How to interpolate intermediate values?

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Jacky Jo
Jacky Jo 2018-6-21
评论： Jacky Jo 2018-6-21
I have an array with 110 values. let say:
M1_allvalues = [1,2,10,-1,-2,..,-10, 1, 2..........,10]
I simply want to make the array to a size of 3600 values in it, by interpolating the values in between each array element. There would be approximately 32-33 values between each element to achieve 3600 values array. For example:
Between 1 and 2 in the given array some 32 values, then the beginning would be:
newArray = [1, 1.03, 1.06, 1.09........,1.97, 2,......, 3,........, 110]
How do I do that? I was thinking of this:
for i=1:length(M1_allvalues) - 1
newArray(i,1)= M1_allvalues(i,1): (32-33 vales): M1_allvalues(i + 1,1);
end
could you me some idea?
  2 个评论
Walter Roberson
Walter Roberson 2018-6-21
Is it required that the existing elements all appear in the output exactly? If equal spacing were used then some elements might only be approximated.
Jacky Jo
Jacky Jo 2018-6-21
编辑：Jacky Jo 2018-6-21
@Walter Roberson Yes. Because they represent a curve. If we consider only the start and end the curve will not be thereafter. In other words, upsampling may be. The array look like as follows. For instance, I would like to have between -0.716515302 and -0.738227694 I would like to have 32/33 values. The similar way for all. Do we have any inbuild function for that?
M1_allvalues = [
-0.716515302
-0.738227694
-0.734134713
-0.795537839
-0.721594973
-0.733554246
-0.633938598
-0.646957164
-0.639987555
-0.650192438
-0.69239733
-0.844869671
-0.891111276
-0.97870424
-0.97912141
-0.87953858
-0.87995575
-0.98037292
-0.98079009
-0.98120726
-0.98162443
-0.9820416
-0.982458771
-0.982875941
-0.983293111
1.014964305
0.98042719
0.974544314
0.688186533
0.658870952
0.595105523
0.390863794
0.199207123
0.215257305
0.10254977
-0.051655171
-0.234441557
-0.306198555
-0.304774991
-0.420750606
-0.559915604
-0.507888071
-0.64621695
-1.025155038
-0.515321439
-0.672936192
-0.620544176
-0.660977732
-0.673011345
-0.785923311
-0.607349152
-0.580663475
-0.63031181
-0.765443108
-0.662154214
-0.646247998
-0.628825981
-0.625081084
-0.618565838
-0.661108932
-0.645737598
-0.606520318
-0.639491009
-0.653300752
-0.670253078
-0.659396223
-0.778644332
-0.870010559
-0.939977168
-0.941245761
-0.942514355
-0.943782948
-0.945051542
-0.946320136
-0.947588729
-0.948857323
-0.950125916
-0.95139451
-0.952663103
-0.953931697
1.102059236
0.800832245
0.878517853
0.771312906
0.682760904
0.524194329
0.845499176
0.780904075
0.185449607
0.08183194
0.005542497
-0.129033531
-0.230487187
-0.316896749
-0.413027888
-0.507425914
-0.599394836
-0.616839434
-0.63656488
-0.669927747
-0.668568359
-0.646474955
-0.646278772
-0.582780115
-0.735585371
-0.772032042
-0.757837306
-0.734415844
-0.740307607
-0.74551905 ]

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采纳的回答

KSSV
KSSV 2018-6-21
编辑：KSSV 2018-6-21
% M1_allvalues = [1,2,3,....,110] ;
iwant = linspace(min(M1_allvalues),max(M1_allvalues ),3600) ;
  3 个评论
显示 1更早的评论
KSSV
KSSV 2018-6-21
Let A be your data.
N = round(3600/length(A)) ;
iwant = zeros(N,length(A)-1) ;
for i = 1:length(A)-1
iwant(:,i) = linspace(A(i),A(i+1),N) ;
end
iwant = iwant(:) ;

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