how to gain minimum value
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Hi everybody,
r=[1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2]
L/200<r
for example L=246 ,
I want minimum r regarding L/200<r
What I should do??
Thanks.
采纳的回答
Mischa Kim
2014-12-11
编辑：Mischa Kim
2014-12-11
Hamid you could use
r_min = r(min(find(r*200>246==1)))
r_min =
1.3000
4 个评论
Hamid
2014-12-11
Way late to the party, but bearing in mind this answer is showing up in another question, this answer is so so wrong!
1.
a>b == 1
is the same
a>b
2.
min(find(x))
is the same as
find(x, 1)
and will simply return the index of the first non-zero element of x in both case.
The above code is just a very convoluted way of finding the first element of r that is greater than L/200. It's only because the elements are ordered that it also happens to be the minimum of the values greater than L/200.
Mischa Kim
2014-12-13
Guillaume, thanks for your interesting contribution.
Are there more efficient ways of solving the problem? Definitely (see your two suggestions). Are there approaches that are didactically more effective? Possibly. I do believe that my solution provides better readability.
As far as correctness of my solution is concerned your post is somewhat inconsistent in itself, and for this reason disputable. If “It's only because the elements are ordered that it also happens to be the minimum of the values”, it seems to me that you are inferring that the solution provides exactly what the requirements are asking for. Hamid had been asking for a robust way of solving a similar but different problem (e.g. an unordered list of numbers) I would have recommended a different approach.
In summary, and in the spirit of fostering a pleasant, productive forum atmosphere I recommend that you are more careful with your wording.
Guillaume
2014-12-13
Hello Mischa,
I must say I was very surprised by your answer. I am standing by my words though, it is very wrong.
First, whenever a student writes a>b==1, it raises alarm bells. Has operator precedence been understood? Did he mean (a>b)==1 or a>(b==1)? So, it's not a good idea to give this sort of constructs to matlab beginners. In any case, in terms of readability, a>b is certainly more readable than a>b==1. For that matter,
r_filtered = r(r*200>246);
filtered_first = r_filtered(1);
would be a lot more readable and more importantly express what you're doing better.
Secondly, from an algorithm point of view min(find(x)) is also wrong. The find has to go over all the element of x to return the indices of the non-zeros, and the min then has to go over all these indices to ultimately return the first one. find(x, 1) only has to go over x until the first non-zero element. This is obviously a lot more efficient.
More importantly, your answer doesn't answer the question asked: "I want minimum r". It never looks for a minimum in r, so I'm afraid it also wrong conceptually.
更多回答（1 个）
For reference, (see comment to Misha's answer), the correct answer should have been:
r=[1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2];
L = 246;
r_min = min(r(r*200 > L))
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