Question

我正在寻找一个内存/ CPU高效的单行程序来对列表中每m个元素中的n进行二次采样。到目前为止我有：

sb = [11,12,21,22,31,32]*4 #stream buffer, e.g. 4 identical frames
ci = 1 #1-indexed channel index
cs = 2 #channel (sample) size
nc = 3 #number of channels in each frame
fs = nc*cs #frame size

[i for l in [sb[j+ci-1:j+ci-1+cs] for j
    in [x*fs+ci-1 for x in xrange(len(sb)/fs)]] for i in l]

Out: [11, 12, 11, 12, 11, 12, 11, 12]

分解我正在创建一个样本列表列表，然后将其展平为包含[i for l in ll for i in l]的一维列表

或者，不是一个班轮，但更容易阅读，我可以做：

os = []
for i in [sb[j+ci-1:j+ci-1+cs] for j in [x*fs+ci-1 for x in xrange(len(sb)/fs)]]: os = os+i

例如，与cs=1特定情况的简单简写：sb[ci-1::fs]进行比较时，两种解决方案看起来都太复杂了。

你能帮我提出一个不错的解决方案吗？

Answer 1

以下内容对我来说似乎相当可读（而且效率也相当高）：

from itertools import chain

sb = [11, 12, 21, 22, 31, 32]*4  # stream buffer, e.g. 4 identical frames

ci = 1      # 1-indexed channel index
cs = 2      # channel size
nc = 3      # number of channels in each frame
fs = nc*cs  # frame size

result = list(chain.from_iterable(sb[i: i+cs] for i in xrange(ci-1, len(sb), fs)))
print(result)  # -> [11, 12, 11, 12, 11, 12, 11, 12]

Answer 2

我将大部分索引移动到range()计算中。然后将索引显示为子列表的速度更快 - 请参阅下面的时间：

sb = [11,12,21,22,31,32]*4 #stream buffer, e.g. 4 identical frames
ci = 1 #1-indexed channel index
cs = 2 #channel size
nc = 3 #number of channels in each frame
fs = nc*cs #frame size

for ci in range(1,4):
    print [x for y in [sb[x:x+cs] for x in range((ci-1)*cs,len(sb),fs)] for x in y]

输出：

[11, 12, 11, 12, 11, 12, 11, 12]
[21, 22, 21, 22, 21, 22, 21, 22]
[31, 32, 31, 32, 31, 32, 31, 32]

我将大部分工作转移到range()调用生成的子列表列表中，其余的是将子列表简单地分解为一个列表。

range((ci-1)*cs,len(sb), fs)
         |         |     |________  frame size, range will use steps the size of the frame
         |         |______________  till end of data
         |________________________  starting at (ci-1) * channel size   

for ci = 1 it starts at 0,   6,12,18,....
for ci = 2 it starts at 2,   8,14,....
for ci = 3 it starts at 4,  10,...  
for ci = 4 it starts at 6,  ...  
    and increases by fs = 6 until end of data. The list comp then gets a sublist of len cs
    and the rest of the list-comp flattens it down from list of list to a simpler list

定时：

import timeit

print timeit.timeit(stmt='''
sb = [11,12,21,22,31,32]*4*5 #stream buffer, e.g. 4 identical frames
ci = 1 #1-indexed channel index
cs = 2 #channel size
nc = 3 #number of channels in each frame
fs = nc*cs #frame size
for ci in range(1,4):
    [x for y in [sb[x:x+cs] for x in range((ci-1)*cs,len(sb),fs)] for x in y] 

''', setup='pass', number=10000)  #  0.588474035263

print timeit.timeit(stmt='''
sb = [11,12,21,22,31,32]*4*5 #stream buffer, e.g. 4 identical frames
ci = 1 #1-indexed channel index
cs = 2 #channel size
nc = 3 #number of channels in each frame
fs = nc*cs #frame size
for ci in range(1,4):
    [i for l in [sb[j+ci-1:j+ci-1+cs] for j in [x*fs+ci-1 for x in xrange(len(sb)/fs)]] for i in l] 

''', setup='pass', number=10000)   # 0.734045982361

Answer 3

<强>代码：

sb = [11,12,21,22,31,32] * 4
ci = 0
cs = 2
nc = 3
fs = cs * nc
result = list(sum(zip(*[sb[i::fs] for i in range(ci, ci+cs)]),()))

<强>输出：

[11, 12, 11, 12, 11, 12, 11, 12]

我建议将ci设置为基于0的索引以匹配python的语法，但如果您坚持，更新func很简单，只需将所有ci替换为ci-1即可。 ci。

它与您原来的方法基本相同，只是更清洁一点，它可以扩展到不同的cs，nc和get_directory_property(MY_LIBRARIES_NAMES BUILDSYSTEM_TARGETS)。

Answer 4

我建议使用更清晰的变量名称而不是注释，不要使用单行。

<强>鉴于

import itertools as it 


stream = [11, 12, 21, 22, 31, 32] * 4
ch_idx = 1
ch_size = 2 
num_chs = 3

<强>代码

使用grouper itertools recipe：

channels = grouper(ch_size, stream)
frames = grouper(num_chs, channels)
list(it.chain.from_iterable(*it.islice(zip(*frames), ch_idx)))
# [11, 12, 11, 12, 11, 12, 11, 12]

作为单行，它看起来如下：

list(it.chain.from_iterable(*it.islice(zip(*grouper(num_chs, grouper(ch_size, stream))), ch_idx)))
# [11, 12, 11, 12, 11, 12, 11, 12]

<强>详情

grouper配方实现如下：

def grouper(iterable, n, fillvalue=None):
    "Collect data into fixed-length chunks or blocks"
    # grouper('ABCDEFG', 3, 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return zip_longest(*args, fillvalue=fillvalue)

另请参阅more_itertools第三方库以了解预先实施的食谱。

高效的单行代码可以从列表中的每个`m`元素中对`i`th`n`元素进行子样本处理

4 个答案: