How to reverse and forward sort NumPy 2D array in Python?

Question

I have the data of the form:

#---------------------
# Data
#---------------------
p   q   r
2   8   14
2   9   22
1   5   19
2   7   19
3   11  13
2   7   20
1   4   15
3   12  17
1   4   14
1   5   20
2   7   17
3   10  13
3   11  20
3   11  14
1   6   18
3   12  16
2   9   21
3   10  19
2   8   13
1   6   22
1   4   13
2   8   15
3   12  15
3   10  16
2   9   16
1   5   16
1   6   21

Now I need to sort this data using NumPy in the following manner:

Ascending order for column p.
Ascending order for column q.
Descending order for column r.

I used the following code, but it does not sort correctly:

import numpy as np

data = open('data.dat', "r")
line = data.readline()
while line.startswith('#'):
    line = data.readline()
data_header = line.split("\t")
data_header[-1] = data_header[-1].strip()

_data_ = np.genfromtxt(data, comments='#', delimiter='\t', names = data_header, dtype = None, unpack = True).transpose()            # Read space-separated values in engine data file.
sorted_index =  np.lexsort((_data_['r'][::-1], _data_['q'], _data_['p']))
_data_ =  _data_[sorted_index]
print (_data_)

Ouptut

What could be possibly wrong in this sorting method?

Max · Accepted Answer · 2018-07-13 23:47:48Z

Actually, you were almost there! Just modify this one line in your code (a minus sign instead of a [::-1]) and it works

sorted_index =  np.lexsort((-_data_['r'], _data_['q'], _data_['p']))

More generically formulated, this minus sign is a bit of a hack but it should work as long as you are only dealing with numerical values

# 2D array will be sorted first by p, then by q (if p is the same), then by r
sortkeys = ['p','q','r']
# 1 is ascending/forward sort, -1 is descending/reverse sort
sortdirection = [1,1,-1]
# need [::-1] as its sorts with last element first...
ind = np.lexsort(tuple([(_data_[skey])*sdir for skey,sdir in
                                     zip(sortkeys[::-1],sortdirection[::-1])]))
_data_ = _data_[ind]
for i in _data_:
    print(i)

Output:

(1, 4, 15)
(1, 4, 14)
(1, 4, 13)
(1, 5, 20)
(1, 5, 19)
(1, 5, 16)
(1, 6, 22)
(1, 6, 21)
(1, 6, 18)
(2, 7, 20)
(2, 7, 19)
(2, 7, 17)
(2, 8, 15)
(2, 8, 14)
(2, 8, 13)
(2, 9, 22)
(2, 9, 21)
(2, 9, 16)
(3, 10, 19)
(3, 10, 16)
(3, 10, 13)
(3, 11, 20)
(3, 11, 14)
(3, 11, 13)
(3, 12, 17)
(3, 12, 16)
(3, 12, 15)

Finwood · Accepted Answer · 2015-04-03 11:55:47Z

0

The problem is, that when you reverse the r column, numpy doesn't know your indices change. A workaround would be a sort in two steps, but it wouldn't be elegant:

pre_sort_index = np.lexsort((_data_['r'],), axis=0)[::-1]
sorted_index =  np.lexsort((_data_[pre_sort_index]['q'], _data_[pre_sort_index]['p']))
_data_ =  _data_[pre_sort_index][sorted_index]

answered Apr 3, 2015 at 11:55

Finwood

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