编辑以澄清输入/输出。我认为无论如何这都会有些缓慢，但到目前为止我还没有真正考虑过我的python脚本中的速度，而我正试图找出加速这类操作的方法。
我的输入是基因组序列的腌制词典。我目前正在研究两种基因组，即芽殖酵母基因组（磁盘上11.5 MB）和人类基因组（磁盘上2.8 GB）。这些词典的形式如下：
seq_d = { 'chr1' : 'ATCGCTCGCTGCTCGCT', 'chr2' : 'CGATCAGTCATGCATGCAT',
'chr3' : 'ACTCATCATCATCATACTGGC' }
我想在基因组的两条链中找到核苷酸的所有单碱基实例。其中+链指的是上述字典中的序列，而-链是序列的反向互补。我的输出是嵌套字典，其中顶级keys是+或-，嵌套的keys是染色体名称，值是0索引位置的列表：
nts = 'T'
test_d = {'+': {'chr3': [2, 5, 8, 11, 14, 17], 'chr2': [3, 7, 10, 14, 18],
'chr1': [1, 5, 9, 12, 16]}, '-': {'chr3': [0, 4, 7, 10, 13, 15],
'chr2': [2, 5, 9, 13, 17], 'chr1': [0]}}
test_d定义了一组位置，以便稍后在脚本中的大型Illumina测序数据集中进行检查。
我的第一次尝试使用enumerate和迭代。
import time
import numpy as np
rev_comps = { 'A' : 'T', 'T' : 'A', 'G' : 'C', 'C' : 'G', 'N' : 'N'}
test_d = { '+' : {}, '-' : {}}
nts = 'T'
s = time.time()
for chrom in seq_d:
plus_pos, minus_pos = [], []
chrom_seq = seq_d[chrom]
for pos, nt in enumerate(chrom_seq):
if nt in nts:
plus_pos.append(pos)
if rev_comps[nt] in nts:
minus_pos.append(pos)
test_d['+'][chrom] = plus_pos
test_d['-'][chrom] = minus_pos
e = time.time()
print 'The serial version took {} minutes...'.format((e-s)/60)
酵母的输出：
The serial version took 0.0455190300941 minutes...
人类的输出：
The serial version took 10.1694028815 minutes...
我尝试使用numpy数组而不是enumerate()和迭代：
s = time.time()
for chrom in seq_d:
chrom_seq = np.array(list(seq_d[chrom]))
nts = list(nts)
rev_nts = [rev_comps[nt] for nt in nts]
plus_pos = list(np.where(np.in1d(chrom_seq, nts) == True)[0])
minus_pos = list(np.where(np.in1d(chrom_seq, rev_nts) == True)[0])
test_d['+'][chrom] = plus_pos
test_d['-'][chrom] = minus_pos
e = time.time()
print 'The numpy version took {} minutes...'.format((e-s)/60)
酵母的输出：
The numpy version took 0.0309354345004 minutes...
人类的输出：
The numpy version took 9.86174853643 minutes...
为什么numpy版本会失去其对更大的人类基因组的性能优势？有更快的方法吗？我尝试使用multiprocessing.Pool实现一个版本，但这比任何一个版本都慢：
def getTestHelper(chrom_seq, nts, rev_comp):
rev_comps = { 'A' : 'T', 'T' : 'A', 'G' : 'C', 'C' : 'G', 'N' : 'N'}
if rev_comp:
nts = [rev_comps[nt] for nt in nts]
else:
nts = list(nts)
chrom_seq = np.array(list(chrom_seq))
mask = np.in1d(chrom_seq, nts)
pos_l = list(np.where(mask == True)[0])
return pos_l
s = time.time()
pool = Pool(4)
plus_pos = pool.map(functools.partial(getTestHelper, nts=nts, rev_comp=False), seq_d.values())
minus_pos = pool.map(functools.partial(getTestHelper, nts=nts, rev_comp=True), seq_d.values())
e = time.time()
print 'The parallel version took {} minutes...'.format((e-s)/60)
我没有在人类基因组上运行，但酵母版本较慢：
The parallel version took 0.652778700987 minutes...
答案 0 :(得分：3)
请尝试str.find或str.index，而不是手动迭代长字符串。不要自己切割字符串，使用这些方法＆＃39;内置切片。
这也会导致enumerate - 虽然不管怎么说都不值钱。
此外，您可以使用set来存储索引，而不是列表 - 添加可能会更快。
您也可以尝试regular expressions做同样的事情（如果您要尝试这个，请尝试两个正则表达式（对于＆＃34; T＆＃34;和＆＃34; A＆＃34;）一个用于＆＃34; T | A＆＃34;）。
此外，而不是做
for chrom in seq_d:
你可以做到
for chromosome_number, chomosome in seq_d.items():
这与性能无关，但使代码更具可读性。
答案 1 :(得分：1)
说实话，我认为你一直在做正确的事情。
但是，您可以对代码进行一些调整。一般来说，当性能是关键时，只在最里面的循环中做最小的。仔细查看代码，在这方面还有一些快速优化：
我猜你的多处理问题归结为这些非常大的字符串的序列化，抵消了并行运行可能带来的任何性能提升。但是，可能还有另一种方法 - 请参阅Parallelizing a Numpy vector operation。我无法验证，因为我在安装numexpr时遇到了困难。
将它们放在一起并尝试其他一些建议，我得到以下结果：
$ python test.py
Original serial took 0.08330821593602498 minutes...
Using sets took 0.09072601397832235 minutes...
Using built-ins took 0.061421032746632895 minutes...
Using regex took 0.11649663050969442 minutes...
Optimized serial took 0.05909080108006795 minutes...
Original numpy took 0.04050511916478475 minutes...
Optimized numpy took 0.03438538312911987 minutes...
我的代码如下。
import time
import numpy as np
from random import choice
import re
# Create single large chromosome for the test...
seq = ""
for i in range(10000000):
seq += choice("ATGCN")
seq_d = {"Chromosome1": seq}
rev_comps = { 'A' : 'T', 'T' : 'A', 'G' : 'C', 'C' : 'G', 'N' : 'N'}
nts = 'T'
# Original serial implementation
def serial():
test_d = { '+' : {}, '-' : {}}
for chrom in seq_d:
plus_pos, minus_pos = [], []
chrom_seq = seq_d[chrom]
for pos, nt in enumerate(chrom_seq):
if nt in nts:
plus_pos.append(pos)
if rev_comps[nt] in nts:
minus_pos.append(pos)
test_d['+'][chrom] = plus_pos
test_d['-'][chrom] = minus_pos
# Optimized for single character tests
def serial2():
test_d = {'+': {}, '-': {}}
rev_nts = rev_comps[nts]
for chrom in seq_d:
plus_pos, minus_pos = [], []
chrom_seq = seq_d[chrom]
for pos, nt in enumerate(chrom_seq):
if nt == nts:
plus_pos.append(pos)
elif nt == rev_nts:
minus_pos.append(pos)
test_d['+'][chrom] = plus_pos
test_d['-'][chrom] = minus_pos
# Use sets instead of lists
def set_style():
test_d = { '+' : {}, '-' : {}}
for chrom in seq_d:
plus_pos, minus_pos = set(), set()
chrom_seq = seq_d[chrom]
for pos, nt in enumerate(chrom_seq):
if nt in nts:
plus_pos.add(pos)
if rev_comps[nt] in nts:
minus_pos.add(pos)
test_d['+'][chrom] = plus_pos
test_d['-'][chrom] = minus_pos
# Use regex to find either nucleotide...
def regex_it():
test_d = { '+' : {}, '-' : {}}
search = re.compile("(T|A)")
for chrom in seq_d:
pos = 0
plus_pos, minus_pos = [], []
chrom_seq = seq_d[chrom]
for sub_seq in search.split(chrom_seq):
if len(sub_seq) == 0:
continue
if sub_seq[0] == 'T':
plus_pos.append(pos)
elif sub_seq[0] == 'A':
minus_pos.append(pos)
pos += len(sub_seq)
test_d['+'][chrom] = plus_pos
test_d['-'][chrom] = minus_pos
# Use str.find instead of iteration
def use_builtins():
test_d = { '+' : {}, '-' : {}}
for chrom in seq_d:
plus_pos, minus_pos = [], []
chrom_seq = seq_d[chrom]
start = 0
while True:
pos = chrom_seq.find("T", start)
if pos == -1:
break
plus_pos.append(pos)
start = pos + 1
start = 0
while True:
pos = chrom_seq.find("A", start)
if pos == -1:
break
minus_pos.append(pos)
start = pos + 1
test_d['+'][chrom] = plus_pos
test_d['-'][chrom] = minus_pos
# Original numpy implementation
def numpy1():
test_d = { '+' : {}, '-' : {}}
for chrom in seq_d:
chrom_seq = np.array(list(seq_d[chrom]))
for_nts = list(nts)
rev_nts = [rev_comps[nt] for nt in nts]
plus_pos = list(np.where(np.in1d(chrom_seq, for_nts) == True)[0])
minus_pos = list(np.where(np.in1d(chrom_seq, rev_nts) == True)[0])
test_d['+'][chrom] = plus_pos
test_d['-'][chrom] = minus_pos
# Optimized for single character look-ups
def numpy2():
test_d = {'+': {}, '-': {}}
rev_nts = rev_comps[nts]
for chrom in seq_d:
chrom_seq = np.array(list(seq_d[chrom]))
plus_pos = np.where(chrom_seq == nts)
minus_pos = np.where(chrom_seq == rev_nts)
test_d['+'][chrom] = plus_pos
test_d['-'][chrom] = minus_pos
for fn, name in [
(serial, "Original serial"),
(set_style, "Using sets"),
(use_builtins, "Using built-ins"),
(regex_it, "Using regex"),
(serial2, "Optimized serial"),
(numpy1, "Original numpy"),
(numpy2, "Optimized numpy")]:
s = time.time()
fn()
e = time.time()
print('{} took {} minutes...'.format(name, (e-s)/60))
答案 2 :(得分：0)
查找最长的染色体字符串，并创建一个空数组，每个染色体一行，列到字典中最长的一列。然后它将每个染色体放入自己的行中，您可以在整个数组上调用np.where
import numpy as np
longest_chrom = max([len(x) for x in seq_d.values()])
genome_data = np.zeros((len(seq_d), dtype=str, longest_chrom)
for index, entry in enumerate(seq_d):
gen_string = list(seq_d[entry]
genome_data[index, :len(gen_string) + 1] = gen_string
nuc_search = "T"
nuc_locs = np.where(genome_data == nuc_search)
使用此方法，对于＆gt;的字符串略有不同1个核苷酸，但我确信它只需要几步即可实现。