I have a list
[[0, 3], [5, 1], [2, 1], [4, 5]]
which I have made into an array using numpy.array:
[[0 3]
[5 1]
[2 1]
[4 5]]
How do I sort this like a table? In particular, I want to sort by the second column in ascending order and then resolve any ties by having the first column sorted in ascending order. Thus I desire:
[[2 1]
[5 1]
[0 3]
[4 5]]
Any help would be greatly appreciated!
See http://docs.scipy.org/doc/numpy/reference/generated/numpy.lexsort.html#numpy.lexsort
Specifically in your case,
import numpy as np
x = np.array([[0,3],[5,1],[2,1],[4,5]])
x[np.lexsort((x[:,0],x[:,1]))]
outputs
array([[2,1],[5,1],[0,3],[4,5]])
You can use numpy.lexsort():
>>> a = numpy.array([[0, 3], [5, 1], [2, 1], [4, 5]])
>>> a[numpy.lexsort(a.T)]
array([[2, 1],
[5, 1],
[0, 3],
[4, 5]])
Another way of doing this - slice out the bits of data you want, get the sort indices using argsort, then use the result of that to slice your original array:
a = np.array([[0, 3], [5, 1], [2, 1], [4, 5]])
subarray = a[:,1] # 3,1,1,5
indices = np.argsort(subarray) # Returns array([1,2,0,3])
result = a[indices]
Or, all in one go:
a[np.argsort(a[:,1])]
If you want to sort using a single column only (e.g., second column), you can do something like:
from operator import itemgetter
a = [[0, 3], [5, 1], [2, 1], [4, 5]]
a_sorted = sorted(a, key=itemgetter(1))
If there are more than one key, then use numpy.lexsort() as pointed out in the other answers.
numpy.array(sorted(a, key=itemgetter(1))) worked perfectly still being the most readable option.