从列表中的每个“m”元素中对“第 i”个“n”元素进行子采样的高效单线

Question

我正在寻找一种 memory/cpu 高效的单行代码来从列表中的每 m 个元素中对 n 进行子采样。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示例列表列表，然后使用 [i for l in ll for i in l]

将其展平为一维列表

或者，不是一行，但更容易阅读，我可以这样做：

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

这两种解决方案在比较时看起来都太复杂了，例如，对于 cs=1 特殊情况的超级简单 shorthand：sb[ci-1::fs].

你能帮我想出一个像样的解决方案吗？

Answer 1

我将大部分索引移动到 range() 计算中。它比将索引显示到子列表中更快 - 请参阅下面的计时：

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]

我将大部分工作移到了 range() 调用中 - 生成子列表列表，其余的是将子列表简单分解为一个列表。

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

时间：

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 2

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)]),()))

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，但如果你坚持，更新函数很简单，只需将所有 ci 替换为ci-1.

它与您原来的方法本质上是一样的，只是更简洁一点，并且它可以扩展到不同的 ci、cs 和 nc。

Answer 3

以下内容对我来说相当可读（并且也相当有效）：

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 4

我建议使用更清晰的变量名而不是注释，不要使用单行代码。

给定

import itertools as it 


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

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]

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 第三方库以获取预实现的方法。

从列表中的每个“m”元素中对“第 i”个“n”元素进行子采样的高效单线

Efficient one-liner to subsample `i`th `n` elements out of every `m` elements on a list

python

arrays

list-comprehension

multiplexing

python-2.7