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 2022-3-23

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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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Arif Hoq 2022-3-23

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as you did not attach your data, you can try this

% A is your cell array
col5=A{:,5}
out=horzcat(A(:,1),A(:,2),A(:,3),A(:,4),col5,A(:,6),A(:,7),A(:,8),A(:,9))

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Answer 1

Voss 2022-3-24

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在 MATLAB Online 中打开

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 2022-3-24

编辑：Voss 2022-3-24

在 MATLAB Online 中打开

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).)

Douglas Anderson 2022-3-24