find index of largest difference from median with numpy

I'm trying to find the index number of the outlier number. based on difference from median
I'm able to get the correct high number, but whenever the low number is the outlier I only get the high number..

import numpy as np



def findoutlier(lis):



  outliermax = np.absolute(np.max(lis) - np.median(lis))

  outliermin = np.absolute(np.min(lis) - np.median(lis))

  if outliermax > outliermin:

     argmax = np.argmax(lis, axis = 1)

     return argmax

  else:

     argmin = np.argmin(lis, axis = 1)

     return argmin



def main():

  Matx = np.array([[10,3,2],[1,2,6]])   

  print(findoutlier(Matx))



  threeMatx = np.array([[1,10,2,8,5],[2,7,3,9,11],[19,2,1,1,5]])

  print(findoutlier(threeMatx))



main()

asked Nov 12 '18 at 23:04

Clueless

275

add a comment |

import numpy as np



def findoutlier(lis):



  outliermax = np.absolute(np.max(lis) - np.median(lis))

  outliermin = np.absolute(np.min(lis) - np.median(lis))

  if outliermax > outliermin:

     argmax = np.argmax(lis, axis = 1)

     return argmax

  else:

     argmin = np.argmin(lis, axis = 1)

     return argmin



def main():

  Matx = np.array([[10,3,2],[1,2,6]])   

  print(findoutlier(Matx))



  threeMatx = np.array([[1,10,2,8,5],[2,7,3,9,11],[19,2,1,1,5]])

  print(findoutlier(threeMatx))



main()

asked Nov 12 '18 at 23:04

Clueless

275

add a comment |

import numpy as np



def findoutlier(lis):



  outliermax = np.absolute(np.max(lis) - np.median(lis))

  outliermin = np.absolute(np.min(lis) - np.median(lis))

  if outliermax > outliermin:

     argmax = np.argmax(lis, axis = 1)

     return argmax

  else:

     argmin = np.argmin(lis, axis = 1)

     return argmin



def main():

  Matx = np.array([[10,3,2],[1,2,6]])   

  print(findoutlier(Matx))



  threeMatx = np.array([[1,10,2,8,5],[2,7,3,9,11],[19,2,1,1,5]])

  print(findoutlier(threeMatx))



main()

asked Nov 12 '18 at 23:04

Clueless

275

import numpy as np



def findoutlier(lis):



  outliermax = np.absolute(np.max(lis) - np.median(lis))

  outliermin = np.absolute(np.min(lis) - np.median(lis))

  if outliermax > outliermin:

     argmax = np.argmax(lis, axis = 1)

     return argmax

  else:

     argmin = np.argmin(lis, axis = 1)

     return argmin



def main():

  Matx = np.array([[10,3,2],[1,2,6]])   

  print(findoutlier(Matx))



  threeMatx = np.array([[1,10,2,8,5],[2,7,3,9,11],[19,2,1,1,5]])

  print(findoutlier(threeMatx))



main()

python numpy functional-programming median argmax

asked Nov 12 '18 at 23:04

Clueless

275

asked Nov 12 '18 at 23:04

Clueless

275

asked Nov 12 '18 at 23:04

Clueless

275

asked Nov 12 '18 at 23:04

Clueless

275

asked Nov 12 '18 at 23:04

Clueless

275

add a comment |

1 Answer
1

active

oldest

votes

You need to specify the axis when using median, max and min:

import numpy as np





def findoutlier(lis):

    omaxs = np.absolute(np.max(lis, axis=1) - np.median(lis, axis=1))

    omins = np.absolute(np.min(lis, axis=1) - np.median(lis, axis=1))



    return [np.argmax(l) if omax > omin else np.argmin(l)  for omax, omin, l in  zip(omaxs, omins, lis)]





def main():

    mat_x = np.array([[10, 3, 2], [1, 2, 6]])

    print(findoutlier(mat_x))



    three_mat_x = np.array([[1, 10, 2, 8, 5], [2, 7, 3, 9, 11], [19, 2, 1, 1, 5]])

    print(findoutlier(three_mat_x))

Output

[0, 2]

[1, 0, 0]

UPDATE

As mentioned by @user3483203 you can use numpy.where:

import numpy as np





def findoutlier(lis):

    omaxs = np.absolute(np.max(lis, axis=1) - np.median(lis, axis=1))

    omins = np.absolute(np.min(lis, axis=1) - np.median(lis, axis=1))



    return np.where(omaxs > omins, np.argmax(lis, axis=1), np.argmin(lis, axis=1))





def main():

    mat_x = np.array([[10, 3, 2], [1, 2, 6]])

    print(findoutlier(mat_x))



    three_mat_x = np.array([[1, 10, 2, 8, 5], [2, 7, 3, 9, 11], [19, 2, 1, 1, 5]])

    print(findoutlier(three_mat_x))



main()

Output

[0 2]

[1 0 0]

edited Nov 13 '18 at 0:36

answered Nov 12 '18 at 23:18

Daniel Mesejo

16.2k21130

Don't mix a list comprehension with vectorized operations. numpy.where is a much faster solution.

– user3483203
Nov 13 '18 at 0:00

add a comment |

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1 Answer
1

active

oldest

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1 Answer
1

active

oldest

votes

You need to specify the axis when using median, max and min:

import numpy as np





def findoutlier(lis):

    omaxs = np.absolute(np.max(lis, axis=1) - np.median(lis, axis=1))

    omins = np.absolute(np.min(lis, axis=1) - np.median(lis, axis=1))



    return [np.argmax(l) if omax > omin else np.argmin(l)  for omax, omin, l in  zip(omaxs, omins, lis)]





def main():

    mat_x = np.array([[10, 3, 2], [1, 2, 6]])

    print(findoutlier(mat_x))



    three_mat_x = np.array([[1, 10, 2, 8, 5], [2, 7, 3, 9, 11], [19, 2, 1, 1, 5]])

    print(findoutlier(three_mat_x))

Output

[0, 2]

[1, 0, 0]

UPDATE

As mentioned by @user3483203 you can use numpy.where:

import numpy as np





def findoutlier(lis):

    omaxs = np.absolute(np.max(lis, axis=1) - np.median(lis, axis=1))

    omins = np.absolute(np.min(lis, axis=1) - np.median(lis, axis=1))



    return np.where(omaxs > omins, np.argmax(lis, axis=1), np.argmin(lis, axis=1))





def main():

    mat_x = np.array([[10, 3, 2], [1, 2, 6]])

    print(findoutlier(mat_x))



    three_mat_x = np.array([[1, 10, 2, 8, 5], [2, 7, 3, 9, 11], [19, 2, 1, 1, 5]])

    print(findoutlier(three_mat_x))



main()

Output

[0 2]

[1 0 0]

edited Nov 13 '18 at 0:36

answered Nov 12 '18 at 23:18

Daniel Mesejo

16.2k21130

Don't mix a list comprehension with vectorized operations. numpy.where is a much faster solution.

– user3483203
Nov 13 '18 at 0:00

add a comment |

You need to specify the axis when using median, max and min:

import numpy as np





def findoutlier(lis):

    omaxs = np.absolute(np.max(lis, axis=1) - np.median(lis, axis=1))

    omins = np.absolute(np.min(lis, axis=1) - np.median(lis, axis=1))



    return [np.argmax(l) if omax > omin else np.argmin(l)  for omax, omin, l in  zip(omaxs, omins, lis)]





def main():

    mat_x = np.array([[10, 3, 2], [1, 2, 6]])

    print(findoutlier(mat_x))



    three_mat_x = np.array([[1, 10, 2, 8, 5], [2, 7, 3, 9, 11], [19, 2, 1, 1, 5]])

    print(findoutlier(three_mat_x))

Output

[0, 2]

[1, 0, 0]

UPDATE

As mentioned by @user3483203 you can use numpy.where:

import numpy as np





def findoutlier(lis):

    omaxs = np.absolute(np.max(lis, axis=1) - np.median(lis, axis=1))

    omins = np.absolute(np.min(lis, axis=1) - np.median(lis, axis=1))



    return np.where(omaxs > omins, np.argmax(lis, axis=1), np.argmin(lis, axis=1))





def main():

    mat_x = np.array([[10, 3, 2], [1, 2, 6]])

    print(findoutlier(mat_x))



    three_mat_x = np.array([[1, 10, 2, 8, 5], [2, 7, 3, 9, 11], [19, 2, 1, 1, 5]])

    print(findoutlier(three_mat_x))



main()

Output

[0 2]

[1 0 0]

edited Nov 13 '18 at 0:36

answered Nov 12 '18 at 23:18

Daniel Mesejo

16.2k21130

Don't mix a list comprehension with vectorized operations. numpy.where is a much faster solution.

– user3483203
Nov 13 '18 at 0:00

add a comment |

You need to specify the axis when using median, max and min:

import numpy as np





def findoutlier(lis):

    omaxs = np.absolute(np.max(lis, axis=1) - np.median(lis, axis=1))

    omins = np.absolute(np.min(lis, axis=1) - np.median(lis, axis=1))



    return [np.argmax(l) if omax > omin else np.argmin(l)  for omax, omin, l in  zip(omaxs, omins, lis)]





def main():

    mat_x = np.array([[10, 3, 2], [1, 2, 6]])

    print(findoutlier(mat_x))



    three_mat_x = np.array([[1, 10, 2, 8, 5], [2, 7, 3, 9, 11], [19, 2, 1, 1, 5]])

    print(findoutlier(three_mat_x))

Output

[0, 2]

[1, 0, 0]

UPDATE

As mentioned by @user3483203 you can use numpy.where:

import numpy as np





def findoutlier(lis):

    omaxs = np.absolute(np.max(lis, axis=1) - np.median(lis, axis=1))

    omins = np.absolute(np.min(lis, axis=1) - np.median(lis, axis=1))



    return np.where(omaxs > omins, np.argmax(lis, axis=1), np.argmin(lis, axis=1))





def main():

    mat_x = np.array([[10, 3, 2], [1, 2, 6]])

    print(findoutlier(mat_x))



    three_mat_x = np.array([[1, 10, 2, 8, 5], [2, 7, 3, 9, 11], [19, 2, 1, 1, 5]])

    print(findoutlier(three_mat_x))



main()

Output

[0 2]

[1 0 0]

edited Nov 13 '18 at 0:36

answered Nov 12 '18 at 23:18

Daniel Mesejo

16.2k21130

You need to specify the axis when using median, max and min:

import numpy as np





def findoutlier(lis):

    omaxs = np.absolute(np.max(lis, axis=1) - np.median(lis, axis=1))

    omins = np.absolute(np.min(lis, axis=1) - np.median(lis, axis=1))



    return [np.argmax(l) if omax > omin else np.argmin(l)  for omax, omin, l in  zip(omaxs, omins, lis)]





def main():

    mat_x = np.array([[10, 3, 2], [1, 2, 6]])

    print(findoutlier(mat_x))



    three_mat_x = np.array([[1, 10, 2, 8, 5], [2, 7, 3, 9, 11], [19, 2, 1, 1, 5]])

    print(findoutlier(three_mat_x))

Output

[0, 2]

[1, 0, 0]

UPDATE

As mentioned by @user3483203 you can use numpy.where:

import numpy as np





def findoutlier(lis):

    omaxs = np.absolute(np.max(lis, axis=1) - np.median(lis, axis=1))

    omins = np.absolute(np.min(lis, axis=1) - np.median(lis, axis=1))



    return np.where(omaxs > omins, np.argmax(lis, axis=1), np.argmin(lis, axis=1))





def main():

    mat_x = np.array([[10, 3, 2], [1, 2, 6]])

    print(findoutlier(mat_x))



    three_mat_x = np.array([[1, 10, 2, 8, 5], [2, 7, 3, 9, 11], [19, 2, 1, 1, 5]])

    print(findoutlier(three_mat_x))



main()

Output

[0 2]

[1 0 0]

edited Nov 13 '18 at 0:36

answered Nov 12 '18 at 23:18

Daniel Mesejo

16.2k21130

edited Nov 13 '18 at 0:36

answered Nov 12 '18 at 23:18

Daniel Mesejo

16.2k21130

answered Nov 12 '18 at 23:18

Daniel Mesejo

16.2k21130

answered Nov 12 '18 at 23:18

Daniel Mesejo

16.2k21130

Don't mix a list comprehension with vectorized operations. numpy.where is a much faster solution.

– user3483203
Nov 13 '18 at 0:00

add a comment |

Don't mix a list comprehension with vectorized operations. numpy.where is a much faster solution.

– user3483203
Nov 13 '18 at 0:00

Don't mix a list comprehension with vectorized operations. numpy.where is a much faster solution.

– user3483203
Nov 13 '18 at 0:00

add a comment |

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