Delete row from a structure array

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bertie
bertie 2015-9-13
评论： Walter Roberson 2015-9-13
For some reason, the solutions given on the internet don't seem to work for me. The most common solution I have seen is to just do what one would do for arrays. If 'a' is a structure array and I want to delete the 9th entry from all fields:
a(9)=[];
I get an error message saying "Matrix index is out of range for deletion."
What am I doing wrong? My strucutre array has both numeric and cell arrays. I tried removing the cell array field entirely and trying the above operation on the new array (which has only numeric fields), that didn't work either.
Thanks.

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回答（1 个）

Matt J
Matt J 2015-9-13
编辑：Matt J 2015-9-13
As far as can be seen, you simply aren't correctly assessing the number of elements in a. When a has 9 elements or more, as in the following example, things should work
>> clear a
>> a(20).numeric=123; a(20).cell={'dog','cat'}
a =
1x20 struct array with fields:
numeric
cell
>> a(9)=[]
a =
1x19 struct array with fields:
numeric
cell
  2 个评论
bertie
bertie 2015-9-13
Thanks, you are right. Turns out my I have a struct variable (ASSET) that is not an array but has fields that are arrays. What I mean is the size of ASSET is 1x1 but I can access individual entries using ASSET.id(9), say.
ASSET =
id: [1x53369 double]
dtnum: [1x53369 double]
lat: [1x53369 double]
lon: [1x53369 double]
type: {1x53369 cell}
sst: [1x53369 double]
sss: [1x53369 double]
I guess the solution is to delete the rows by accessing each individual field:
ASSET.id(rows to delete)=[] ;
ASSET.dtnum(rows to delete)=[]
.
. and so on.
Not the most elegant way to do things but will work.
Walter Roberson
Walter Roberson 2015-9-13
You might want to consider putting all of the numeric information together into one array, 6 x 53369 double, and accessing by rows when using the data. For example,
as_id = 1;
as_dtnum = 2;
as_lat = 3;
as_lon = 4;
as_sst = 5;
as_sss = 6;
ASSET.data(as_id,431) %equivalent to old ASSET.id(431)
and then you can delete all of the information for one asset by
ASSET.data(:,9) = [];
ASSET.type(9) = [];
(the cell needs to be maintained separately from the numeric values if you want easy numeric access to the numeric data.)

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