生成器在输入迭代器的开始和结束处保留一个占位符 [英] Generator which leaves a placeholder at the beginning and at the end of the input iterator intact
问题描述
$ $ $ $ $ $ c $ a = [255,255,1,255,255,255, 1,2,255,255,2,255,255,3,255,3,255,255,255]
255
是一个特殊值。这是一个占位符。
我制作了一个生成器,它替换列表中的一些占位符。它符合预期。
但我不需要处理开始占位符 [255,255
和结束占位符 255,255,255]
并完好保留。
所以,我试图修改生成器来工作out:
Python 2.7
from __future__ import print_function $ b $ from itertools import tee,izip,ifilterfalse
def replace(iterable,placeholder = 255):
it = enumerate(可迭代)#需要位置列表中间的逻辑
it = ifilterfalse(lambda x:x [1] ==占位符,它)#创建一个迭代器,删除所有占位符
for i,(left,right)在枚举(window(it,2))中:#通过带有2个元素窗口的过滤列表
if i == 0:#将开始占位符完整
替换为范围内的j(left [ 0]):
产生占位符
#Some中间的逻辑LIST(效果很好)
#需要完整地保留尾随占位符。
将临时值转换为列表以简化对代码的理解:
>>>迭代式
[255,1,255,255,1,255,255,255,2,2,255,255,255,2,2,3,255,255,255,3,255,255]
$ b $ gt;>> it =枚举(可迭代)
[(0,255),(1,1),(2,255),(3,255),(4, 1),(5,255),(6,255),(7,255),(8,2),(9,2),(10,255),(11,255),(12,255) ,(13,2),(14,2),(15,3),(16,255),(17,255),(18,255),(19,3),(20,255),(
>>> it = ifilterfalse(lambda x:x [1] ==占位符,它)
[(1,1),(4, 1),(8,2),(9,2),(13,2),(14,2),(15,3),(19,3)]
>> ((1,1)),(1,((4,1),(8))列表(枚举(窗口(it,2)))
[ ,(2,((8,2),(9,2))),(3,((9,2),(13,2))),(4,((13,2) ),(14,2))),(5,((14,2),(15,3))),(6,((15,3),(19,3)))]
因此,您可以看到,列表(枚举(window(it,2)) )
包含前导非占位符值的索引(0,((** 1 **,1),(4,1))),
,但它不包含这个信息有多少个初始迭代器的尾随占位符: list(enumerate(window(it,2)))
以这个值结尾(6,( (15,3),(** 19 **,3)))
,它只有最后一个非占位符值的索引,它没有给出信息剩下多少个占位符。
我通过依靠 it = enumerate(iterable)
来设法处理主要占位符,初始迭代器值,它在第一次产生的值中由 ifilterfalse
。
保持不变但我花了很多时间尝试弄清楚如何用尾随占位符做同样的事情。问题是, 问题 更正的最佳方法是什么此代码为它来处理尾随占位符? 由于目标不是通过一切手段创建代码(我已经),然后使用一种不同的技术来完成它的工作。 ,我想通过修改代码来解决任务,而不是完全重写它。 这不仅仅是一项训练,而是一项真正的任务。 其他信息 我的代码几乎与这个通过@ nye17回答。但是在这段代码中,作者对初始列表进行了原位修改。我想创建一个生成器,它将产生与该代码中的结果列表相同的值。 ifilterfalse
只是吞下枚举(iterable)
的最后一个占位符值,我看不到它们的访问方式(自从第一次生成的枚举值(可迭代)的索引后,前导占位符是可能的$ c
window
是此处的代码。
这不是我必须在生活中解决的真正任务。这只是一个培训。
完整代码
http://codepad.org/9UJ9comY
我想出了最简单的解决方案是通过一个生成器来处理 it =枚举(可迭代)
,它只保存最后返回的值。
<因此,我在
it = enumerate(iterable)(在 replace
函数内)添加了以下代码: def save_last(可迭代):
for iterable:
yield i
替换。 last_index = i [0]#保存最后一个值
it = save_last(it)
在 iterable
用尽之后,生成器的最后一个操作符保存已赋值的索引(即 i [0]
as enumerate
将它保存在tupele的位置 0
)作为替换
属性(因为替换
函数是一个类的实例,它可以有局部变量)。 code> it
包装在新创建的生成器 save_last
中。 在函数的最后,我添加了使用 replace.last_index
变量中已保存索引的代码。
if right [0]< replace.last_index:
for i in range(replace.last_index-right [0]):
产生占位符
完整的代码:
from __future__ import print_function $ b $ from itertools import tee,izip,ifilterfalse
def window(iterable,n):
els = tee(iterable,n)
为我,el为枚举(els):
为_范围(i):
next(el,None)
return izip * els)
def replace(iterable,placeholder = 255):
it =枚举(可迭代)
def save_last(可迭代的):
for i in iterable:
yield i
replace.last_index = i [0]#保存最后一个值
it = save_last(it)
it = ifilterfalse(lambda x:x [1] ==占位符,它)
在枚举(窗口(it,2))中为i,(左,右):
if i = = 0:
为范围内的j(左[0]):
产生占位符
产生剩余[1]
如果正确[0]> left [0] +1 :
if left [1] == right [1]:
for _ in range(right [0] -left [0] -1):
yield left [1]
else:
for _ in range(right [0] -left [0] -1):
产生占位符
产权[1]
if right [0] < replace.last_index:
for i in range(replace.last_index-right [0]):
产生占位符
a = [255,1,255,255,1,255,255,255 ,2,2,255,255,255,2,2,3,255,255,255,3,255,255]
print('\\\
Input:{} .format(a))
output = list(replace(a))
print('Proram output:{}'。format(output))
print('目标输出:{} '.format([255,1,1,1,1,255,255,255,2,2,2,2,2,2,2,3,3,3,3,3,255,255]))
$ c $输入 :[255,1,255,255,1,255,255,255,2,2,25,255,255,2,2,3,255,255,255,3,255,255]
Proram输出:[255,1,1,1,1,25,255,255,2,2,2,2,2,2,2,3,3,3,3,25,255]
目标输出:[255,1,1,1,1,25,255,255,2,2,2,2,2,2,2,3,3,3,3,3,5,25,255]
我唯一不喜欢的是以Python编写的C
ifilterfalse
和save_last
非常高效的组合。Let's take a list as an example:
a = [255, 255, 1, 255, 255, 255, 1, 2, 255, 255, 2, 255, 255, 3, 255, 3, 255, 255, 255]
255
is a special value in it. It's a placeholder.I've made a generator which replaces some of the placeholder inside the list. It works as expected.
But I need not to process the beginning placeholders
[255, 255
and the ending placeholders255, 255, 255]
and yield them intact.So, I tried to modify the generator to work it out:
Python 2.7
from __future__ import print_function from itertools import tee, izip, ifilterfalse def replace(iterable,placeholder=255): it = enumerate(iterable) #the position is needed for the logic for the middle of the list it = ifilterfalse(lambda x: x[1]==placeholder, it) #create an iterator that deletes all the placeholders for i,(left,right) in enumerate(window(it,2)): #Slide through the filtered list with the window of 2 elements if i==0: #Leaving the beginning placeholders intact for j in range(left[0]): yield placeholder #SOME LOGIC FOR THE MIDDLE OF THE LIST (it works well) #Need to leave the trailing placeholders intact.
The interim values converted to list just to ease the comprehension of the code:
>>>iterable [255,1,255,255,1,255,255,255,2,2,255,255,255,2,2,3,255,255,255,3,255,255] >>>it = enumerate(iterable) [(0, 255), (1, 1), (2, 255), (3, 255), (4, 1), (5, 255), (6, 255), (7, 255), (8, 2), (9, 2), (10, 255), (11, 255), (12, 255), (13, 2), (14, 2), (15, 3), (16, 255), (17, 255), (18, 255), (19, 3), (20, 255), (21, 255)] >>>it = ifilterfalse(lambda x: x[1]==placeholder, it) [(1, 1), (4, 1), (8, 2), (9, 2), (13, 2), (14, 2), (15, 3), (19, 3)] >>>list(enumerate(window(it,2))) [(0, ((1, 1), (4, 1))), (1, ((4, 1), (8, 2))), (2, ((8, 2), (9, 2))), (3, ((9, 2), (13, 2))), (4, ((13, 2), (14, 2))), (5, ((14, 2), (15, 3))), (6, ((15, 3), (19, 3)))]
So, as you can see, the
list(enumerate(window(it,2)))
contains the index of the leading non-placeholder value(0, ((**1**, 1), (4, 1))),
, but it doesn't contain the information how many trailing placeholder the initial iterator had:list(enumerate(window(it,2)))
ends in this value(6, ((15, 3), (**19**, 3)))
which has only the index of the last non-placeholder value, which doesn't give the information how many placeholders are left.I managed to process the leading placeholders by relying on
it = enumerate(iterable)
which yields the position of the initial iterator value which persists in the first yielded value byifilterfalse
.But I spent quite a lot of time trying to figure out how to do the same thing with the trailing placeholders. The problem is that
ifilterfalse
just swallows the last placeholder values ofenumerate(iterable)
and I see no way to access them (it was possible for the leading placeholders since the first generated value ofifilterfalse
contained the index of the value of theenumerate(iterable)
).Question
What is the best way to correct this code for it to process the trailing placeholders?
As the goal is not to create a code by all means (I have already done it using a different technique), I want to solve the task by tinkering a bit wit the code, not a complete rewriting it.
It's more of a training than a real task.
Additional information
window
is the code from here.My code does nearly the same as in this answer by @nye17. But in this code the author make inplace modifications of the initial list. And I want to create a generator which will be yielding the same values as the resultant list in that code.
Furthermore, I want my generator to accept any iterables as a parameter, not only lists (for example it may accept the iterator which reads the values from file one by one). With having only lists as a parameter, the task becomes simpler, since we can scan the list from the end.
This is not a real task I have to solve in life. It's just for a training.
Full code http://codepad.org/9UJ9comY
解决方案The simplest solution I came up with is to process
it = enumerate(iterable)
through one more generator which just saves the last returned value by it.So, I added the following code after
it = enumerate(iterable)
(inside thereplace
function):def save_last(iterable): for i in iterable: yield i replace.last_index = i[0] #Save the last value it = save_last(it)
After
iterable
is exhausted, the last operator of the generator saves the index of the yielded value (which isi[0]
asenumerate
stores it at the position0
of tupele) as thereplace
attribute (sincereplace
function is a instance of a class, which can have local variables).The
it
is wrapped in the newly created generatorsave_last
.At the very end of the function I added the code which uses the saved index in
replace.last_index
variable.if right[0]<replace.last_index: for i in range(replace.last_index-right[0]): yield placeholder
The full code:
from __future__ import print_function from itertools import tee, izip, ifilterfalse def window(iterable,n): els = tee(iterable,n) for i,el in enumerate(els): for _ in range(i): next(el, None) return izip(*els) def replace(iterable,placeholder=255): it = enumerate(iterable) def save_last(iterable): for i in iterable: yield i replace.last_index = i[0] #Save the last value it = save_last(it) it = ifilterfalse(lambda x: x[1]==placeholder, it) for i,(left,right) in enumerate(window(it,2)): if i==0: for j in range(left[0]): yield placeholder yield left[1] if right[0]>left[0]+1: if left[1]==right[1]: for _ in range(right[0]-left[0]-1): yield left[1] else: for _ in range(right[0]-left[0]-1): yield placeholder yield right[1] if right[0]<replace.last_index: for i in range(replace.last_index-right[0]): yield placeholder a = [255,1,255,255,1,255,255,255,2,2,255,255,255,2,2,3,255,255,255,3,255,255] print('\nInput: {}'.format(a)) output = list(replace(a)) print('Proram output: {}'.format(output)) print('Goal output : {}'.format([255,1,1,1,1,255,255,255,2,2,2,2,2,2,2,3,3,3,3,3,255,255]))
Which works as expected:
Input: [255, 1, 255, 255, 1, 255, 255, 255, 2, 2, 255, 255, 255, 2, 2, 3, 255, 255, 255, 3, 255, 255] Proram output: [255, 1, 1, 1, 1, 255, 255, 255, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 255, 255] Goal output : [255, 1, 1, 1, 1, 255, 255, 255, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 255, 255]
The only thing that I don't like is the combination of very efficient written in C
ifilterfalse
andsave_last
written in Python.这篇关于生成器在输入迭代器的开始和结束处保留一个占位符的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!