Python:优雅而有效的屏蔽列表的方法 [英] Python: Elegant and efficient ways to mask a list

问题描述

示例:

from __future__ 导入师将 numpy 导入为 npn = 8"""屏蔽列表"""lst = 范围(n)打印列表# 掩码(过滤器)msk = [(el>3) and (el<=6) for el in lst]打印 msk# 面膜的使用打印 [lst[i] for i in xrange(len(lst)) 如果 msk[i]]"""屏蔽数组"""ary = np.arange(n)印刷厂# 掩码(过滤器)msk = (ary>3)&(ary<=6)打印 msk# 面膜的使用print ary[msk] # 非常优雅

结果是:

<预><代码>>>>[0, 1, 2, 3, 4, 5, 6, 7][假，假，假，假，真，真，真，假][4, 5, 6][0 1 2 3 4 5 6 7]【假假假假真假真假】[4 5 6]

如您所见，与列表相比，对数组进行掩码的操作更加优雅.如果您尝试使用列表中的数组屏蔽方案，您将收到一个错误:

<预><代码>>>>lst[msk]回溯(最近一次调用最后一次):文件<交互式输入>"，第 1 行，在 <module> 中类型错误:只有一个元素的整数数组才能转换为索引

问题是为 list 找到一个优雅的掩码.

更新:
jamylak 的回答因引入 compress 而被接受，但是 Joel Cornett 提到的要点使解决方案完全符合我的兴趣.

<预><代码>>>>mlist = MaskableList>>>mlist(lst)[msk]>>>[4, 5, 6]

解决方案

如果你使用的是 numpy:

<预><代码>>>>将 numpy 导入为 np>>>a = np.arange(8)>>>掩码 = np.array([假，假，假，假，真，真，真，假]，dtype=np.bool)>>>一张面具]数组([4, 5, 6])

如果您不使用 numpy，您正在寻找 itertools.compress

<预><代码>>>>从 itertools 导入压缩>>>a = 范围(8)>>>掩码 = [假，假，假，假，真，真，真，假]>>>列表(压缩(a，掩码))[4, 5, 6]

Example:

from __future__ import division
import numpy as np

n = 8
"""masking lists"""
lst = range(n)
print lst

# the mask (filter)
msk = [(el>3) and (el<=6) for el in lst]
print msk

# use of the mask
print [lst[i] for i in xrange(len(lst)) if msk[i]]

"""masking arrays"""
ary = np.arange(n)
print ary

# the mask (filter)
msk = (ary>3)&(ary<=6)
print msk

# use of the mask
print ary[msk]                          # very elegant  

and the results are:

>>> 
[0, 1, 2, 3, 4, 5, 6, 7]
[False, False, False, False, True, True, True, False]
[4, 5, 6]
[0 1 2 3 4 5 6 7]
[False False False False  True  True  True False]
[4 5 6]

As you see the operation of masking on array is more elegant compared to list. If you try to use the array masking scheme on list you'll get an error:

>>> lst[msk]
Traceback (most recent call last):
  File "<interactive input>", line 1, in <module>
TypeError: only integer arrays with one element can be converted to an index

The question is to find an elegant masking for lists.

Updates:
The answer by jamylak was accepted for introducing compress however the points mentioned by Joel Cornett made the solution complete to a desired form of my interest.

>>> mlist = MaskableList
>>> mlist(lst)[msk]
>>> [4, 5, 6]

解决方案

If you are using numpy:

>>> import numpy as np
>>> a = np.arange(8)
>>> mask = np.array([False, False, False, False, True, True, True, False], dtype=np.bool)
>>> a[mask]
array([4, 5, 6])

If you are not using numpy you are looking for itertools.compress

>>> from itertools import compress
>>> a = range(8)
>>> mask = [False, False, False, False, True, True, True, False]
>>> list(compress(a, mask))
[4, 5, 6]

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