) output <- output_sum/NROW(actual) return(output)

– Sourabh
Jan 4 at 11:57

This is also working. Also, if we add na.rm = T in mean function, this will not consider those NA's at all, therefore the mean would be wrong as the length of df will change after excluding NA's. To deal with this, I used na.rm = T is sum function and then diving this sum with length of dataframe.

– Sourabh
Jan 4 at 11:48

Adding to my previous comment, Here is the code: output_sum<- sum(abs(actual-predicted)/(abs(actual)+abs(predicted)), na.rm = T) output <- output_sum/NROW(actual) return(output)

– Sourabh
Jan 4 at 11:57

add a comment |

The problem is that actual == 0 & predicted == 0 is a logical vector, while if you put a logical vector as the condition of if sentence, only the first element will be judged.

Use sum(abs(actual))+sum(abs(predicted)) == 0 instead in the condition.

answered Jan 3 at 9:43

Xinz

543

This worked. Using sum function allows R to go the else condition. Thanks

– Sourabh
Jan 4 at 11:47

add a comment |

The problem is that actual == 0 & predicted == 0 is a logical vector, while if you put a logical vector as the condition of if sentence, only the first element will be judged.

Use sum(abs(actual))+sum(abs(predicted)) == 0 instead in the condition.

answered Jan 3 at 9:43

Xinz

543

This worked. Using sum function allows R to go the else condition. Thanks

– Sourabh
Jan 4 at 11:47

add a comment |

The problem is that actual == 0 & predicted == 0 is a logical vector, while if you put a logical vector as the condition of if sentence, only the first element will be judged.

Use sum(abs(actual))+sum(abs(predicted)) == 0 instead in the condition.

answered Jan 3 at 9:43

Xinz

543

The problem is that actual == 0 & predicted == 0 is a logical vector, while if you put a logical vector as the condition of if sentence, only the first element will be judged.

Use sum(abs(actual))+sum(abs(predicted)) == 0 instead in the condition.

answered Jan 3 at 9:43

Xinz

543

answered Jan 3 at 9:43

Xinz

543

answered Jan 3 at 9:43

Xinz

543

answered Jan 3 at 9:43

Xinz

543

This worked. Using sum function allows R to go the else condition. Thanks

– Sourabh
Jan 4 at 11:47

add a comment |

This worked. Using sum function allows R to go the else condition. Thanks

– Sourabh
Jan 4 at 11:47

This worked. Using sum function allows R to go the else condition. Thanks

– Sourabh
Jan 4 at 11:47

add a comment |

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