Imputing outliers in tibble after identifying them












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I've written a function using this article (http://www.questionflow.org/2017/12/26/combined-outlier-detection-with-dplyr-and-ruler/) I found online which flags any outliers within my tbl based on either being outside of z-score threshold, or outside of 1.5*IQR. I want to enhance my function by adding an "impute" argument which can be set to mean, median, or mode, (or none, in which case the output is just a tbl of logicals) which will set the values that were flagged to the input argument for that column. How can I go about doing that? My code so far:



# Detecting outliers based on two outlier metrics (3 out of z-score or 1.5 out of IQR)
# Defining the functions to detect outliers
isnt_out_z <- function(x, thres = 3, na.rm = TRUE) {
abs(x - mean(x, na.rm = na.rm)) <= thres * sd(x, na.rm = na.rm)
}

isnt_out_IQR <- function(x, k = 1.5, na.rm = TRUE) {
quar <- quantile(x, probs = c(0.25, 0.75), na.rm = na.rm)
iqr <- diff(quar)
(quar[1] - k * iqr <= x) & (x <= quar[2] + k * iqr)
}

# Column-based non-outlier rows: row is not an outlier based on some column if it doesn't contain outlier (computed based on target column) on the intersection with that column

## Useable functions
find_Outliers <- function(data, method = 'z', impute = none) {
if (method == 'z') {
data %>%
transmute_if(is.numeric, isnt_out_z)
} else {
data %>%
transmute_if(is.numeric, isnt_out_IQR)
}
}









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    0















    I've written a function using this article (http://www.questionflow.org/2017/12/26/combined-outlier-detection-with-dplyr-and-ruler/) I found online which flags any outliers within my tbl based on either being outside of z-score threshold, or outside of 1.5*IQR. I want to enhance my function by adding an "impute" argument which can be set to mean, median, or mode, (or none, in which case the output is just a tbl of logicals) which will set the values that were flagged to the input argument for that column. How can I go about doing that? My code so far:



    # Detecting outliers based on two outlier metrics (3 out of z-score or 1.5 out of IQR)
    # Defining the functions to detect outliers
    isnt_out_z <- function(x, thres = 3, na.rm = TRUE) {
    abs(x - mean(x, na.rm = na.rm)) <= thres * sd(x, na.rm = na.rm)
    }

    isnt_out_IQR <- function(x, k = 1.5, na.rm = TRUE) {
    quar <- quantile(x, probs = c(0.25, 0.75), na.rm = na.rm)
    iqr <- diff(quar)
    (quar[1] - k * iqr <= x) & (x <= quar[2] + k * iqr)
    }

    # Column-based non-outlier rows: row is not an outlier based on some column if it doesn't contain outlier (computed based on target column) on the intersection with that column

    ## Useable functions
    find_Outliers <- function(data, method = 'z', impute = none) {
    if (method == 'z') {
    data %>%
    transmute_if(is.numeric, isnt_out_z)
    } else {
    data %>%
    transmute_if(is.numeric, isnt_out_IQR)
    }
    }









    share|improve this question

























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      I've written a function using this article (http://www.questionflow.org/2017/12/26/combined-outlier-detection-with-dplyr-and-ruler/) I found online which flags any outliers within my tbl based on either being outside of z-score threshold, or outside of 1.5*IQR. I want to enhance my function by adding an "impute" argument which can be set to mean, median, or mode, (or none, in which case the output is just a tbl of logicals) which will set the values that were flagged to the input argument for that column. How can I go about doing that? My code so far:



      # Detecting outliers based on two outlier metrics (3 out of z-score or 1.5 out of IQR)
      # Defining the functions to detect outliers
      isnt_out_z <- function(x, thres = 3, na.rm = TRUE) {
      abs(x - mean(x, na.rm = na.rm)) <= thres * sd(x, na.rm = na.rm)
      }

      isnt_out_IQR <- function(x, k = 1.5, na.rm = TRUE) {
      quar <- quantile(x, probs = c(0.25, 0.75), na.rm = na.rm)
      iqr <- diff(quar)
      (quar[1] - k * iqr <= x) & (x <= quar[2] + k * iqr)
      }

      # Column-based non-outlier rows: row is not an outlier based on some column if it doesn't contain outlier (computed based on target column) on the intersection with that column

      ## Useable functions
      find_Outliers <- function(data, method = 'z', impute = none) {
      if (method == 'z') {
      data %>%
      transmute_if(is.numeric, isnt_out_z)
      } else {
      data %>%
      transmute_if(is.numeric, isnt_out_IQR)
      }
      }









      share|improve this question














      I've written a function using this article (http://www.questionflow.org/2017/12/26/combined-outlier-detection-with-dplyr-and-ruler/) I found online which flags any outliers within my tbl based on either being outside of z-score threshold, or outside of 1.5*IQR. I want to enhance my function by adding an "impute" argument which can be set to mean, median, or mode, (or none, in which case the output is just a tbl of logicals) which will set the values that were flagged to the input argument for that column. How can I go about doing that? My code so far:



      # Detecting outliers based on two outlier metrics (3 out of z-score or 1.5 out of IQR)
      # Defining the functions to detect outliers
      isnt_out_z <- function(x, thres = 3, na.rm = TRUE) {
      abs(x - mean(x, na.rm = na.rm)) <= thres * sd(x, na.rm = na.rm)
      }

      isnt_out_IQR <- function(x, k = 1.5, na.rm = TRUE) {
      quar <- quantile(x, probs = c(0.25, 0.75), na.rm = na.rm)
      iqr <- diff(quar)
      (quar[1] - k * iqr <= x) & (x <= quar[2] + k * iqr)
      }

      # Column-based non-outlier rows: row is not an outlier based on some column if it doesn't contain outlier (computed based on target column) on the intersection with that column

      ## Useable functions
      find_Outliers <- function(data, method = 'z', impute = none) {
      if (method == 'z') {
      data %>%
      transmute_if(is.numeric, isnt_out_z)
      } else {
      data %>%
      transmute_if(is.numeric, isnt_out_IQR)
      }
      }






      r dataframe outliers tbl






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      asked Nov 21 '18 at 22:38









      NinaNina

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