How to code "while" loop for the following problem?

조회 수: 7 (최근 30일)
Mohammad Sulaiman Stanekzai
Mohammad Sulaiman Stanekzai 2019년 7월 14일
편집: Image Analyst 2019년 7월 14일
hello friends. I want to write "while" loop for the follwing problem but it don't give the answer which i want. ii starts from 0. The system which i want is like that.
3.749893 , 3.749893 + ii(0) = ii1
5.217726 , ii1 + 5.217726 = ii2
0.412081 , ii2 + 0.412081 = ii3
0.113385 , ii3 + 0.113385 = ii4
1.27062 , ii4 + 1.27062 = ii5
2.740936 , ii5 + 2.740936 = ii6
2.208374 , ii6 + 2.208374= ii7
so on..
I cant code this proble in while loop.
Thanks in advance.
ii = 0;
for sun = 1:1:p ;
if load(sun,1) < pv (sun,1);
h = pv(sun,1)-load (sun,1);
n(sun,1) = double (h); % (PV'den artan Enerji)
deger = sum (n); % (Artan Enerjinin toplam degeri)
Q = find (n > 0);
answer = n(Q);
WW= answer / 4;
k = 1;
while ii < deger ;
g2 = ii + WW;
ii = g2;
end
end
  댓글 수: 4
KALYAN ACHARJYA
KALYAN ACHARJYA 2019년 7월 14일
편집: KALYAN ACHARJYA 2019년 7월 14일
Sorry, still the question is not clear for me.
Mohammad Sulaiman Stanekzai
Mohammad Sulaiman Stanekzai 2019년 7월 14일
편집: Mohammad Sulaiman Stanekzai 2019년 7월 14일
3.749893
5.217726
0.412081
0.113385
1.27062
2.740936
2.208374
above values are belong to WW varaible in ''while'' loop.
I want to add them like;
ii + 3.749893 = ii1 here the first value of ii is '0'
ii1 + 5.217726 = ii2 here the value of ii1 is (3.749893 + 0)
ii2 + 0.412081 = ii3 here the value of ii2 is (ii1 + 5.217726)
ii3 + 0.113385 = ii4 here the value of ii3 is (ii2 + 0.412081)
ii4 + 1.27062 = ii5 here the value of ii4 is ( ii3 + 0.113385)
ii5 + 2.740936 = ii6 here the value of ii5 is ( ii4 + 1.27062)
ii6 + 2.208374= ii7 here the value of ii6 is ( ii5 + 2.740936)
and so on.
I hope you understand what i mean.

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채택된 답변

KALYAN ACHARJYA
KALYAN ACHARJYA 2019년 7월 14일
편집: KALYAN ACHARJYA 2019년 7월 14일
ww=[3.749893
5.217726
0.412081
0.113385
1.27062
2.740936
2.208374];
% Here considering sample ww only
count=1;
result=0;
while count<=length(ww)
result=ww(count)+result;
count=count+1;
end
disp(result);
  댓글 수: 12
Mohammad Sulaiman Stanekzai
Mohammad Sulaiman Stanekzai 2019년 7월 14일
편집: Image Analyst 2019년 7월 14일
Hello Sir, i have a question if could help me please,
i have values array like
Y = [24.1013520547196
24.1013520547196
24.1013520547196
24.1013520547196
5.56899020519836
70.5900447343365
74.5811543813954 ]
This time ii = 11698.56
ii - 24.1013520547196 = ii1 here the first value of 'ii' is '11698.56'
ii1 - 24.1013520547196 = ii2 here the value of ii1 is (ii - 24.1013520547196)
ii2 - 24.1013520547196 = ii3 here the value of ii2 is ( ii1 - 24.1013520547196)
ii3 - 24.1013520547196= ii4 here the value of ii3 is (ii2 - 24.1013520547196)
ii4 - 5.56899020519836 = ii5 here the value of ii4 is ( ii3 - 24.1013520547196)
ii5 - 70.5900447343365 = ii6 here the value of ii5 is ( ii4 - 5.56899020519836)
ii6 - 74.5811543813954= ii7 here the value of ii6 is ( ii5 - 70.5900447343365 )
and so on.
how to code it in while loop?
Thanks in advance.
Image Analyst
Image Analyst 2019년 7월 14일
편집: Image Analyst 2019년 7월 14일
How about
ii = 11698.56
index = 2;
while index <= length(Y)
ii(index) = ii(index - 1) - Y(index - 1)
index = index + 1;
end

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