有没有办法可以使用numpy函数在mergesort中进行合并?
像merge这样的函数:
a = np.array([1,3,5])
b = np.array([2,4,6])
c = merge(a, b) # c == np.array([1,2,3,4,5,6])
我希望通过numpy
可以获得大数据的高性能答案 0 :(得分:10)
您可以使用
c = concatenate((a,b))
c.sort(kind='mergesort')
我担心你不能做得比这更好,除非你把你自己的排序函数写成python扩展,àlacython
。
请参阅this问题以了解类似问题,但仅保留合并数组中的唯一值。基准和评论也很有见地。
答案 1 :(得分:5)
sortednp包实现了已排序的numpy数组的有效合并:
import numpy as np
import sortednp
a = np.array([1,3,5])
b = np.array([2,4,6])
c = sortednp.merge(a, b) # c == np.array([1,2,3,4,5,6])
受桑德(Sander)的启发,我测量了numpy的mergesort(v1.17.4),桑德的答案和sortednp(v0.2.1),使用以下代码针对不同的数组大小以及a和b之间的大小比率进行了测量:
from timeit import timeit as t
import sortednp as snp
import numpy as np
def numpy_mergesort(a, b):
c = np.concatenate((a,b))
c.sort(kind='mergesort')
return c
def sanders_merge(a, b):
if len(a) < len(b):
b, a = a, b
c = np.empty(len(a) + len(b), dtype=a.dtype)
b_indices = np.arange(len(b)) + np.searchsorted(a, b)
a_indices = np.ones(len(c), dtype=bool)
a_indices[b_indices] = False
c[b_indices] = b
c[a_indices] = a
return c
results = []
for size_factor in range(3):
for max_digits in range(3, 8):
size = 10**max_digits
# size difference of a factor 10 here makes the difference!
a = np.arange(size // 10**size_factor, dtype=np.int)
b = np.arange(size, dtype=np.int)
assert np.array_equal(numpy_mergesort(a, b), sanders_merge(a, b))
assert np.array_equal(numpy_mergesort(a, b), snp.merge(a, b))
classic = t(lambda: numpy_mergesort(a, b), number=10)
sanders = t(lambda: sanders_merge(a, b), number=10)
snp_result = t(lambda: snp.merge(a, b), number=10)
results.append((size_factor, max_digits, classic, sanders, snp_result))
text_format = " ".join(["{:<18}"] * 5)
print(text_format.format("log10(size factor)", "log10(max size)", "np mergesort", "Sander's merge", "sortednp"))
table_format = " ".join(["{:.5f}"] * 5)
for result in results:
print(table_format.format(*result))
结果表明,sortednp始终是最快的实现:
log10(size factor) log10(max size) np mergesort Sander's merge sortednp
0.00000 3.00000 0.00016 0.00062 0.00005
0.00000 4.00000 0.00135 0.00469 0.00029
0.00000 5.00000 0.01160 0.03813 0.00292
0.00000 6.00000 0.14952 0.54160 0.03527
0.00000 7.00000 2.00566 5.91691 0.67119
1.00000 3.00000 0.00005 0.00017 0.00002
1.00000 4.00000 0.00019 0.00058 0.00014
1.00000 5.00000 0.00304 0.00633 0.00137
1.00000 6.00000 0.03743 0.06893 0.01828
1.00000 7.00000 0.62334 1.01523 0.38732
2.00000 3.00000 0.00004 0.00015 0.00002
2.00000 4.00000 0.00012 0.00028 0.00013
2.00000 5.00000 0.00217 0.00275 0.00122
2.00000 6.00000 0.03457 0.03205 0.01524
2.00000 7.00000 0.51307 0.50120 0.34335
答案 2 :(得分:2)
当一个数组比另一个数组大得多时,可以通过执行np.searchorted获得不错的速度(在我的电脑上是5倍),这主要是通过搜索较小数组的插入索引来限制速度的: / p>
ItemsControl
计时给出:
import numpy as np
def classic_merge(a, b):
c = np.concatenate((a,b))
c.sort(kind='mergesort')
return c
def new_merge(a, b):
if len(a) < len(b):
b, a = a, b
c = np.empty(len(a) + len(b), dtype=a.dtype)
b_indices = np.arange(len(b)) + np.searchsorted(a, b)
a_indices = np.ones(len(c), dtype=bool)
a_indices[b_indices] = False
c[b_indices] = b
c[a_indices] = a
return c
当a和b的长度大致相等时,差异较小:
from timeit import timeit as t
results = []
for size_digits in range(2, 8):
size = 10**size_digits
# size difference of a factor 10 here makes the difference!
a = np.arange(size // 10, dtype=np.int)
b = np.arange(size, dtype=np.int)
classic = t(lambda: classic_merge(a, b), number=10)
new = t(lambda: new_merge(a, b), number=10)
results.append((size_digits, classic, new))
if True:
text_format = " ".join(["{:<15}"] * 3)
print(text_format.format("log10(size)", "Classic", "New"))
table_format = " ".join(["{:.5f}"] * 3)
for result in results:
print(table_format.format(*result))
log10(size) Classic New
2.00000 0.00009 0.00027
3.00000 0.00021 0.00030
4.00000 0.00233 0.00082
5.00000 0.02827 0.00601
6.00000 0.33322 0.06059
7.00000 4.40571 0.86764