这样做必须有一个简单的矩阵方法。 我有双管齐下,我想检查所有要素是否平等。 随意强迫这样做是容易的,但寻求更有效的办法: 这些要素是分类账。
我认为,这既简单又简单。
if all(v == v(1))
另一种方法是:
if range(v) == 0
如何:max(v) = min(v)?
另一种解决办法:
numel(unique(v))==1
另一种解决办法:
sum(abs(diff(v))) == 0
You don t even need the all in the solution above, just if v == v(1).
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