I have a fairly simple question about how to sort an entire array/recarray by a given column. For example, given the array:
import numpy as np
data = np.array([[5,2], [4,1], [3,6]])
I would like to sort data by the first column to return:
array([[3,6], [4,1], [5,2]])
Use data[np.argsort(data[:, 0])] where the 0 is the column index on which to sort:
In [27]: import numpy as np
In [28]: data = np.array([[5,2], [4,1], [3,6]])
In [29]: col = 0
In [30]: data=data[np.argsort(data[:,col])]
Out[30]:
array([[3, 6],
[4, 1],
[5, 2]])
np.lexsort .you are looking for operator.itemgetter
>>> from operator import itemgetter, attrgetter
>>> sorted(student_tuples, key=itemgetter(2))
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]
>>> sorted(student_objects, key=attrgetter('age'))
[('dave', 'B', 10), ('jane', 'B', 12), ('john', 'A', 15)]
i.e.
In [7]: a
Out[7]: [[5, 2], [4, 1], [3, 6]]
In [8]: sorted(a, key=operator.itemgetter(0))
Out[8]: [[3, 6], [4, 1], [5, 2]]
This is somewhat tricky:
data[data[:,0].argsort()]
# data[:,n] -- get entire column of index n
# argsort() -- get the indices that would sort it
# data[data[:,n].argsort()] -- get data array sorted by n-th column
I found this recipe here:
To sort on the second column use itemgetter
>>> from operator import itemgetter
>>> data = [[5,2], [4,1], [3,6]]
>>> sorted(data)
[[3, 6], [4, 1], [5, 2]]
>>> sorted(data,key=itemgetter(1))
[[4, 1], [5, 2], [3, 6]]
>>>
Here's an extension that works with slices:
import numpy as np
x = np.array([[9, 1, 2],
[5, 3, 4],
[0, 5, 6]])
Sorting by rows:
x[:, x[1,:].argsort()] # Sort by second row
array([[1, 2, 9]
[3, 4, 5]
[5, 6, 0]])
Sorting by columns:
x[x[:,0].argsort(), :] # Sort by first column
array([[0, 5, 6],
[5, 3, 4],
[9, 1, 2]])
