find max value of structure

I have a small problem today.

I have gotten a struct that's 1x30 and it's just 30 values in the one column. It shoudn't be so hard but I need to find the max values of these. But somehow I cant do this.

Simple question, but today I can't do anything.

Answer 1

[Image Analyst]

Try using the brackets trick to extract the values out of the fields and concatenate into a vector:

allValues = [yourStructure.yourFieldName]
maxValue = max(allValues)

If your field is a row vector instead of a scalar, use vertcat():

allValues = vertcat(yourStructure.yourFieldName)
columnMaxValues = max(allValues, 1) % Max going down all the columns
rowMaxValues = max(allValues, 2) % Max going across all the rows.

Answer 2

[Johannes Rebling]

max([Struct(:).yourData])

[Tracy]

Thank you! had thesame problem and surprisingly this works perfectly. cant beleive it was such a simple fix.

[Image Analyst]

@Tracy yes, and it could be simpler. Note in my Accepted Answer above I don't use (:) because it's unnecessary if the values are simply scalars. And you will want to assign the output of max() to some variable (like I did above). Otherwise it's just basically thrown away and you can't use it. Above I essentially did

maxValue = max([yourStructure.yourFieldName])

Or for his code...

Struct(1).yourData = 1;
Struct(2).yourData = 2;
maxValue = max([Struct.yourData]) % No (:) needed.  And save the max value for use in subsequent code.

Answer 3

[Peter O]

If I'm understanding correctly, you have a struct array, s, of size 1x30, with a single field (let's call it x). Something like

s(1).x= 10
s(2).x = 11
...
s(30).x = 1.123

Use arrayfun to access the field of each struct in the array. Because there's a uniform output, you'll get a vector output. Then take the max:

max(arrayfun(@(z) z.x, s))

Alternatively, use the colon operator to access each element, turn it into a single array using the brackets, and then take the max:

max([s(:).x])

Hope this helps!

[Image Analyst]

Like I showed in my answer, the (:) is unneeded.