How to move array data to additional columns based on repeats in a vector or column
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Stephen Devlin
am 25 Jul. 2017
Kommentiert: Stephen Devlin
am 25 Jul. 2017
I have an array like A (below) with duplicates in the second column, for example there are three 1's,three 6's and four 7's. What I would like to do is either have a separate array for each block (to include all the multiples) or preferably be able to move the data for the multiples in the 3rd and 4th columns to the same row as the first element but in extra columns I could preallocate for? i have some code to move the duplicates out onto the same row and replace the original data with nans but cannot get them all on one row.
A=[ 1 1 1.0277 0.56932
2 1 1.0426 0.4202
3 1 1.048 0.42017
4 2 1.1466 0.48309
5 6 1.4731 0.4024
6 6 1.5244 0.62436
7 6 1.5295 0.40456
8 7 1.55 0.58695
9 7 1.5658 0.41037
10 7 1.6107 0.4693
11 7 1.6248 0.62088
12 8 1.7182 0.48]
for k=2:size(A)
if A(k,2)==A(k-1,2)
A(k-1,5)=A(k,3)
A(k-1,6)=A(k,4)
A(k,3)=nan
A(k,4)=nan
end
end
Best regards
Steve
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Akzeptierte Antwort
Stephen23
am 25 Jul. 2017
Bearbeitet: Stephen23
am 25 Jul. 2017
Creating new columns would not be trivially simple without expanding the array continuously. A simpler solution is to put the groups into their own cells of one cell array: one of the most versatile is to use accumarray:
>>[U,~,X] = unique(A(:,2));
>> fun = @(r){A(r,1:4)};
>> Z = accumarray(X,A(:,1),[],fun);
>> Z{:}
ans =
1.00000 1.00000 1.02770 0.56932
2.00000 1.00000 1.04260 0.42020
3.00000 1.00000 1.04800 0.42017
ans =
4.00000 2.00000 1.14660 0.48309
ans =
5.00000 6.00000 1.47310 0.40240
6.00000 6.00000 1.52440 0.62436
7.00000 6.00000 1.52950 0.40456
ans =
8.00000 7.00000 1.55000 0.58695
9.00000 7.00000 1.56580 0.41037
10.00000 7.00000 1.61070 0.46930
11.00000 7.00000 1.62480 0.62088
ans =
12.00000 8.00000 1.71820 0.48000
You could easily reshape the data within each group, or select only the columns that you want: simply change the function to suit your needs, e.g.:
fun = @(r){A(r,3:4)};
or
fun = @(r){reshape(A(r,3:4).',1,[])};
to produce:
>> Z {:}
ans =
1.02770 0.56932 1.04260 0.42020 1.04800 0.42017
ans =
1.14660 0.48309
ans =
1.47310 0.40240 1.52440 0.62436 1.52950 0.40456
ans =
1.55000 0.58695 1.56580 0.41037 1.61070 0.46930 1.62480 0.62088
ans =
1.71820 0.48000
You might also find X useful, e.g.:
>> U(X)
ans =
1
1
1
2
6
6
6
7
7
7
7
8
3 Kommentare
Stephen23
am 25 Jul. 2017
Bearbeitet: Stephen23
am 25 Jul. 2017
[U,~,X] = unique(A(:,2));
I use unique to get the vector X: this is basically a sequential enumeration of the input values, or you could think of it as a mapping from whatever the input values are to 1,2,3,.... The input values here are A(:,2), as you specified that this defines the groups. Thus each group gets a number!
fun = @(r){A(r,1:4)};
fun is an anonymous function that collects columns 1:4 of rows r of array A and puts that sub-array into a scalar cell array.
accumarray(X,A(:,1),[],fun);
uses the values of A(:,1) as inputs to fun (because the first column of A seems to contain the rows of A, and fun requires a row index input). fun then returns those rows of A and accumarray puts them into the output cell array: their positions in the output are determined by X (the group enumeration 1,2,3,...): thus each row gets placed into the appropriate cell array for that group.
I know that accumarray is a bit of a mindflip at first, but it is well worth the time and effort learning how it works. Practice.
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