Separate one column of in a multi-column, (which is itself a cell array) into two columns in the existing cell array

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Douglas Anderson
Douglas Anderson 2022-3-23
评论： Douglas Anderson 2022-3-24
I have an application that reads data (from someone else's code), in a cell array with 8 columns. Here is a snipped of the first 9 rows:
9×8 cell array
Columns 1 through 6
{'E100M000.txt'} {'6109'} {'23-Sep-2020 14:…'} {[0.04]} {1×2 cell} {[4]}
{'E100M001.txt'} {'6109'} {'23-Sep-2020 14:…'} {[0.05]} {1×2 cell} {[4]}
{'E100M002.txt'} {'4781'} {'23-Sep-2020 15:…'} {[5.08]} {1×2 cell} {[0]}
{'E100M003.txt'} {'4717'} {'23-Sep-2020 15:…'} {[3.36]} {1×2 cell} {[0]}
{'E100M004.txt'} {'4781'} {'23-Sep-2020 15:…'} {[5.04]} {1×2 cell} {[0]}
{'E100M005.txt'} {'6051'} {'23-Sep-2020 16:…'} {[5.08]} {1×2 cell} {[0]}
{'E100M006.txt'} {'4167'} {'23-Sep-2020 15:…'} {[ 3.6]} {1×2 cell} {[0]}
{'E100M007.txt'} {'6022'} {'23-Sep-2020 15:…'} {[0.26]} {1×2 cell} {[0]}
{'E100M008.txt'} {'6002'} {'23-Sep-2020 16:…'} {[2.92]} {1×2 cell} {[0]}
Columns 7 through 8
{[2971.6]} {'Pleasant Valley…'}
{[3523.4]} {'Pleasant Valley…'}
{[ 0]} {'Peckham Propert…'}
{[ 0]} {'251 Creek Road' }
{[ 0]} {'Peckham Propert…'}
{[ 0]} {'23 Plateau Road' }
{[ 0]} {'Peckham Propert…'}
{[ 0]} {'Peckham Propert…'}
Column 5 is itself a cell array, and I would like to replace this with two columns in a new cell array of everything. Column 5 has a letter in the first column of the array, and a number in the second. Simple to split this? reshape? split? ???
Wish I could get it in the right format in the first place, no such luck.
Thanks!
Doug Anderson

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Voss
Voss 2022-3-24
Ran in:
C = { ...
'E100M000.txt' '6109' '23-Sep-2020 14:…' 0.04 {'A' 1} 4 2971.6 'Pleasant Valley…'; ...
'E100M001.txt' '6109' '23-Sep-2020 14:…' 0.05 {'B' 2} 4 3523.4 'Pleasant Valley…'; ...
'E100M002.txt' '4781' '23-Sep-2020 15:…' 5.08 {'C' 3} 0 0 'Peckham Propert…'}
C = 3×8 cell array
{'E100M000.txt'} {'6109'} {'23-Sep-2020 14:…'} {[0.0400]} {1×2 cell} {[4]} {[2.9716e+03]} {'Pleasant Valley…'} {'E100M001.txt'} {'6109'} {'23-Sep-2020 14:…'} {[0.0500]} {1×2 cell} {[4]} {[3.5234e+03]} {'Pleasant Valley…'} {'E100M002.txt'} {'4781'} {'23-Sep-2020 15:…'} {[5.0800]} {1×2 cell} {[0]} {[ 0]} {'Peckham Propert…'}
C_new = [C(:,1:4) vertcat(C{:,5}) C(:,6:end)]
C_new = 3×9 cell array
{'E100M000.txt'} {'6109'} {'23-Sep-2020 14:…'} {[0.0400]} {'A'} {[1]} {[4]} {[2.9716e+03]} {'Pleasant Valley…'} {'E100M001.txt'} {'6109'} {'23-Sep-2020 14:…'} {[0.0500]} {'B'} {[2]} {[4]} {[3.5234e+03]} {'Pleasant Valley…'} {'E100M002.txt'} {'4781'} {'23-Sep-2020 15:…'} {[5.0800]} {'C'} {[3]} {[0]} {[ 0]} {'Peckham Propert…'}
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Voss
Voss 2022-3-24
编辑：Voss 2022-3-24
Ran in:
Actually, this is not correct:
col5=A{:,5}
That assigns only A{1,5} to col5 (because col5 is only one variable). Observe:
A = { ...
'E100M000.txt' '6109' '23-Sep-2020 14:…' 0.04 {'A' 1} 4 2971.6 'Pleasant Valley…'; ...
'E100M001.txt' '6109' '23-Sep-2020 14:…' 0.05 {'B' 2} 4 3523.4 'Pleasant Valley…'; ...
'E100M002.txt' '4781' '23-Sep-2020 15:…' 5.08 {'C' 3} 0 0 'Peckham Propert…'};
col5 = A{:,5}
col5 = 1×2 cell array
{'A'} {[1]}
% try assigning to 2 output variables:
[col51,col52] = A{:,5}
col51 = 1×2 cell array
{'A'} {[1]}
col52 = 1×2 cell array
{'B'} {[2]}
So you would need to call vertcat() at that point as well:
col5 = vertcat(A{:,5})
col5 = 3×2 cell array
{'A'} {[1]} {'B'} {[2]} {'C'} {[3]}
Then, the explicit call to horzcat():
out = horzcat(A(:,1),A(:,2),A(:,3),A(:,4),col5,A(:,6),A(:,7),A(:,8))
out = 3×9 cell array
{'E100M000.txt'} {'6109'} {'23-Sep-2020 14:…'} {[0.0400]} {'A'} {[1]} {[4]} {[2.9716e+03]} {'Pleasant Valley…'} {'E100M001.txt'} {'6109'} {'23-Sep-2020 14:…'} {[0.0500]} {'B'} {[2]} {[4]} {[3.5234e+03]} {'Pleasant Valley…'} {'E100M002.txt'} {'4781'} {'23-Sep-2020 15:…'} {[5.0800]} {'C'} {[3]} {[0]} {[ 0]} {'Peckham Propert…'}
is the same as using [ ] to do the horizontal concatenation:
out = [A(:,1),A(:,2),A(:,3),A(:,4),col5,A(:,6),A(:,7),A(:,8)] % same
out = 3×9 cell array
{'E100M000.txt'} {'6109'} {'23-Sep-2020 14:…'} {[0.0400]} {'A'} {[1]} {[4]} {[2.9716e+03]} {'Pleasant Valley…'} {'E100M001.txt'} {'6109'} {'23-Sep-2020 14:…'} {[0.0500]} {'B'} {[2]} {[4]} {[3.5234e+03]} {'Pleasant Valley…'} {'E100M002.txt'} {'4781'} {'23-Sep-2020 15:…'} {[5.0800]} {'C'} {[3]} {[0]} {[ 0]} {'Peckham Propert…'}
And you don't need to take each individual column separately; you can take the first 4 together and the last 3 together:
out = [A(:,1:4),col5,A(:,6:end)] % same
out = 3×9 cell array
{'E100M000.txt'} {'6109'} {'23-Sep-2020 14:…'} {[0.0400]} {'A'} {[1]} {[4]} {[2.9716e+03]} {'Pleasant Valley…'} {'E100M001.txt'} {'6109'} {'23-Sep-2020 14:…'} {[0.0500]} {'B'} {[2]} {[4]} {[3.5234e+03]} {'Pleasant Valley…'} {'E100M002.txt'} {'4781'} {'23-Sep-2020 15:…'} {[5.0800]} {'C'} {[3]} {[0]} {[ 0]} {'Peckham Propert…'}
So the idea is essentially the same, yes; I just did it without the temporary variable col5. (And I don't think anything is transposed or needs to be transposed - you do vertcat on the elements of column 5, then horzcat (maybe using [ ]) that result with all the other columns (in the right order of course).)

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