在列表中识别连续重复项最多的Pythonic方法是什么? [英] What's the most Pythonic way to identify consecutive duplicates in a list?
问题描述
例如,如果我有一个列表,如:
0,0,0,3,3,2,5,2,6,6]
希望结果为:
[(0,3),(3,2),(2,1),( 5,1),(2,1),(6,2)]
em>公平地使用for循环,临时表和计数器执行此操作的简单方法:
result_list = []
current = source_list [0]
count = 0
在source_list中的值:
如果value == current:
count + = 1
else:
result_list.append((current,count))
current = value
count = 1
result_list.append((current,count))
但我真的很喜欢python的fu指令编程习语,我想用一个简单的生成器表达式来做到这一点。但是,在使用发电机时,我发现很难保留分计数。我有一个两步的过程可能会让我在那里,但现在我被困了。
有没有一个特别优雅/ pythonic的方式来做到这一点,特别是与发生器?
>>>来自itertools import groupby
>>>> L = [0,0,0,3,3,2,5,2,6,6]
>>>对于组(L)中的k,g,分组_L= [(k,i代表i中的1))
>>>> #or(k,len(list(g))),但是创建一个中间列表
>>>分组_
[(0,3),(3,2),(2,1),(5,1),(2,1),(6,2)]
电池包括,正如他们所说。
建议使用 sum
和来自JBernardo的生成器表达式;看到评论。
I've got a list of integers and I want to be able to identify contiguous blocks of duplicates: that is, I want to produce an order-preserving list of duples where each duples contains (int_in_question, number of occurrences).
For example, if I have a list like:
[0, 0, 0, 3, 3, 2, 5, 2, 6, 6]
I want the result to be:
[(0, 3), (3, 2), (2, 1), (5, 1), (2, 1), (6, 2)]
I have a fairly simple way of doing this with a for-loop, a temp, and a counter:
result_list = []
current = source_list[0]
count = 0
for value in source_list:
if value == current:
count += 1
else:
result_list.append((current, count))
current = value
count = 1
result_list.append((current, count))
But I really like python's functional programming idioms, and I'd like to be able to do this with a simple generator expression. However I find it difficult to keep sub-counts when working with generators. I have a feeling a two-step process might get me there, but for now I'm stumped.
Is there a particularly elegant/pythonic way to do this, especially with generators?
>>> from itertools import groupby
>>> L = [0, 0, 0, 3, 3, 2, 5, 2, 6, 6]
>>> grouped_L = [(k, sum(1 for i in g)) for k,g in groupby(L)]
>>> # Or (k, len(list(g))), but that creates an intermediate list
>>> grouped_L
[(0, 3), (3, 2), (2, 1), (5, 1), (2, 1), (6, 2)]
Batteries included, as they say.
Suggestion for using sum
and generator expression from JBernardo; see comment.
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