How to calculate moving average
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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
2 commentaires
Quin Silk
le 11 Déc 2019
Can someone write code to do it without using movmean?
Star Strider
le 11 Déc 2019
@Quin Silk — See the Moving Average Filter link in my Answer.
Réponses (6)
Bandar
le 11 Déc 2019
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().
3 commentaires
Quin Silk
le 11 Déc 2019
How do you do this without using movmean?
Bandar
le 11 Déc 2019
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
le 8 Sep 2020
@Bandar Aldhafeeri: Thank you! Your answer was helpful.
KL
le 16 Nov 2017
If you have 2016b or later, use movmean,
mean_arr = movmean(your_array,3)
Star Strider
le 16 Nov 2017
1 vote
1 commentaire
Pallavi Bharati
le 25 Nov 2020
How to do centered moving average ...three point centred moving average in matlab without the movmean command as i have matlab 2015
dpb
le 16 Nov 2017
0 votes
Sudhakar Rayabarapu
le 8 Sep 2018
Modifié(e) : Sudhakar Rayabarapu
le 8 Sep 2018
0 votes
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
le 24 Nov 2020
0 votes
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 .
2 commentaires
Pallavi Bharati
le 24 Nov 2020
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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