can any one tell me how this code is working??

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can any one tell me how this code is working ..I cant able to understand the logic ..pls help.(Please note that locs_Rwave will be always higher than locs_Qwave)
q=0;
k=1;
locs_Rwave = [116 110 57]';
locs_Qwave=[110 210 356 457 575 313]';
for j=k:size(locs_Rwave)
for i=1:numel(locs_Qwave)
if (i== numel(locs_Qwave))
q=[q locs_Qwave(i)];
break;
end
if( locs_Qwave(i)>locs_Rwave(j))
q=[q locs_Qwave(i-1)];
break;
end
end
end
q
Thanks in advance

回答(1 个)

Roger Stafford
Roger Stafford 2014-9-15
This does not seem like good code to me. It is likely that what it does is not what its author intended.
For example, if one of the 'locs_Rwave' elements happens to be less than the first element of 'locs_Qwave' (which is not true in this particular case,) then when j indexes that element of 'locs_Rwave' and when the inner loop sets i = 1, there would be an attempted access to locs_Qwave(0) which would result in an error message. One can only speculate as to what was intended here.
Also the line
for j=k:size(locs_Rwave)
is strange because size(locs_Rwave) is a two-element vector. Fortunately for the author the colon operator apparently ignores the second part of it, but why didn't the author write a more straightforward size(locs_Rwave,1) or numel(locs_Rwave)?
Furthermore, I would guess that q was meant to be empty at the beginning instead of being set to the scalar quantity zero. If so, the proper code should have been "q = [];". That zero will persist as locs_Qwave elements are appended to q.
  2 个评论
Abhishek sadasivan
Abhishek sadasivan 2014-9-15
locs_Rwave will always a higher value than locs_Qwave..(R wave of ECG data).
Roger Stafford
Roger Stafford 2014-9-15
编辑:Roger Stafford 2014-9-15
That's not true in the example you have given above:
locs_Rwave = [116 110 57]';
locs_Qwave = [110 210 356 457 575 313]';
For these values, only the fact that 57 is the last element in locs_Rwave saves this code from issuing an error message about locs_Qwave(0).

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