What is going wrong with this code

wves=xlsread('Data_all.xlsx','Wave_data');
c=wves(3:end,Waves);
disp(c)
WH01=0;
WH23=0;
WH45=0;
WH67=0;
WH89=0;
WH1011=0;
WH1213=0;
for i = 0:1:498
if (c(i) == 0) || (c(i) == 1)
WH01=WH01+1;
elseif (c(i) == 2) || (c(i) == 3)
WH23=WH23+1;
elseif (c(i)==4) || (c(i)==5)
WH45=WH45+1;
elseif (c(i)==6) || (c(i)==7)
WH67=WH67+1;
elseif (c(i)==8) || (c(i)==9 )
WH89=WH89+1;
elseif (c(i)==10) || (c(i)==11)
WH1011=WH1011+1;
elseif (c(i)==12) || (c(i)==13 )
WH1213=WH1213+1;
end
end

4 Kommentare

Sriram Tadavarty
Sriram Tadavarty am 25 Apr. 2020
Replace c(i) with c(i+1), or start i in the for loop from 1 rather than 0. MATLAB indexing is 1 based.
Themistoklis Pytharides
Themistoklis Pytharides am 25 Apr. 2020
Thanks a lot
Walter Roberson
Walter Roberson am 25 Apr. 2020
Also Waves does not appear to be defined.
Themistoklis Pytharides
Themistoklis Pytharides am 25 Apr. 2020
The Waves are defined in the beggining of the complete code

Antworten (2)

dpb
dpb am 25 Apr. 2020

0 Stimmen

>> n=arrayfun(@(i1,i2) sum(iswithin(c,i1,i2)),0:2:13,1:2:13)
n =
14 15 10 14 14 17 16
>> type iswithin
function flg=iswithin(x,lo,hi)
% returns T for values within range of input
% SYNTAX:
% [log] = iswithin(x,lo,hi)
% returns T for x between lo and hi values, inclusive
flg= (x>=lo) & (x<=hi);
>>
Presuming the magic number 498 is numel(c)-1
This should be a job for one of the histogram functions, but there's no way Io make one of them use the inclusive bin edges to count the two end points within the bin even for nonoverlapping data

2 Kommentare

h = histcounts(c, 0:2:14)
The first bin corresponds to [0, 2) which for integer c data means it captures c = 0 and c = 1. The last bin corresponds to [12, 14] which captures c = 12 and c = 13 (and would capture c = 14, but the question implies that c only contains integer values between 0 and 13 inclusive.)
dpb
dpb am 25 Apr. 2020
I thought I had done that, Steven...[search command history....] Oh! I had tried to use 1 to capture inclusive. I don't see why my attempt at using a tolerance -- oh, I do now.
I just wasn't thinking about the [0, 2) definition correctly.
Thanks for setting me straight...
dpb
dpb am 25 Apr. 2020

0 Stimmen

Alternate solution:
ix=kron([1:numel(0:13)/2].',ones(2,1));
n=accumarray(ix(c+1),ones(size(c)));
>> n
n =
14
15
10
14
14
17
16
>>

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am 20 Aug. 2021

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