排序多个列表的最快方法 - Python

Question

I have two lists, x and y, and I want to sort x and permute y by the permutation of x-sorting. For example, given

x = [4, 2, 1, 3]
y = [40, 200, 1, 30]

I want to get

x_sorted = [1,2,3,4]
y_sorted = [1, 200, 30, 40]

As discussed in past questions, a simple way to solve this is

x_sorted, y_sorted = zip(*sorted(zip(x,y)))

Here is my question: What is the FASTEST way to do this?

---

I have three methods to do the task.

import numpy as np
x = np.random.random(1000)
y = np.random.random(1000)

method 1:

x_sorted, y_sorted = zip(*sorted(zip(x,y))) #1.08 ms 

method 2:

foo = zip(x,y)
foo.sort()
zip(*foo)       #1.05 ms

method 3;

ind = range(1000)
ind.sort(key=lambda i:x[i])
x_sorted = [x[i] for i in ind]
y_sorted = [y[i] for i in ind]  #934us

Is there a better method, which executes faster than above three methods?

---

Additional questions.

1. Why method 2 is not faster than method 1 though it uses sort method?
2. If I execute method 2 separately, it is faster. In IPython terminal,

I have

%timeit foo = zip(x,y)   #1000 loops, best of 3: 220 us per loop
%timeit foo.sort()       #10000 loops, best of 3: 78.9 us per loop
%timeit zip(*foo)        #10000 loops, best of 3: 73.8 us per loop

Answer 1

Using numpy.argsort :

>>> import numpy as np
>>> x = np.array([4,2,1,3])
>>> y = np.array([40,200,1,30])
>>> order = np.argsort(x)
>>> x_sorted = x[order]
>>> y_sorted = y[order]
>>> x_sorted
array([1, 2, 3, 4])
>>> y_sorted
array([  1, 200,  30,  40])

---

>>> timeit('order = np.argsort(x); x_sorted = x[order]; y_sorted = y[order]', 'from __main__ import x, y, np', number=1000)
0.030632019043

NOTE

This makes sense if input data are already numpy arrays.

Answer 2

>>> x = [4, 2, 1, 3]
>>> y = [40, 200, 1, 30]    
>>> x_sorted, y_sorted = zip(*sorted(zip(x, y), key=lambda a:a[0]))
>>> x_sorted
(1, 2, 3, 4)
>>> y_sorted
(1, 200, 30, 40)

Performance:

>>> timeit('foo = zip(x,y); foo.sort(); zip(*foo)', 'from __main__ import x, y', number=1000)
1.0197240443760691
>>> timeit('zip(*sorted(zip(x,y)))', 'from __main__ import x, y', number=1000)
1.0106219310922597
>>> timeit('ind = range(1000); ind.sort(key=lambda i:x[i]); x_sorted = [x[i] for i in ind]; y_sorteds = [y[i] for i in ind]', 'from __main__ import x, y', number=1000)
0.9043525504607857
>>> timeit('zip(*sorted(zip(x, y), key=lambda a:a[0]))', 'from __main__ import x, y', number=1000)
0.8288150863453723

To see the full picture:

>>> timeit('sorted(x)', 'from __main__ import x, y', number=1000)
0.40415491505723367            # just getting sorted list from x
>>> timeit('x.sort()', 'from __main__ import x, y', number=1000)
0.008009909448446706           # sort x inplace

@falsetru method - fastest for np.arrays

>>> timeit('order = np.argsort(x); x_sorted = x[order]; y_sorted = y[order]', 'from __main__ import x, y, np', number=1000)
0.05441799872323827

As @AshwiniChaudhary suggested in comments, for lists there's a way to speed it up by using itertools.izip instead of zip :

>>> timeit('zip(*sorted(izip(x, y), key=itemgetter(0)))', 'from __main__ import x, y;from operator import itemgetter;from itertools import izip', number=1000)
0.4265049757161705

Answer 3

You're not timing this properly

%timeit foo.sort()

After the 1st loop, it's already sorted for the remainder. Timsort is very efficient for presorted lists.

I was a little surprised that @Roman's use of a key function was so much faster. You can improve on this further by using itemgetter

from operator import itemgetter
ig0 = itemgetter(0)
zip(*sorted(zip(x, y), key=ig0))

This is about 9% faster than using a lambda function for lists of 1000 elements