blocks|key|1703036|text|您只需要这样的条件列表理解：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1703037|l+=+[2000,2200,5000,2350]
n+=+2000++++
a+=+[+(x)+for+x+in+l+if+((n+-250)+<+x+<+(n+%2B+250))+]|code-block|syntax|javascript|1703038|然后你可以用普通的|1703039|np.mean(a)|1703040|或者任何你喜欢的方法。|1703041|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You just need a conditional list comprehension like this:
<pre><code>l = [2000,2200,5000,2350]
n = 2000 
a = [ (x) for x in l if ((n -250) &lt; x &lt; (n + 250)) ]
</code></pre>
Then you can average with
<pre><code>np.mean(a)
</code></pre>
or whatever method you prefer.

blocks|key|1703060|text|你似乎在寻找一种聚类算法--类似于K--意味着。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1703061|该算法在学习程序包中实现。|offset|length|1703062|在您找到您的K均值之后，您可以计算出您的数据中有多少是用这个平均值进行聚类的，并进行计算。|1703063|但是，在您的情况下，不清楚K是什么。在得到约束(均值之间的n%2B500距离)之前，您可以尝试运行几个K值的算法。|style|CODE|1703064|entityMap|0|LINK|mutability|MUTABLE|url|https://scikit-learn.org/stable/modules/generated/sklearn.cluster.KMeans.html^0|0|4|5|0|0|0|T|5|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@$D|V|E|W|1|X]]|A|$]]|$1|F|3|G|5|6|7|Y|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|Z|8|@$D|10|E|11|J|K]]|9|@]|A|$]]|$1|L|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|M|$N|$5|O|P|Q|A|$R|S]]]]

It seems that you look for a clustering algorithm - something like K-means.
This algorithm is implemented in <a href="https://scikit-learn.org/stable/modules/generated/sklearn.cluster.KMeans.html" rel="nofollow noreferrer">scikit-learn package</a>
After you find your K means, you can count how many of your data were clustered with that mean, and make your computations.
However, it's not clear in your case what is K. You can try and run the algorithm for several K values until you get your constraints (the <code>n+500</code> distance between the means)

blocks|key|858054|text|您可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|858055|import+numpy+as+np
l+=+np.array([2000,2200,5000,2350])

#+find+similar+numbers+(that+are+within+each+500+fold)
similar+=+l+//+500

#+for+each+similar+group+get+the+average+and+convert+it+to+integer+(as+in+the+desired+output)
new_list+=+[np.average(l[similar+==+num]).astype(int)+for+num+in+np.unique(similar)]

print(new_list)|code-block|syntax|javascript|858056|输出：|858057|[2183,+5000]|858058|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You can use:
<pre><code>import numpy as np
l = np.array([2000,2200,5000,2350])

# find similar numbers (that are within each 500 fold)
similar = l // 500

# for each similar group get the average and convert it to integer (as in the desired output)
new_list = [np.average(l[similar == num]).astype(int) for num in np.unique(similar)]

print(new_list)
</code></pre>
Output:
<pre><code>[2183, 5000]
</code></pre>

blocks|key|850912|text|步骤1：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|850913|list+=+[5620.77978515625,
+7388.43017578125,
+7683.580078125,
+8296.6513671875,
+8320.82421875,
+8557.51953125,
+8743.5,
+9163.220703125,
+9804.7939453125,
+9913.86328125,
+9940.1396484375,
+9951.74609375,
+10074.23828125,
+10947.0419921875,
+11048.662109375,
+11704.099609375,
+11958.5,
+11964.8232421875,
+12335.70703125,
+13103.0,
+13129.529296875,
+16463.177734375,
+16930.900390625,
+17712.400390625,
+18353.400390625,
+19390.96484375,
+20089.0,
+34592.15625,
+36542.109375,
+39478.953125,
+40782.078125,
+41295.26953125,
+42541.6796875,
+42893.58203125,
+44578.27734375,
+45077.578125,
+48022.2890625,
+52535.13671875,
+58330.5703125,
+61597.91796875,
+62757.12890625,
+64242.79296875,
+64863.09765625,
+66930.390625]|code-block|syntax|javascript|850914|第2步：|850915|seen+=+[]+#to+log+used+indices+pairs
diff_dic+=+{}+#to+record+indices+and+diff
for+i,a+in+enumerate(list):
++for+j,b+in+enumerate(list):
++++if+i!=j+and+(i,j)[::-1]+not+in+seen:
++++++seen.append((i,j))
++++++diff_dic[(i,j)]+=+abs(a-b)
++++++

keys+=+[]
for+ind,+diff+in+diff_dic.items():
++if+diff+<=+850:
++++keys.append(ind)
uniques_k+=+[]+#to+record+unique+indices
for+pair+in+keys:
++for+key+in+pair:
++++if+key+not+in+uniques_k:
++++++uniques_k.append(key)

import+numpy+as+np
list_arr+=+np.array(list)
nearest_avg+=+np.mean(list_arr[uniques_k])
list_arr+=+np.delete(list_arr,+uniques_k)
list_arr+=+np.append(list_arr,+nearest_avg)
list_arr|850916|产出：|850917|array([+5620.77978516,+34592.15625,+36542.109375,+39478.953125,+48022.2890625,+52535.13671875,+58330.5703125+,+61597.91796875,+62757.12890625,+66930.390625++,+20566.00205365])|850918|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Step 1:
<pre><code>list = [5620.77978515625,
 7388.43017578125,
 7683.580078125,
 8296.6513671875,
 8320.82421875,
 8557.51953125,
 8743.5,
 9163.220703125,
 9804.7939453125,
 9913.86328125,
 9940.1396484375,
 9951.74609375,
 10074.23828125,
 10947.0419921875,
 11048.662109375,
 11704.099609375,
 11958.5,
 11964.8232421875,
 12335.70703125,
 13103.0,
 13129.529296875,
 16463.177734375,
 16930.900390625,
 17712.400390625,
 18353.400390625,
 19390.96484375,
 20089.0,
 34592.15625,
 36542.109375,
 39478.953125,
 40782.078125,
 41295.26953125,
 42541.6796875,
 42893.58203125,
 44578.27734375,
 45077.578125,
 48022.2890625,
 52535.13671875,
 58330.5703125,
 61597.91796875,
 62757.12890625,
 64242.79296875,
 64863.09765625,
 66930.390625]
</code></pre>
Step 2:
<pre><code>seen = [] #to log used indices pairs
diff_dic = {} #to record indices and diff
for i,a in enumerate(list):
 for j,b in enumerate(list):
 if i!=j and (i,j)[::-1] not in seen:
 seen.append((i,j))
 diff_dic[(i,j)] = abs(a-b)
 

keys = []
for ind, diff in diff_dic.items():
 if diff &lt;= 850:
 keys.append(ind)
uniques_k = [] #to record unique indices
for pair in keys:
 for key in pair:
 if key not in uniques_k:
 uniques_k.append(key)

import numpy as np
list_arr = np.array(list)
nearest_avg = np.mean(list_arr[uniques_k])
list_arr = np.delete(list_arr, uniques_k)
list_arr = np.append(list_arr, nearest_avg)
list_arr
</code></pre>
output:
<pre><code>array([ 5620.77978516, 34592.15625, 36542.109375, 39478.953125, 48022.2890625, 52535.13671875, 58330.5703125 , 61597.91796875, 62757.12890625, 66930.390625 , 20566.00205365])
</code></pre>

<img src="https://i.stack.imgur.com/CCVym.png" alt="image" />
I currently have the numbers above in a list. How would you go about adding similar numbers (by nearest 850) and finding average to make the list smaller.
For example I have the list
<pre><code>l = [2000,2200,5000,2350]
</code></pre>
In this list, i want to find numbers that are similar by n+500
So I want all the numbers similar by n+500 which are 2000,2200,2350 to be added and divided by the amount there which is 3 to find the mean. This will then replace the three numbers added. so the list will now be <code>l = [2183,5000]</code>
As the image above shows the numbers in the list. Here I would like the numbers close by n+850 to all be selected and the mean to be found

Finding similar numbers in a list and getting the average

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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​​我现在把上面的数字列在一张清单上。您将如何添加相似的数字(以最近的850)，并找到平均值，以使列表更小。例如，我有一个列表l = [2000,2200,5000,2350]在这个列表中，我希望找到与n+500相似的数字所以，我想让所有与n+500相似的数字，也就是2000,2200,2350，被加起来，除以3的数值...

问在列表中找到相似的数字并得到平均值
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