Finding a noninteger value from an array

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SAZZAD HOSSAIN
SAZZAD HOSSAIN 2012-12-20
评论： Matthew Murphy 2016-7-6
I have an array which looks like:
x = linspace(0,1,101);
To find the index of the maximum point from an array, I can write:
[val,ind] = max(x);
However, I need the index of a particular point, let's say 0.70. How should i write the commands to find this?
My second question is that I have a specific value y = 0.703. I need to do my calculation using the array x, but I have to start from the value in x that is closest to y. Which means I am looking for the index of the point 0.70 in x, but i need to express it in terms of y in my code. How can I do that?

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采纳的回答

Matt J
Matt J 2012-12-20
Both questions have the same answer
[~,loc]=min(abs(x-y))

更多回答（2 个）

Kye Taylor
Kye Taylor 2012-12-20
编辑：Kye Taylor 2012-12-20
To do (1) and (2) try
y = 0.7; % or y = 0.73;
[~,idx] = min(abs(x-y));
% idx contains the index of the value in x closest to y
Matthew Murphy
Matthew Murphy 2016-7-6
Given the array x, you could also use the find command.
find(x == .7) --> ans = 71
For the second part, you could try a slightly different way with the same command.
Given some plus/minus window of .01, you could do this:
find(x < .703 + .01 & x > .703 - .01 ) --> ans = 71 72
Now, there are multiple values, so you could use a while loop with smaller and smaller margins until one value is found (with a catch to widen the margin if no values are found).
[I know this is an old post, but I still wanted to follow up with an approach that worked for me.]
  2 个评论
Matt J
Matt J 2016-7-6
编辑：Matt J 2016-7-6
Using find() without floating point tolerances can be hit or miss, for example,
>> x=0:0.3:1
x =
0 0.3000 0.6000 0.9000
>> find(x==(1-.7))
ans =
Empty matrix: 1-by-0
That is why I recommended the use of min() for both scenarios raised by the OP,
>> [~,loc]=min(abs(x-(1-.7)))
loc =
2
Matthew Murphy
Matthew Murphy 2016-7-6
I think my version would work if you had the explicit value (eg .3) used in find.
x=0:0.3:1
x =
0 0.3000 0.6000 0.9000
find(x == .3)
ans =
2
You are definitely right with the tolerance issue, no problem there. In my application, I had points to search for, so I had exact values.

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