How to extract data from a multi layered structure
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Hi,
I have a multi-layered structure A that include 3 structures (a, b and c) and and all of them contain 10 fieldnames with 10 rows and every cell further contains 10 double numbers mixed with nans.
Doing
A.b.fieldname1
yields the content of the first cell (10 double), but I cant seem to to be able to extract the entire fielname. Doing
c = cell2mat(struct2cell(A.b.fieldname1(1,1)))
yields the following error: Expected one output from a curly brace or dot indexing expression, but there were 10 results.
So how can I extract the data from this multi-layered structure ?
Thank you,
6 commentaires
Stephen23
le 22 Août 2019
Modifié(e) : Stephen23
le 22 Août 2019
Your structure is non-scalar:
>> load('PMAN_climatology.mat')
>> size(pman_climatology)
ans =
1 14
That explains the error message you give in your question.
Not only that, but none of its fields contain nested structures:
>> F = @(s)any(structfun(@isstruct,s));
>> arrayfun(F,pman_climatology)
ans =
0 0 0 0 0 0 0 0 0 0 0 0 0 0
"How can i calculate the minimum of a given field (min_temp for example)?"
You don't define "the minimum" for mutliple vectors: do you want the global minimum, or the minimum of each vector, or the minimum of each column when the vectors are concatenated into one matrix, or some other minimum?
Réponses (1)
Stephen23
le 22 Août 2019
Modifié(e) : Stephen23
le 23 Août 2019
"How can i calculate the minimum of a given field (min_temp for example) ?"
You do not define what "the minimum" means for multiple vectors, which makes it hard to guesss what output you expect...
Method one: comma-separated list and cellfun:
For example:
>> vec = cellfun(@min,{pman_climatology.min_temp})
vec =
1.0000 4.6000 3.0000 3.0000 2.1000 2.5000 2.0000 2.0000 6.5000 1.5000 1.0000 1.5000 4.1000 1.0000
That gives the minimum value of the field min_temp for each element of the structure pman_climatology. Compare with the first few elements of pman_climatology:
>> min(pman_climatology(1).min_temp)
ans =
1
>> min(pman_climatology(2).min_temp)
ans =
4.6000
>> min(pman_climatology(3).min_temp)
ans =
3
Method two: anonymous function and arrayfun:
>> F = @(s)min(s.min_temp);
>> arrayfun(F,pman_climatology)
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