1x1矩阵的第一个元素与1x1矩阵的所有元素之间的差异 [英] Difference between first element of 1x1 matrix and all elements of 1x1 matrix

问题描述

在中找到一些逻辑之后处理了结构，我想检查一下如何将其推广到矩阵.

After finding some logic in how empty structs are dealt with, I wanted to check how this generalized to matrices.

在这里，我注意到以下内容:

Here I noticed the following:

如果您具有1x1矩阵，则将其分配给第一个元素.这与分配给所有元素不同.

If you have a 1x1 matrix, and assign to the first element. It is not the same as assigning to all elements.

这让我感到很惊讶，因为第一个元素实际上与本例中的所有元素相同.这是我的观察结果:

This rather surprised me as the first element is really the same as all the elements in this case. Here are my observations:

x = 1;
y = 1;
z = 1;

x(:) = []; % Evaluates to [] as I expected
y(1) = []; % Evaluates to Empty matrix: 1-by-0, rather than []
z(1,1) = []; %Throws an error: 'Subscripted assignment dimension mismatch.' even though size(z)  gives [1 1];
z(1,:) = []; % Evaluates to Empty matrix: 0-by-1, just like z(:,:) = []

看到此问题后，我的问题是:

After seeing this my question is:

为什么以不同的方式分配给同一事物会导致四个不同的结果?

推荐答案

这似乎是一种一致性.

It just seems like a consistency thing.

让我们考虑一个更大的矩阵，看看行为是否与1 -by- 1矩阵一致(在我看来，扰流板警报):

Lets consider a bigger matrix and see if the behaviour is consistent with a 1-by-1 matrix (spoiler alert, it is in my opinion):

X = rand(3);

案例1:

X(1,1) = []

要使其正常工作是没有意义的.我们无法保留形状并放下单个元素，因此会出现尺寸不匹配误差，这与您的观察一致.尺寸失配也是一个适当的错误，因为我们试图将0 -by- 0矩阵强制插入到1 -by- 1插槽中. (顺便说一句，您说size(z)给您[1 1]，但size(z, 3)也给您1，size(z,7)也给您size(z,7)等，所以实际上它是一个[1 1 1 ...矩阵)

It would make no sense for this to work. We can't preserve the shape and drop a single element hence we get a dimension mismatch error, which is consistent with your observation. Also dimension mismatch is an appropriate error as we're trying to force a 0-by-0 matrix into a 1-by-1 slot. (BTW on a side note you say size(z) gives you [1 1] but size(z, 3) also gives you 1 and so does size(z,7) etc so actually it's a [1 1 1 ... matrix)

案例2:

X(1) = []

这将导致X，使得size(X)是1 -by- 8，如果您指定线性索引，MATLAB似乎很乐意线性化您的矩阵.这对我来说很有意义，并且再次与1 -by- 1情况一致，因为它会导致1 -by- numel(X)-1矩阵(即1 -by- 0对于X = 1)

This results in an X such that size(X) is 1-by-8, MATLAB seems happy to linearize your matrix if you specify a linear index. This makes sense to me, and again is consistent with the 1-by-1 case since it results in a 1-by-numel(X)-1 matrix (i.e. 1-by-0 for X = 1)

案例3:

X(1,:) = []

这很简单，删除第一行，所以现在我们有了一个n-1 -by- m矩阵.因此，3 -by- 3变为2 -by- 3，所以我对1 -by- 1变为0 -by- 1感到满意这个案例.请注意，X(:,1) = []在相似的静脉中也是一致的.

That's pretty straight forward, remove the first row so now we have an n-1-by-m matrix. So a 3-by-3 becomes a 2-by-3 so I'm happy with a 1-by-1 becoming a 0-by-1 in this case. Note that X(:,1) = [] is also consistent in a similar vein.

案例4:

X(:) = []

这很有意义，重新分配整个矩阵.没有问题.没有歧义.

This one just makes sense, re-assign the entire matrix. No issues. No ambiguity.

因此，即使可以肯定，它们都可以毫无疑问地表示同一件事.我认为，对于MATLAB而言，与对单个元素矩阵进行相同操作相比，获得与大型矩阵一致的不同结果是完全合理的设计选择.

So even though sure, they all could unambiguously mean the same thing. I think it's a perfectly reasonable design choice for MATLAB to have different results that are consistent with larger matrices than always do the same thing for a single element matrix.

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