操作“内部"&“内部"清单 [英] Operations "intra" & "inter" lists
问题描述
请考虑以下内容:
lesDisques={{14.2065, 10.609, 0.974938}, {19.5653, 6.92721, 0.974938},
{30.4607,17.4802, 0.974938}, {27.4621, 10.0393, 0.974938},
{15.915, 20.4278,0.974938}, {28.6921, 5.2132, 1.53205},
{27.0317, 24.8346,1.53205}, {20.8853, 18.8588, 1.53205}}
where lesDisques[[#]] is {X,Y,R}
frmCorner = {{6.5946, 1.5946`}, {6.5946, 28.4054`},
{60.2162`,28.4054`}, {33.4054`, 28.4054`}}
cog = {23.91871026577044`, 15.010499627383863`}
scrCenter = {20, 15}
frmXY={{6.5946, 1.5946}, {33.4054, 28.4054}}
Graphics[{
White, EdgeForm[Thick],
Rectangle @@ frmXY,
Red, PointSize[.04],
Point@cog,
Black, Disk @@@ (lesDisques /. {a_, b_, c_} :> {{a, b}, c})},
ImageSize -> 600]
对于8个磁盘中的每一个，
For each of the 8 disks,
我想计算其边缘和:之间的最小距离.
I would like to compute the minimum distance between its edge and :
-其他所有磁盘的边缘(7个值)-每个框架角{4个值}，
-The edge of every other disks (7 values) -Each frame corner {4 values},
然后我将获得8个包含11个值的列表.
I would then obtain 8 lists of 11 values.
以下内容使我可以指出"磁盘的周边:
The following enables me to "pointize" the disks perimeter :
pointize[{{x_,y_},r_},size_:12]:=Table[{x+r Cos[i ((2\[Pi])/size)],
y+r Sin[i ((2\[Pi])/size)]},{i,0,size}]
这样我可以找到2个磁盘的2个最接近的点并计算距离，但是我觉得这可能不是正确的方法.
With this I could find the 2 closest points for 2 disks and compute de distance, but I feel this might not be the right way to do it.
推荐答案
尝试一下，
Outer[ Norm[#1[[;;2]] - #2[[;;2]]] - #1[[3]] - #2[[3]]&, #, #, 1]& @ lesDisques
通过计算光盘中心之间的距离 Norm [#1 [[;; 2]]-#2 [[;; 2]] 进行工作，然后减去它们对所有光盘的半径.但是，对于较大的列表，这可能不是最快的方法，因为它会两次计算所有值，但这很简单.
It works by computing the distance between the centers of the discs, Norm[#1[[;;2]] - #2[[;;2]]], and then subtracting off their radii for all pairs of discs. For a large list, though, this may not be the speediest as it computes all the values twice, but it is straightforward.
为了加快速度，首先我们需要确定要计算的对.一种简单的方法是使用
For a speed up, first we need to determine what the pairs we want to calculate are. A straightforward way is to determine all unique pairs with
Subsets[Range[Length@lesDisques], {2}]
返回
{{1, 2}, {1, 3}, {1, 4}, {1, 5}, {1, 6}, {1, 7}, {1, 8}, {2, 3},
{2, 4}, {2, 5}, {2, 6}, {2, 7}, {2, 8}, {3, 4}, {3, 5}, {3, 6},
{3, 7}, {3, 8}, {4, 5}, {4, 6}, {4, 7}, {4, 8}, {5, 6}, {5, 7},
{5, 8}, {6, 7}, {6, 8}, {7, 8}}
此外，它不与光盘配对，与外部不同.我会这样使用它
Also, it doesn't pair up the discs with themselves, unlike Outer. I'd use it like this
With[{ps = lesDisques[[ # ]]},
Norm[#1[[;;2]] - #2[[;;2]]] - #1[[3]] - #2[[3]]& @@ ps ]& /@
Subsets[Range[Length@lesDisques], {2}]
编辑:我不喜欢在多个地方使用 lesDisques 之类的变量，因为这样会使以后更改变得更加困难.因此，这里是一个重写:
Edit: I have a pathological dislike for using variables like lesDisques in more than one place as it makes it more difficult to change later. So, here's a rewrite:
With[{ps = #},
Norm[Subtract @@ ps[[#1,;;2]]] - Plus @@ ps[[#1,3]]& /@
Subsets[Range[Length@ps], {2}]
]& @ lesDisques
编辑: Subsets 版本具有 Outer 版本不存在的缺陷，如您所写，您无法分辨出哪对磁盘正在比较.这是改写的版本，
Edit: The Subsets version has a flaw that the Outer version does not, as written you can't tell which pair of disks are being compared. Here's a rewritten version,
With[{ps = #},
Rule[#1,Norm[Subtract @@ ps[[#1,;;2]]] - Plus @@ ps[[#1,3]]]& /@
Subsets[Range[Length@ps], {2}]
]& @ lesDisques
返回
{{1, 2} -> 4.55184, {1, 3} -> 15.697, {1, 4} -> 11.318,
{1, 5} -> 8.01646, {1, 6} -> 12.9509, {1, 7} -> 16.6464,
{1, 8} -> 8.10742, {2, 3} -> 13.2184, {2, 4} -> 6.53803,
{2, 5} -> 12.0355, {2, 6} -> 6.77936, {2, 7} -> 16.8946,
{2, 8} -> 9.4974, {3, 4} -> 6.0725, {3, 5} -> 12.8915,
{3, 6} -> 9.88685, {3, 7} -> 5.60752, {3, 8} -> 7.16714,
{4, 5} -> 13.5826, {4, 6} -> 2.47339, {4, 7} -> 12.2946,
{4, 8} -> 8.49473, {5, 6} -> 17.361, {5, 7} -> 9.45131,
{5, 8} -> 2.70508, {6, 7} -> 16.6274, {6, 8} -> 12.6569,
{7, 8} -> 5.50844}
在我看来，我从未回答过您问题的第二部分，而是找到了光盘和框架角之间的最小距离.最好使用外部完成此任务，因为没有任何多余的计算.所以，这就是我要做的
It occurs to me that I never answered the second part of your question, find the minimum distances between the discs and the corners of the frame. This task is best done using Outer as there aren't any redundant calculations. So, this is what I'd do
Outer[ Norm[#1 - #2[[;;2]]]- #2[[3]]&, #1, #2, 1]& @@ {frmCorner, lesDisques}
这只是对原始代码的微小修改.请注意，在外部生成的矩阵中，行对应于第一个输入(在这种情况下为 frmCorner )，列对应于第二个输入，如下所示
which is only a slight modification of the original code. Note, in the matrix generated by Outer, the rows correspond to the first input (frmCorner in this case) and the columns to the second input, as follows
{{10.8234, 13.0492, 27.6946, 21.5365, 20.0384, 20.8598, 29.4159, 20.8795},
{18.381, 24.1159, 25.2729, 26.8237, 11.2934, 30.502, 19.2147, 15.654},
{48.3566, 45.0012, 30.7229, 36.577, 44.0388, 37.6042, 31.844, 38.9409},
{25.2035, 24.5762, 10.3402, 18.3289, 18.2489, 22.1342, 5.77375, 14.2125}}
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