WP7 ApplicationBarIcon可见性(WP7 ApplicationBarIcon Visibility)
感谢WP7 ApplicationBarIcon不是标准控件。
我需要能够以编程方式隐藏它(我需要隐藏而不是禁用)
1 /除了添加/删除图标之外,还有其他方法可以做到这一点
2 /假设我必须添加和删除它,如何将事件与我添加的控件相关联?
- 谢谢
Appreciate that the WP7 ApplicationBarIcon is not standard control as such.
I need to be able to hide this programatically (I need to hide rather than disable)
1/ is there any other way I can do this other than adding/removing the icon
2/ assuming that I have to add and remove it, how do I associate an event to the control that I am adding?
- thanks
原文:https://stackoverflow.com/questions/6702866
最满意答案
你可能需要这个:
% generate small dummy data nxs = 2; nls = 3; nms = 4; nks = 5; MS = rand(nxs, nls, nms); KS = rand(nxs, nls, nks); R = sum(abs(bsxfun(@minus,MS,permute(KS,[1,2,4,3]))),4)
这将产生一个大小为
[2,3,4]
的矩阵,即[nxs,nls,nms]
。 每个元素[k1,k2,k3]
将对应于R(k1,k2,k3) == sum_k abs(MS(k1,k2,k3) - KS(k1,k2,k))
例如,在我的随机运行中
R(2,1,3) ans = 1.255765020150647 >> sum(abs(MS(2,1,3)-KS(2,1,:))) ans = 1.255765020150647
诀窍是引入具有
permute
单一维度:permute(KS,[1,2,4,3])
的大小为[nxs,nls,1,nks]
,而大小为[nxs,nls,nms]
MS
是隐式的也是大小[nxs,nls,nms,1]
:假设MATLAB中的每个数组都具有可数无限数量的尾随单例维度。 从这里可以很容易地看到如何将大小分别为[nxs,nls,nms,1]
和[nxs,nls,1,nks]
数组bsxfun
组合在一起,以获得大小为[nxs,nls,nms,nks]
。 沿着维度4求和可以达成交易。
我在评论中指出,将求和指数置于首位可能会更快。 事实证明,这本身会使代码运行得更慢 。 但是,通过重新整形数组以减小尺寸大小,整体性能会提高(由于最佳内存访问)。 比较一下:
% generate larger dummy data nxs = 20; nls = 30; nms = 40; nks = 500; MS = rand(nxs, nls, nms); KS = rand(nxs, nls, nks); MS2 = permute(MS,[4 3 2 1]); KS2 = permute(KS,[3 4 2 1]); R3 = permute(squeeze(sum(abs(bsxfun(@minus,MS2,KS2)),1)),[3 2 1]);
我所做的是将求和
nks
维度放在第一位,然后按降序排列其余维度。 这可以自动完成,我只是不想让这个例子过于复杂。 在您的使用案例中,您可能无论如何都知道尺寸的大小。具有以上两个代码的运行时:原始的0.07028秒,重新排序的一个的0.051162秒(最好的5个)。 不幸的是,现在更大的例子不适合我。
You might need this:
% generate small dummy data nxs = 2; nls = 3; nms = 4; nks = 5; MS = rand(nxs, nls, nms); KS = rand(nxs, nls, nks); R = sum(abs(bsxfun(@minus,MS,permute(KS,[1,2,4,3]))),4)
This will produce a matrix of size
[2,3,4]
, i.e.[nxs,nls,nms]
. Each element[k1,k2,k3]
will correspond toR(k1,k2,k3) == sum_k abs(MS(k1,k2,k3) - KS(k1,k2,k))
For instance, in my random run
R(2,1,3) ans = 1.255765020150647 >> sum(abs(MS(2,1,3)-KS(2,1,:))) ans = 1.255765020150647
The trick is to introduce singleton dimensions with
permute
:permute(KS,[1,2,4,3])
is of size[nxs,nls,1,nks]
, whileMS
of size[nxs,nls,nms]
is implicitly also of size[nxs,nls,nms,1]
: every array in MATLAB is assumed to possess a countably infinite number of trailing singleton dimensions. From here it's easy to see how you canbsxfun
together arrays of size[nxs,nls,nms,1]
and[nxs,nls,1,nks]
, respectively, to obtain one with size[nxs,nls,nms,nks]
. Summing along dimension 4 seals the deal.
I noted in a comment, that it might be faster to
permute
the summing index to be in the first place. Turns out that this by itself makes the code run slower. However, by reshaping the arrays to have decreasing dimension sizes, the overall performance increases (due to optimal memory access). Compare this:% generate larger dummy data nxs = 20; nls = 30; nms = 40; nks = 500; MS = rand(nxs, nls, nms); KS = rand(nxs, nls, nks); MS2 = permute(MS,[4 3 2 1]); KS2 = permute(KS,[3 4 2 1]); R3 = permute(squeeze(sum(abs(bsxfun(@minus,MS2,KS2)),1)),[3 2 1]);
What I did was put the summing
nks
dimension into first place, and order the rest of the dimensions in decreasing order. This could be done automatically, I just didn't want to overcomplicate the example. In your use case you'll probably know the magnitude of the dimensions anyway.Runtimes with the above two codes: 0.07028 s for the original, 0.051162 s for the reordered one (best out of 5). Larger examples don't fit into memory for me now, unfortunately.
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