blocks|key|4896023|text|data.reduce([data.first])+do+%7Cresult,+item%7C
++result.last.last+==+item.last+?+result+:+result+%2B+[item]
end|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4896024|这并不能精确地产生您想要的输出-它跳过了运行的最后一项。但好消息是您不需要最后一项，因为您知道它的值与您的第一项相同，并且您知道它的时间戳比下一项小一。(如果您的时间戳不是连续的，那么这是不好的)。如果最后一个条目也不在Time.now上，最简单的做法就是在最后手动添加它。|unstyled|offset|length|style|CODE|4896025|它的用途：|4896026|4896027|使用第一个值初始化结果。|unordered-list-item|4896028|4896029|​|4896030|我已经编写了它，以便每次迭代都会生成一个带有result+%2B+[item]的新result数组，这是使用reduce的函数风格和首选方式，但这会产生许多不必要的中间数组。您可以只通过实际追加(<<)来创建一个新数组。|4896031|entityMap^0|0|32|8|0|0|0|0|0|0|M|F|13|6|1G|6|2O|2|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|Y|8|@$G|Z|H|10|I|J]]|9|@]|A|$]]|$1|K|3|L|5|F|7|11|8|@]|9|@]|A|$]]|$1|M|3|-4|5|F|7|12|8|@]|9|@]|A|$]]|$1|N|3|O|5|P|7|13|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|F|7|14|8|@]|9|@]|A|$]]|$1|R|3|S|5|F|7|15|8|@]|9|@]|A|$]]|$1|T|3|U|5|F|7|16|8|@$G|17|H|18|I|J]|$G|19|H|1A|I|J]|$G|1B|H|1C|I|J]|$G|1D|H|1E|I|J]]|9|@]|A|$]]|$1|V|3|-4|5|F|7|1F|8|@]|9|@]|A|$]]]|W|$]]

<pre><code>data.reduce([data.first]) do |result, item|
 result.last.last == item.last ? result : result + [item]
end
</code></pre>

This doesn't produce exactly your desired output - it skips the last item of the run. But the good news is you don't need the last item, because you know its value is the same as your first item, and you know its timestamp is one less than the next item. (If your timestamps aren't consecutive, then this is no good). If the last entry also isn't at <code>Time.now</code>, the simplest thing to do is just manually tack it on at the end.

What it does:

<ul>
<li>Initializes the result with the first value. This is simply to avoid a <code>nil</code> case at the beginning.</li>
<li>For each <code>item</code> in <code>data</code>

<ul>
<li>If the value in <code>item.last</code> is the same as the last entry currently in <code>result</code>, do nothing</li>
<li>If the value in <code>item.last</code> is different, append it to <code>result</code></li>
</ul></li>
</ul>

I've written it so that each iteration produces a new <code>result</code> array with <code>result + [item]</code>, which is the functional style and preferred way to use <code>reduce</code>, but that produces a lot of unnecessary intermediate arrays. You can create just one new array by actually appending (<code>&lt;&lt;</code>) instead.

blocks|key|48525|text|这不是一个优雅的解决方案，但它是有效的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|48526|data+=+['8:00',+4],['8:01',+4],['8:02',+4],['8:03',+7],['8:04',+7],['8:05',+8],['8:06',+9],['8:07',+13],['8:08',+13],['8:09',+13]

def+clean_array(data)
++++item_to_delete+=+[]

++++(0..(data.count-3)).each+do+%7Ci%7C
++++++++if+data[i][1].eql?(data[i%2B2][1])
++++++++++++item_to_delete+<<+data[i%2B1]
++++++++end
++++end

++++data+-+item_to_delete
end

new_data+=+clean_array(data)|code-block|syntax|javascript|48527|正如预期的那样，输出是|48528|=>+[["8:00",+4],+["8:02",+4],+["8:03",+7],+["8:04",+7],+["8:05",+8],+["8:06",+9],+["8:07",+13],+["8:09",+13]]|48529|编辑|48530|另一种方法|48531|data+=+['8:00',+4],['8:01',+4],['8:02',+4],['8:03',+7],['8:04',+7],['8:05',+8],['8:06',+9],['8:07',+13],['8:08',+13],['8:09',+13]++++
new_data+=+[]

data.each+{+%7Citem%7C+(new_data[-2]+and+item[1].eql?(new_data[-2][1]))+?+new_data[-1]+=+item+:+new_data+<<+item+}

new_data

#+=>+=>+[["8:00",+4],+["8:02",+4],+["8:03",+7],+["8:04",+7],+["8:05",+8],+["8:06",+9],+["8:07",+13],+["8:09",+13]]|48532|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|X|8|@]|9|@]|A|$]]|$1|O|3|P|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

That's not an elegant solution, but it works.

<pre><code>data = ['8:00', 4],['8:01', 4],['8:02', 4],['8:03', 7],['8:04', 7],['8:05', 8],['8:06', 9],['8:07', 13],['8:08', 13],['8:09', 13]

def clean_array(data)
 item_to_delete = []

 (0..(data.count-3)).each do |i|
 if data[i][1].eql?(data[i+2][1])
 item_to_delete &lt;&lt; data[i+1]
 end
 end

 data - item_to_delete
end

new_data = clean_array(data)
</code></pre>

The output, as expected is

<pre><code>=&gt; [["8:00", 4], ["8:02", 4], ["8:03", 7], ["8:04", 7], ["8:05", 8], ["8:06", 9], ["8:07", 13], ["8:09", 13]]
</code></pre>

Edit

Another approach

<pre><code>data = ['8:00', 4],['8:01', 4],['8:02', 4],['8:03', 7],['8:04', 7],['8:05', 8],['8:06', 9],['8:07', 13],['8:08', 13],['8:09', 13] 
new_data = []

data.each { |item| (new_data[-2] and item[1].eql?(new_data[-2][1])) ? new_data[-1] = item : new_data &lt;&lt; item }

new_data

# =&gt; =&gt; [["8:00", 4], ["8:02", 4], ["8:03", 7], ["8:04", 7], ["8:05", 8], ["8:06", 9], ["8:07", 13], ["8:09", 13]]
</code></pre>

blocks|key|5698456|text|我正在发布我自己问题的解决方案。我从Kristján的解决方案开始，它使用了reduce。请注意，我的解决方案无法生成最终的采样时间，但我选择接受此行为，因为我的示例只是用于模拟流。所以8:09样本并不是最终的值。下一个传入的样本将确定是否存储该8:09值。因此，我的原始帖子的细节可以更好地解释。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5698457|samples+=+[['8:00',+4],['8:01',+4],['8:02',+4],['8:03',+7],['8:04',+7],['8:05',+8],['8:06',+9],['8:07',+13],['8:08',+13],['8:09',+13]].lazy

prev+=+[]
compressed+=+samples.reduce([samples.first])+do+%7Ckeepers,+sample%7C
++keepers+<<+prev+<<+sample+if+keepers.last.last+!=+sample.last
++prev+=+sample
++keepers
end
puts+compressed.inspect

#+=>+[["8:00",+4],+["8:02",+4],+["8:03",+7],+["8:04",+7],+["8:05",+8],+["8:05",+8],+["8:06",+9],+["8:06",+9],+["8:07",+13]]|code-block|syntax|javascript|5698458|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I am posting a solution to my own question. I started with Kristján's solution which used reduce. Note that my solution fails to produce the final sample time, but I am choosing to accept this behavior because my example was just meant to be a simulated stream. So that 8:09 sample is not meant to be the final value. The next incoming sample will determine whether that 8:09 value gets stored. So that detail of my original post could have been better explained. 

<pre><code>samples = [['8:00', 4],['8:01', 4],['8:02', 4],['8:03', 7],['8:04', 7],['8:05', 8],['8:06', 9],['8:07', 13],['8:08', 13],['8:09', 13]].lazy

prev = []
compressed = samples.reduce([samples.first]) do |keepers, sample|
 keepers &lt;&lt; prev &lt;&lt; sample if keepers.last.last != sample.last
 prev = sample
 keepers
end
puts compressed.inspect

# =&gt; [["8:00", 4], ["8:02", 4], ["8:03", 7], ["8:04", 7], ["8:05", 8], ["8:05", 8], ["8:06", 9], ["8:06", 9], ["8:07", 13]]
</code></pre>

I have a stream that I would like to read from a sensor. The stream never ends. Most of the time the values repeat over time. So I would like to identify runs of values and just keep the first and last of each run, and keep their timestamps too.

Here is an example of 10 minutes of data:

[['8:00', 4],['8:01', 4],['8:02', 4],['8:03', 7],['8:04', 7],['8:05', 8],['8:06', 9],['8:07', 13],['8:08', 13],['8:09', 13]].lazy 

I want to compress this data to this:
[['8:00', 4],['8:02', 4],['8:03', 7],['8:04', 7],['8:05', 8],['8:06', 9],['8:07', 13],['8:09', 13]]

I've been trying to accomplish this through enumerable functions such as chunk, each_cons, each_with_object. This problem, though, seems inherently functional. Can I accomplish this using lazy enumerator in ruby?

Use ruby to compress a stream of time series data

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个想要从传感器读取的流。这条小溪永远不会结束。大多数时候，这些值会随着时间的推移而重复。因此，我想确定值的运行，只保留每次运行的第一次和最后一次，并保留它们的时间戳。['8:00'，4，'8:01'，4，'8:02'，4，'8:03'，7，'8:04'，7，'8:05'，8，'8:06'，9，'8:07'，13，...

问使用ruby压缩时间序列数据流
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