Could anyone suggest me is there any way of choosing maximum and minimum values together

1 回表示 (過去 30 日間)
jaah navi
jaah navi 2019 年 10 月 17 日
コメント済み: Jos (10584) 2019 年 10 月 17 日
A = [23 42 37 18 52];
With respect to command M = max(A) gives 52
min(A) gives 18
Is there any way so that i can get both 52 and 18 together(maximum and minimum together)
  1 件のコメント
dpb 2019 年 10 月 17 日
mnmx=[min(A) max(A)];
or write your own "syntactic sugar" routine minmax()--
function mnmx=minmax(x)
mnmx=[min(x(:)) max(x(:))];
The latter should probably also have the ability to return the locations to mimic builtin interfaces.....and maybe treat arrays the same as them also in working on column-basis by default with the dimension as the optional many choices to make.


回答 (3 件)

Walter Roberson
Walter Roberson 2019 年 10 月 17 日

Jos (10584)
Jos (10584) 2019 年 10 月 17 日
Here is a nice trick that also allows you to combine functions in a single call, which also can return the other outputs of these functions:
minmaxFun = @(x) cellfun(@(F) F(x), {@min, @max}) ;
[V, I] = minmaxFun([2 1 3 5 2])
% V = [1 5], I = [2 4]
You can take this a step further:
mapF = @(val, fcns) cellfun(@(f) f(val{:}), fcns) ;
data = randi(10, [1 10])
[minmax, minmaxIDX] = mapF({data}, {@min @max})
S = mapF({data}, {@mean @median @std})
You can learn much more about this at Loren's Blog:

Daniel M
Daniel M 2019 年 10 月 17 日
編集済み: Daniel M 2019 年 10 月 17 日
A = [23 42 37 18 52];
minmaxVals = prctile(A,[0 100])
ans =
18 52
  3 件のコメント
Walter Roberson
Walter Roberson 2019 年 10 月 17 日
As I have been telling you: use the simple ways first and get them working, and do not worry about fancy ways until you have significant performance problems.
So just call min() and max() instead of spending a lot of time looking for a function that will return the values and indices. Or write a small function yourself that uses min and max internally to return what you want.
Jos (10584)
Jos (10584) 2019 年 10 月 17 日
I second that, Walter! +1



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