用Python排序的最快方法（没有cython）

Question

I have a problem where I've to sort a very big array(shape - 7900000X4X4) with a custom function. I used sorted but it took more than 1 hour to sort. My code was something like this.

def compare(x,y):
    print('DD '+str(x[0]))
    if(np.array_equal(x[1],y[1])==True):
        return -1
    a = x[1].flatten()
    b = y[1].flatten()
    idx = np.where( (a>b) != (a<b) )[0][0]
    if a[idx]<0 and b[idx]>=0:
        return 0
    elif b[idx]<0 and a[idx]>=0:
        return 1
    elif a[idx]<0 and b[idx]<0:
        if a[idx]>b[idx]:
            return 0
        elif a[idx]<b[idx]:
            return 1
    elif a[idx]<b[idx]:
        return 1
    else:
        return 0
def cmp_to_key(mycmp):
    class K:
        def __init__(self, obj, *args):
            self.obj = obj
        def __lt__(self, other):
            return mycmp(self.obj, other.obj)
    return K
tblocks = sorted(tblocks.items(),key=cmp_to_key(compare))

This worked but I want it to complete in seconds. I don't think any direct implementation in python can give me the performance I need, so I tried cython. My Cython code is this, which is pretty simple.

cdef int[:,:] arrr
cdef int size

cdef bool compare(int a,int b):
    global arrr,size
    cdef int[:] x = arrr[a]
    cdef int[:] y = arrr[b]
    cdef int i,j
    i = 0
    j = 0
    while(i<size):
        if((j==size-1)or(y[j]<x[i])):
            return 0
        elif(x[i]<y[j]):
            return 1
        i+=1
        j+=1
    return (j!=size-1)

def sorted(np.ndarray boxes,int total_blocks,int s):
    global arrr,size
    cdef int i
    cdef vector[int] index = xrange(total_blocks)
    arrr = boxes
    size = s
    sort(index.begin(),index.end(),compare)
    return index

This code in cython took 33 seconds! Cython is the solution, but I am looking for some alternate solutions which can run directly on python. For example numba. I tried Numba, but I didn't get satisfying results. Kindly help!

Answer 1

It is hard to give an answer without a working example. I assume, that arrr in your Cython code was a 2D-array and I assume that size was size=arrr.shape[0]

Numba Implementation

import numpy as np
import numba as nb
from numba.targets import quicksort


def custom_sorting(compare_fkt):
  index_arange=np.arange(size)

  quicksort_func=quicksort.make_jit_quicksort(lt=compare_fkt,is_argsort=False)
  jit_sort_func=nb.njit(quicksort_func.run_quicksort)
  index=jit_sort_func(index_arange)

  return index

def compare(a,b):
    x = arrr[a]
    y = arrr[b]
    i = 0
    j = 0
    while(i<size):
        if((j==size-1)or(y[j]<x[i])):
            return False
        elif(x[i]<y[j]):
            return True
        i+=1
        j+=1
    return (j!=size-1)


arrr=np.random.randint(-9,10,(7900000,8))
size=arrr.shape[0]

index=custom_sorting(compare)

This gives 3.85s for the generated testdata. But the speed of a sorting algorithm heavily depends on the data....

Simple Example

import numpy as np
import numba as nb
from numba.targets import quicksort

#simple reverse sort
def compare(a,b):
  return a > b

#create some test data
arrr=np.array(np.random.rand(7900000)*10000,dtype=np.int32)
#we can pass the comparison function
quicksort_func=quicksort.make_jit_quicksort(lt=compare,is_argsort=True)
#compile the sorting function
jit_sort_func=nb.njit(quicksort_func.run_quicksort)
#get the result
ind_sorted=jit_sort_func(arrr)

This implementation is about 35% slower than np.argsort, but this is also common in using np.argsort in compiled code.

Answer 2

If I understand your code correctly then the order you have in mind is the standard order, only that it starts at 0 wraps around at +/-infinity and maxes out at -0 . On top of that we have simple left-to-right lexicographic order.

Now, if your array dtype is integer, observe the following: Because of complement representation of negatives view-casting to unsigned int makes your order the standard order. On top of that, if we use big endian encoding, efficient lexicographic ordering can be achieved by view-casting to void dtype.

The code below shows that using a 10000x4x4 example that this method gives the same result as your Python code.

It also benchmarks it on a 7,900,000x4x4 example (using array, not dict). On my modest laptop this method takes 8 seconds.

import numpy as np

def compare(x, y):
#    print('DD '+str(x[0]))
    if(np.array_equal(x[1],y[1])==True):
        return -1
    a = x[1].flatten()
    b = y[1].flatten()
    idx = np.where( (a>b) != (a<b) )[0][0]
    if a[idx]<0 and b[idx]>=0:
        return 0
    elif b[idx]<0 and a[idx]>=0:
        return 1
    elif a[idx]<0 and b[idx]<0:
        if a[idx]>b[idx]:
            return 0
        elif a[idx]<b[idx]:
            return 1
    elif a[idx]<b[idx]:
        return 1
    else:
        return 0
def cmp_to_key(mycmp):
    class K:
        def __init__(self, obj, *args):
            self.obj = obj
        def __lt__(self, other):
            return mycmp(self.obj, other.obj)
    return K

def custom_sort(a):
    assert a.dtype==np.int64
    b = a.astype('>i8', copy=False)
    return b.view(f'V{a.dtype.itemsize * a.shape[1]}').ravel().argsort()

tblocks = np.random.randint(-9,10, (10000, 4, 4))
tblocks = dict(enumerate(tblocks))

tblocks_s = sorted(tblocks.items(),key=cmp_to_key(compare))

tblocksa = np.array(list(tblocks.values()))
tblocksa = tblocksa.reshape(tblocksa.shape[0], -1)
order = custom_sort(tblocksa)
tblocks_s2 = list(tblocks.items())
tblocks_s2 = [tblocks_s2[o] for o in order]

print(tblocks_s == tblocks_s2)

from timeit import timeit

data = np.random.randint(-9_999, 10_000, (7_900_000, 4, 4))

print(timeit(lambda: data[custom_sort(data.reshape(data.shape[0], -1))],
             number=5) / 5)

Sample output:

True
7.8328493310138585