blocks|key|1174372|text|像这样计算平均值和距离：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1174373|(r+%2B+g+%2B+b)+/+3+=+average
(r+-+average)+%2B+(g+-+average)+%2B+(b+-+average)+=+distance|code-block|syntax|javascript|1174374|这应该会让你对最接近的值有一个很好的了解。|1174375|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Calculate both the average and the distance like this:

<pre><code>(r + g + b) / 3 = average
(r - average) + (g - average) + (b - average) = distance
</code></pre>

This should give you a good idea of the closest value.

blocks|key|1174461|text|下面的代码完全符合您的描述：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1174462|select+(abs(my_R+-+t.r)+%2B+abs(my_G+-+t.g)+%2B+abs(my_B+-+t.b))+/+3+as+difference,+t.*
from+RGBtable+t
order+by+difference+desc;|code-block|syntax|javascript|1174463|然而，使用非线性的东西可能会得到更好的结果。在“取平均值”方法中，如果您的目标颜色是(25，25，25)，则颜色(45，25，25)将比(35，35，35)更接近。然而，我打赌第二个实际上会看起来更近，因为它也是灰色的。|1174464|有几个想法浮现在脑海中:你可以试着在求平均值之前，先求出差异的平方。或者你可以做一些复杂的事情，找到不同值之间具有最接近比率的颜色。找到最接近的比例会让你最接近正确的色调，但不会考虑饱和度(如果我没记错的话…)|1174465|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

The following does exactly what you describe:

<pre><code>select (abs(my_R - t.r) + abs(my_G - t.g) + abs(my_B - t.b)) / 3 as difference, t.*
from RGBtable t
order by difference desc;
</code></pre>

However, you might get better results with something that was non-linear. In the "take the averages" approach, if you goal color is (25, 25, 25) the color (45, 25, 25) would be closer than (35, 35, 35). However, I bet the second would actually look closer, since it would also be gray. 

A few ideas come to mind: you could try squaring the differences before you average them. Or you could do something complicated with finding the color with the closest ratio between the different values. Finding the closest ratios would get you closest to the right hue, but won't account for saturation (if I'm remembering the terms right...)

blocks|key|1389974|text|从the+Wikipedia+page+on+Color+difference的角度来看，其想法是将RGB颜色视为三维中的点。两种颜色之间的差异与两点之间的距离相同：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1389975|difference+=+sqrt((red1+-+red2)%5E2+%2B+(green1+-+green2)%5E2+%2B+(blue1+-+blue2)%5E2)|code-block|syntax|javascript|1389976|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Color_difference^0|1|12|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

From looking at <a href="http://en.wikipedia.org/wiki/Color_difference" rel="nofollow noreferrer">the Wikipedia page on Color difference</a>, the idea is to treat RGB colours as points in three dimensions. The difference between two colours is the same as the distance between two points:

<pre><code>difference = sqrt((red1 - red2)^2 + (green1 - green2)^2 + (blue1 - blue2)^2)
</code></pre>

blocks|key|1390032|text|让数据库为您完成这项工作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1390033|select+top+1
++c.r,
++c.b,
++c.g
from
++color+c
order+by
++(square(c.r+-+@r)+%2B+square(c.g+-+@g)+%2B+square(c.b+-+@b))|code-block|syntax|javascript|1390034|其中@r、@g和@b是要搜索的颜色的r、g、b值(SQL+Server参数语法，因为您没有指定数据库)。请注意，这仍然需要进行表扫描，因为order+by中有一个函数调用。|offset|length|style|CODE|1390035|注意，额外的平方根调用实际上并不是必需的，因为它是一个单调函数。这可能不是很重要，但仍然很重要。|1390036|entityMap^0|0|0|2|2|5|2|8|2|1X|8|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]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|P|$]]

Let the database do it for you:

<pre><code>select top 1
 c.r,
 c.b,
 c.g
from
 color c
order by
 (square(c.r - @r) + square(c.g - @g) + square(c.b - @b))
</code></pre>

Where <code>@r</code>, <code>@g</code>, and <code>@b</code> are the r,g,b values of the color that you're searching for (SQL Server parameter syntax, since you didn't specify a database). Note that this is still going to have to do a table scan since the <code>order by</code> has a function call in it.

Note that the extra square root call isn't actually required since it's a monotonic function. Not that it would probably matter very much, but still.

blocks|key|1174628|text|比平均值更好的一步是最近的平方根：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1174629|((delta+red)%5E2+%2B+(delta+green)%5E2+%2B+(delta+blue)%5E2)%5E0.5|code-block|syntax|javascript|1174630|这最小化了3D颜色空间中的距离。|1174631|因为根是严格递增的，所以您可以搜索平方的最大值。如何在SQL中表达这一点将取决于您使用的RDBMS。|1174632|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

One step better than average is nearest square root:

<pre><code>((delta red)^2 + (delta green)^2 + (delta blue)^2)^0.5
</code></pre>

This minimizes the distance in 3D color space.

Since a root is strictly increasing, you can search for the maximum of the square instead. How you express this in SQL would depend on which RDBMS you're using.

blocks|key|1390164|text|颜色相似度(+Euclidean+distance+difference+=+sqrt(sqr(red1+-+red2)+%2B+sqr(green1+-+green2)+%2B+sqr(blue1+-+blue2))+)是确定两种颜色相似度的standard+way。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1390165|然而，如果你在一个简单的列表中有你的颜色，那么找到最接近的颜色需要计算列表中每种颜色到新颜色的距离。这是一个O(n)运算。|1390166|sqrt()是一个昂贵的操作，如果您只是比较两个距离，那么您可以简单地省略sqrt()。|1390167|如果你有一个非常大的颜色调色板，它可能会更快地将颜色组织到一个kd+tree+(或一个alternatives)中，以减少需要计算的差异的数量。|1390168|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Euclidean_distance|1|http://en.wikipedia.org/wiki/Color_difference|2|http://en.wikipedia.org/wiki/Kd-tree|3|http://en.wikipedia.org/wiki/Nearest_neighbor_search#Space_partitioning^0|Q|27|7|I|0|3A|C|1|0|0|0|6|11|6|0|V|7|2|17|C|3|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@$9|12|A|13|1|14]|$9|15|A|16|1|17]]|E|$]]|$1|F|3|G|5|6|7|18|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|1E|8|@]|D|@$9|1F|A|1G|1|1H]|$9|1I|A|1J|1|1K]]|E|$]]|$1|L|3|-4|5|6|7|1L|8|@]|D|@]|E|$]]]|M|$N|$5|O|P|Q|E|$R|S]]|T|$5|O|P|Q|E|$R|U]]|V|$5|O|P|Q|E|$R|W]]|X|$5|O|P|Q|E|$R|Y]]]]

The <a href="http://en.wikipedia.org/wiki/Euclidean_distance" rel="nofollow noreferrer">Euclidean distance</a> <code>difference = sqrt(sqr(red1 - red2) + sqr(green1 - green2) + sqr(blue1 - blue2))</code> is the <a href="http://en.wikipedia.org/wiki/Color_difference" rel="nofollow noreferrer">standard way</a> to determine the similarity of two colours.

However, if you have your colours in a simple list, then to find the nearest colour requires computing the distance from the new colour with every colour in the list. This is an O(n) operation.

The <code>sqrt()</code> is an expensive operation, and if you're just comparing two distances then you can simply omit the <code>sqrt()</code>.

If you have a very large palette of colours, it is potentially quicker to organise the colours into a <a href="http://en.wikipedia.org/wiki/Kd-tree" rel="nofollow noreferrer">kd tree</a> (or one of the <a href="http://en.wikipedia.org/wiki/Nearest_neighbor_search#Space_partitioning" rel="nofollow noreferrer">alternatives</a>) so as to reduce the number of diffreences that require computing.

I have a rgb value and if it doesn't exist in the color table in my database I need to find the closest color. I was thinking of comparing all values and finding the difference(in red,green,and blue) then take the average. The lowest average deviation should be the closest color. There seems to me like there should be a better way. Any ideas?

Given an RGB value what would be the best way to find the closest match in the database?

SQL Server

数据库

我有一个RGB值，如果它不存在于我的数据库的颜色表中，我需要找到最接近的颜色。我在考虑比较所有的值并找出差异(红色，绿色和蓝色)，然后取平均值。最小的平均偏差应该是最接近的颜色。在我看来，应该有更好的方法。有什么想法吗？

问给定一个RGB值，在数据库中查找最接近的匹配项的最佳方法是什么？
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问给定一个RGB值，在数据库中查找最接近的匹配项的最佳方法是什么？EN

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问给定一个RGB值，在数据库中查找最接近的匹配项的最佳方法是什么？
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