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How do I avoid using a lot of if statements in a row?

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Samuele Bolotta
Samuele Bolotta 2020-4-1
关闭： MATLAB Answer Bot 2021-8-20
Hi all! I have to execute the content of the following code, which I'm sure can be written in a much more elegant and efficient way. The meaning is clear: for each of the 11 events I'm interested in, if the matrix dffx that corresponds to that event is empty, then it becomes [dffx,num_dff]; otherwise, it becomes [dff1;num_dff]. I am sure that such a trivial work can be executed in way less lines of code. Can you help? Thank you!
if code_events(i) == 1
if isempty(dff1)
dff1=[dff1,num_dff];
else
dff1=[dff1;num_dff];
end
elseif code_events(i) == 2
if isempty(dff2)
dff2=[dff2,num_dff];
else
dff2=[dff2;num_dff];
end
elseif code_events(i) == 3
if isempty(dff3)
dff3=[dff3,num_dff];
else
dff3=[dff3;num_dff];
end
elseif code_events(i) == 4
if isempty(dff4)
dff4=[dff4,num_dff];
else
dff4=[dff4;num_dff];
end
elseif code_events(i) == 5
if isempty(dff5)
dff5=[dff5,num_dff];
else
dff5=[dff5;num_dff];
end
elseif code_events(i) == 6
if isempty(dff6)
dff6=[dff6,num_dff];
else
dff6=[dff6;num_dff];
end
elseif code_events(i) == 7
if isempty(dff7)
dff7=[dff7,num_dff];
else
dff7=[dff7;num_dff];
end
elseif code_events(i) == 8
if isempty(dff8)
dff8=[dff8,num_dff];
else
dff8=[dff8;num_dff];
end
elseif code_events(i) == 9
if isempty(dff9)
dff9=[dff9,num_dff];
else
dff9=[dff9;num_dff];
end
elseif code_events(i) == 10
if isempty(dff10)
dff10=[dff10,num_dff];
else
dff10=[dff10;num_dff];
end
elseif code_events(i) == 11
isempty(dff11)
dff11=[dff11,num_dff];
else
dff11=[dff11;num_dff];
end
end
end
  5 个评论
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Stephen23
Stephen23 2020-4-1
C = {dff1,dff2,...};
for k = 1:numel(C)
C{k} = [C{k};num_dff];
end
Using numbered variables is usually a sign that your code will be complex and/or inefficient.
Samuele Bolotta
Samuele Bolotta 2020-4-1
Thanks! This is almost what I was looking for, except for the fact that, as you can see in the first code I uploaded, I have 11 code events (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11) which correspond to 11 behavioural events and 11 dffs. So I have to find a way to link each dff in C with its code event (dff1 with code event 1, dff2 with code event 2...).

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回答（2 个）

Star Strider
Star Strider 2020-4-1
Another option is switch,case,otherwise. You will need to determine if that is an improvement over the multiple if blocks.
Steven Lord
Steven Lord 2020-4-1
Do you need to have eleven individually numbered variables, or are they all the same size and type (and therefore "stackable" as rows / columns / pages / etc. of a larger array)?
x1 = [ 1; 2; 3; 4];
x2 = [ 5; 6; 7; 8];
x3 = [ 9; 10; 11; 12];
x4 = [13; 14; 15; 16];
x = reshape(1:16, [4 4]); % [x1; x2; x3; x4]
With those examples, it would be much easier to iterate over the columns of x than the individual variables x1, x2, x3, and x4.
If they're not all the same size or not all the same type, consider a cell array.
  1 个评论
Samuele Bolotta
Samuele Bolotta 2020-4-1
Thanks for your answer!
No, the eleven variables are all numerical but have different sizes; what I did now was:
switch code_events(i)
case 1
dff1=[dff1;num_dff];
case 2
dff2=[dff2;num_dff];
case 3
dff3=[dff3;num_dff];
case 4
dff4=[dff4;num_dff];
case 5
dff5=[dff5;num_dff];
case 6
dff6=[dff6;num_dff];
case 7
dff7=[dff7;num_dff];
case 8
dff8=[dff8;num_dff];
case 9
dff9=[dff9;num_dff];
case 10
dff10=[dff10;num_dff];
case 11
dff11=[dff11;num_dff];
end
But I'm still convinced it coul be improved, in the end the task is very simple and the only difference between the 11 cases is the number. I have 11 code events (1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11) which correspond to 11 behavioural events and 11 dffs. And for each of those 11 code events, its dff gives rise to the matrix with that same structure. I'm looking for an elegant and efficient way to carry out this task!

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