CODEWITHSUNDEEP
pythonpython-2.7sortingnumpy
Jonathan Lindgren
Jonathan Lindgren

Reputation: 1252

Sorting list according to other list in python not working when second list contains numpy arrays

The standard way to sort a list a based on list b, is to zip them together and use sorted, like

sorted(zip(b,a))

However, if two elements in b are equal, it will automatically proceed to try to sort according to a, and that sometimes causes problems. Consider for example

import numpy as np
a=[np.array([4,5]),np.array([3,4])]
b=[1,1]
e=zip(b,a)
sorted(e)

this causes the error

Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: The truth value of an array with more than one element is ambiguous. Use a.any() or a.all()

So when two elements in b are equal, it seems to proceed to sort according to a, but since a contains numpy arrays, the sorting algorithm complain when it tries to compare them.

Anyway, is this expected behaviour of sorted? In any case, I just want to sort according to b. If two elements are equal, the order does not matter. Any ideas on how to fix this without writing a sorting algorithm myself?

Upvotes: 1

Views: 126

Answers (1)

akuiper
akuiper

Reputation: 214927

The error shows up because numpy.array is not comparable in the sense of sorting, i.e. it doesn't return a True or False value (but another boolean array) when comparing two arrays, which is different from lists comparison.

If you only want to sort according to b, provide a key function to sorted:

sorted(e, key=lambda x: x[0])
# [(1, array([4, 5])), (1, array([3, 4]))]

Upvotes: 1

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