Question

import distance
from collections import defaultdict

my_list = ['ACAA', 'TCAA','TCAT','TGAT','TCGA','TGGA','GCGA','AAAA','GGGG','GGGC']
counts = {'ACAA':60, 'TCAA':3,'TCAT':30,'TGAT':8,'TCGA':1,'TGGA':1,'GCGA':8,'AAAA':5,'GGGG':8,'GGGC':1}

adj_list = defaultdict(list)
for strng1 in my_list:
    for strng2 in my_list:
        if distance.hamming(strng1, strng2) == 1 and counts[strng1] >= (counts[strng2]*2):
            adj_list[strng1].append(strng2)

我有这个实现来获取方向性邻接列表。预期结果：

ACAA: TCAA
TCAA: TCGA
TCAT: TCAA, TGAT
TGAT
TCGA: TGGA
TGGA: TCGA
GCGA: TCGA
AAAA
GGGG: GGGC
GGGC

是否有更快的实施？对于大型数据集，这变得非常慢。在cython中重写这个会加速吗？如果是，有人可以帮助我开始使用cython吗？

Answer 1

我不了解Cython，但您可以避免在内循环中访问dict项目：

adj_list = defaultdict(list)
for strng1 in my_list:
    a1 = adj_list[strng1]
    c1 = counts[strng1]
    for strng2 in my_list:
        if distance.hamming(strng1, strng2) == 1 and c1 >= (counts[strng2]*2):
            a1.append(strng2)

你甚至可以通过仅在下半部分迭代并执行对称附加来削减更多。这样你可以节省50％的距离计算，因为它是对称的。你只在上面的矩阵三角形上执行它（对角线排除，我认为一个字符串与自身的距离是0）而不是完整的矩阵。

for i,strng1 in enumerate(my_list):
    ...
    for j in range(i+1,len(my_list)):

我的尝试，我不确定，但应该关闭：

adj_list = defaultdict(list)
for i,strng1 in enumerate(my_list):
    a1 = adj_list[strng1]
    c1 = counts[strng1]
    for j in range(i+1,len(my_list)):
        strng2 = my_list[j]
        if distance.hamming(strng1, strng2) == 1:
            c2 = counts[strng2]
            if c1 >= (c2*2):
                a1.append(strng2)
            if c2 >= (c1*2):
                adj_list[strng1].append(strng2)

crysis405编辑：

原件：

def adj_lst(my_list, counts):
    adj_list = defaultdict(list)
    for strng1 in my_list:
        a1 = adj_list[strng1]
        c1 = counts[strng1]
        for strng2 in my_list:
            if distance.hamming(strng1, strng2) == 1 and c1 >= (counts[strng2]*2):
                adj_list[strng1].append(strng2)

建议的改进：

def adj_lst_fast(my_list, counts):
    adj_list_fast = defaultdict(list)
    for i,strng1 in enumerate(my_list):
        a1 = adj_list_fast[strng1]
        c1 = counts[strng1]
        for j in range(i+1,len(my_list)):
            strng2 = my_list[j]
            if distnace.hamming(strng1, strng2:
                c2 = counts[strng2]
                if c1 >= (c2*2):
                    adj_list_fast[strng1].append(strng2)
                elif c2 >= (c1*2):
                    adj_list_fast[strng2].append(strng1)

性能：

print(timeit.timeit('adj_lst(my_list, counts)', number = 10000, 
setup="from __main__ import adj_lst, my_list, counts"))

1.2892486669989012

print(timeit.timeit('adj_lst_fast(my_list, counts)', number = 10000, 
setup="from __main__ import adj_lst_fast, my_list, counts"))

0.6437049919986748

定向邻接列表快速实现

1 个答案: