Question

我想知道是否有一个python“核心”语法来处理基于范围的选择的补充。

说那个

 a = [0,1,2,3,4,5,6]

然后，例如，

 offset = 1
 step = 3
 a[offset::step] = [1,4].

我的问题是：

“我可以这样做吗

 a[~(offset::step)] == [0,2,3,5,6]

不使用ifs？“

或者，“最多 pythonic 处理这个问题的方法是什么？”

附录：

说我必须对可变大小的数千个列表（实际上是粒子的轨迹）进行这种子采样操作（即可变时间长度的轨迹）。所以我无法预先计算出正确的索引集。

Answer 1

您必须生成索引，然后使用列表推导来选择与这些索引不匹配的所有值。使用range()对象来测试索引的有效方法（在python 2中xrange()）：

indices = range(offset, len(a), step)
[v for i, v in enumerate(a) if i not in indices]

Python 3中的range()对象（Python 2中的xrange()）仅保存起始值，结束值和步长值，如果测试值为in，则只需进行快速计算部分范围。

演示：

>>> a = [0, 1, 2, 3, 4, 5, 6]
>>> offset, step = 1, 3
>>> indices = range(offset, len(a), step)
>>> indices
range(1, 7, 3)
>>> [v for i, v in enumerate(a) if i not in indices]
[0, 2, 3, 5, 6]

是的，这仍然需要使用if语句，但测试很便宜，可以根据需要合并到生成器表达式中：

for i in (v for i, v in enumerate(a) if i not in range(offset, len(a), step)):

Answer 2

即使你没有提前填充索引，

集合（通常）也会快一个数量级：

r100 = range(100)
r2 = range(3, 40, 3)

# Find indices in r100 that aren't in r2.
# This is a set difference (or symmetric difference)
## Set methods
# Precalculated is fastest:
sr100 = set(r100)
sr2 = set(r2)
%timeit sr100 - sr2
100000 loops, best of 3: 3.84 us per loop

# Non-precalculated is still faster:
%timeit set(range(100)) ^ set(range(3,40,3))
100000 loops, best of 3: 9.76 us per loop
%timeit set(xrange(100)) ^ set(xrange(3,40,3))
100000 loops, best of 3: 8.84 us per loop

# Precalculating the original indices still helps, if you can hold it in memory:
%timeit sr100 ^ set(xrange(3,40,3))
100000 loops, best of 3: 4.87 us per loop

# This is true even including converting back to list, and sorting (if necessary):
%timeit [x for x in sr100 ^ set(xrange(3,40,3))]
100000 loops, best of 3: 9.02 us per loop
%timeit sorted(x for x in sr100 ^ set(xrange(3,40,3)))
100000 loops, best of 3: 15 us per loop


## List comprehension:

# Precalculated indices
%timeit [x for x in r100 if x not in r2]
10000 loops, best of 3: 30.5 us per loop

# Non-precalculated indices, using xrange
%timeit [x for x in xrange(100) if x not in xrange(3, 40, 3)]
10000 loops, best of 3: 65.8 us per loop

# The cost appears to be in the second xrange?
%timeit [x for x in r100 if x not in xrange(3, 40, 3)]
10000 loops, best of 3: 64.3 us per loop
%timeit [x for x in xrange(100) if x not in r2]
10000 loops, best of 3: 29.9 us per loop
# xrange is not really any faster than range here - uses less memory, but still have
# to walk through entire list
%timeit [x for x in range(100) if x not in range(3, 40, 3)]
10000 loops, best of 3: 63.5 us per loop

Answer 3

在使用if的同时，以下是单步列表理解，我相信为您提供所需的答案：

>>> offset = 1
>>> step = 3
>>> a = [0,1,2,3,4,5,6]
>>> [v for i, v in enumerate(a) if not i%step == offset]
[0, 2, 3, 5, 6]
>>>

我不知道的是，这是否比使用范围构造超过mod更有效率。

Python - 基于范围的子列表选择的补充

3 个答案: