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Preallocate memory for a cell of structures

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John on 10 Jan 2019
Edited: Stephen Cobeldick on 10 Jan 2019
What is the correct syntax to preallocate memory for the cell x in the following?
N = 10;
for n = 1:N
numRows = ceil(100*rand);
x{n}.field1 = 1*ones(numRows,1);
x{n}.field2 = 2*ones(numRows,1);
x{n}.field3 = 3*ones(numRows,1);

  1 Comment

Stephen Cobeldick
Stephen Cobeldick on 10 Jan 2019
@John: is there a specific requirement to use a cell array of structures? From the code that you have shown, a single non-scalar structure would likely be a better choice:

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Answers (1)

Guillaume on 10 Jan 2019
Just preallocating the cell array:
x = cell(1, N);
for ...
There wouldn't be much point preallocating the scalar structures inside each cell, particularly if you did it naively using repmat as they would be shared copy which would need deduplicating at each step of the loop. You could preallocate the structures inside the loop. For a structure with 3 fields, there wouldn't be much benefit:
x = cell(1, N);
for n = 1:N
x{n} = struct('field1', [], 'field2', [], 'field3', [])
x{n}.field1 = ...
But you may as well fill the structure directly with:
x = cell(1, N);
for n = 1:N
numRows = ceil(100*rand);
x{n} = struct('field1', 1*ones(numRows,1), 'field2', 2*ones(numRows,1), 'field3', 3*ones(numRows,1))
However, instead of a cell array of scalar structures you would be better off using a structure array:
x = struct('field1', cell(1, N), 'field2', [], 'field3', []) %creates a 1xN structure with 3 empty fields
for n = 1:N
x(n).field1 = ...
x(n).field2 = ...
x(n).field3 = ...


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