How to calculate moving average
이전 댓글 표시
Hi,
I have following data:
0.10
0.24
0.30
0.25
0.33
0.35
0.46
0.47
0.48
0.51
0.52
0.53
0.53
0.57
0.58
0.58
0.58
0.59
0.63
0.64
0.66
0.72
I want to take 3-points moving average,
Please help some one
댓글 수: 2
Quin Silk
2019년 12월 11일
Can someone write code to do it without using movmean?
Star Strider
2019년 12월 11일
@Quin Silk — See the Moving Average Filter link in my Answer.
답변 (6개)
Bandar
2019년 12월 11일
Take a look this data
file=[1 2 3 4];
movmean(file,3)
which returns
1.5000 2.0000 3.0000 3.5000
The filter works as follows:
1 2 (1+2)/2 = 1.5 when k points at 1
1 2 3 (1+2+3)/3 = 2.0 when k points at 2
2 3 4 (2+3+4)/3 = 3.0 when k points at 3
3 4 (3+4)/2 = 3.5 when k points at 4
Now it is easy to convert it to a logical code or merely use movmean().
댓글 수: 3
Quin Silk
2019년 12월 11일
How do you do this without using movmean?
Bandar
2019년 12월 11일
A=[1 2 3 4 5 6];
m=ones(1,length(A));
for i=1:length(A)
if i == 1
m(i) = (A(i)+A(i+1))/2;
elseif i == length(A)
m(i) = (A(i-1)+A(i))/2;
else
m(i) = (A(i-1)+A(i)+A(i+1))/3;
end
end
movmean(A,3)
m
Micheal Omojola
2020년 9월 8일
@Bandar Aldhafeeri: Thank you! Your answer was helpful.
KL
2017년 11월 16일
If you have 2016b or later, use movmean,
mean_arr = movmean(your_array,3)
Star Strider
2017년 11월 16일
1 개 추천
댓글 수: 1
Pallavi Bharati
2020년 11월 25일
How to do centered moving average ...three point centred moving average in matlab without the movmean command as i have matlab 2015
dpb
2017년 11월 16일
0 개 추천
Sudhakar Rayabarapu
2018년 9월 8일
편집: Sudhakar Rayabarapu
2018년 9월 8일
0 개 추천
Name your data array A; and keep as a column; then use this command
OutPut = tsmovavg(A, 's', 3, 1)
this will give you the simple moving average
Pallavi Bharati
2020년 11월 24일
0 개 추천
Can anyone help me to compute three point moving average of a 5 year data.I used the filter command but the result are erroneous .I am using MATLAB 2015.And I have a huge data 5 year day wise data and i have to compute three point moving average for each month .
댓글 수: 2
Pallavi Bharati
2020년 11월 24일
Using looping can be erroneous for so large data please suggest something else
The results aren't "erroneous" (presuming you defined the coefficients correctly), they just depend on how one treats the end conditions...examine the following:
>> x=1:10;
>> movmean(x,3)
ans =
1.5000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 9.5000
>> filter(ones(1,3)/3,1,x)
ans =
0.3333 1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000
>> conv(x,ones(1,3)/3,'same')
ans =
1.0000 2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000 6.3333
>> conv(x,ones(1,3)/3,'valid')
ans =
2.0000 3.0000 4.0000 5.0000 6.0000 7.0000 8.0000 9.0000
>>
movmean special-cases the end positions to change the divisor to match the number of elements from 2:N on each end; filter and conv do not.
It's up to you to determine the solution you wish to use; none is theoretically any more or less correct than any other; just different assumptions are made.
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