Find a minimum value in a column element in a numpy array, and store corresponding row as a new array

Question

I have a numpy array like below, and I would like to make a new numpy array which has a row that has the minimum value in the last column which is element of xx[:,4].

xx =
[[6.18167195e-02 -3.20902583e-01  7.96803103e+00   5.69096614e+00  6.82949858e+00]
[1.14139479e-02 -2.93352490e-02  5.17031336e+00   2.50552347e+01  1.51127740e+01]
[8.84761009e-03 -2.93352490e-02  1.84764173e+01   2.50552347e+01  2.17658260e+01]
[1.96567549e-03 -2.93352490e-02  8.18878876e+01   2.50552347e+01  5.34715612e+01]
[3.54827629e-01 -1.88511194e+00  4.70728062e-01   1.95791971e-01  3.33260017e-01]
[3.53146766e-01 -1.88511194e+00  9.42210619e-01   1.95791971e-01  5.69001295e-01]
[6.64244146e-02 -3.20902583e-01  1.10815151e+00   5.69096614e+00  3.39955882e+00]
[6.18167195e-02 -3.20902583e-01  7.96803103e+00   5.69096614e+00  6.82949858e+00]
[1.95005819e-02 -1.40482917e-01  2.64188251e+00   1.63546768e+00  2.13867510e+00]
...
...]

I know that we can use np.min(xx[:,4]) to get 3.33260017e-01 in the last column. However, I need to extract all values in the corresponding row, which will be this

[3.54827629e-01 -1.88511194e+00 4.70728062e-01 1.95791971e-01 3.33260017e-01].

In a multidimensional array, such as xx.shape = (1000,4), how can I get the element locations of the minimum value?

My question is either one of them.

Make a new numpy array only with the row that has the minimum value in the last column.
Delete every rows except the row that has the minimum value in the last column.

xx[xx[:, -1].argmin()]?

javidcf
– javidcf

2019-11-20 18:31:17 +00:00
Commented Nov 20, 2019 at 18:31 — javidcf
– javidcf, Commented Nov 20, 2019 at 18:31

Brad Campbell · Accepted Answer · 2019-11-20 19:46:13Z

2

argmin will give you the index where the minimum value is.

>>> xx=np.array([[2,1,3],[1,0,2],[3,3,1]])
>>> xx
array([[2, 1, 3],
       [1, 0, 2],
       [3, 3, 1]])
>>> column = 1
>>> i = xx[:,column].argmin()
1
>>> xx[i,:]  # get the row where column 1 has the minimum
array([1, 0, 2])

edited Nov 20, 2019 at 19:46

answered Nov 20, 2019 at 18:21

Brad Campbell

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1 Comment

javidcf Over a year ago

This does more work than necessary, you only need to do argmin on the relevant column (xx[xx[:, i].argmin()]).

Kris · Accepted Answer · 2019-11-20 18:38:18Z

1

One solution:

>>> import numpy as np
>>>
>>> xx = [[6.18167195e-02,-3.20902583e-01,7.96803103e+00,5.69096614e+00,6.82949858e+00],
... [1.14139479e-02,-2.93352490e-02,5.17031336e+00,2.50552347e+01,1.51127740e+01],
... [8.84761009e-03,-2.93352490e-02,1.84764173e+01,2.50552347e+01,2.17658260e+01],
... [1.96567549e-03,-2.93352490e-02,8.18878876e+01,2.50552347e+01,5.34715612e+01],
... [3.54827629e-0,-1.88511194e+00,4.70728062e-01,1.95791971e-01,3.33260017e-01],
... [3.53146766e-01,-1.88511194e+00,9.42210619e-01,1.95791971e-01,5.69001295e-01],
... [6.64244146e-02,-3.20902583e-01,1.10815151e+00,5.69096614e+00,3.39955882e+00],
... [6.18167195e-02,-3.20902583e-01,7.96803103e+00,5.69096614e+00,6.82949858e+00],
... [1.95005819e-02,-1.40482917e-01,2.64188251e+00,1.63546768e+00,2.13867510e+00]]
>>>
>>> xx = np.asarray(xx)
>>>
>>> def MinRow(array):
...     low = np.min(array[:,4])
...     for idx, el in enumerate(array):
...         if el[-1] <= low:
...             index = idx
...     newarray = array[index, :]
...     return newarray
...
>>> xx2 = MinRow(xx)
>>> print(xx2)
[ 3.54827629 -1.88511194  0.47072806  0.19579197  0.33326002]

Pure Numpy Solution:

>>> import numpy as np
>>>
>>> xx = [[6.18167195e-02,-3.20902583e-01,7.96803103e+00,5.69096614e+00,6.82949858e+00],
... [1.14139479e-02,-2.93352490e-02,5.17031336e+00,2.50552347e+01,1.51127740e+01],
... [8.84761009e-03,-2.93352490e-02,1.84764173e+01,2.50552347e+01,2.17658260e+01],
... [1.96567549e-03,-2.93352490e-02,8.18878876e+01,2.50552347e+01,5.34715612e+01],
... [3.54827629e-0,-1.88511194e+00,4.70728062e-01,1.95791971e-01,3.33260017e-01],
... [3.53146766e-01,-1.88511194e+00,9.42210619e-01,1.95791971e-01,5.69001295e-01],
... [6.64244146e-02,-3.20902583e-01,1.10815151e+00,5.69096614e+00,3.39955882e+00],
... [6.18167195e-02,-3.20902583e-01,7.96803103e+00,5.69096614e+00,6.82949858e+00],
... [1.95005819e-02,-1.40482917e-01,2.64188251e+00,1.63546768e+00,2.13867510e+00]]
>>>
>>> xx = np.asarray(xx)
>>> idx = np.where(xx[:,4]==np.min(xx[:,4]))
>>> xx2 = xx[idx]
>>> print(xx2)
[[ 3.54827629 -1.88511194  0.47072806  0.19579197  0.33326002]]

edited Nov 20, 2019 at 18:38

answered Nov 20, 2019 at 18:13

Kris

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3 Comments

Aaron Over a year ago

Please post all your code as text rather than screenshots of text. Images are not consistently formatted correctly across devices, and are impossible for assistive technologies such as screen readers to interpret.

Kris Over a year ago

Thanks @Aaron. Will fix.

mario119 Over a year ago

@Kris Your answer was very helpful! and it worked! Thank you!

Ben Taylor · Accepted Answer · 2019-11-21 08:46:00Z

1

You're nearly there, you want np.where

idx = np.where(xx[:, -1] == np.min(xx[:,4]))
output = xx[idx]

edited Nov 21, 2019 at 8:46

answered Nov 20, 2019 at 18:06

Ben Taylor

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3 Comments

Aaron Over a year ago

np.argmin(xx, axis=-2): more concise, and less computation.

Kris Over a year ago

need to change to: idx=np.where(xx[:,4] == np.min(xx[:,4]))

Ben Taylor Over a year ago

Ah, you're right, my test I used -1 there, not sure how I converted to a zero here...

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