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

5 次查看（过去 30 天）
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

请先登录，再回答此问题。

采纳的回答

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…'}
  5 个评论
显示 3更早的评论
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).)

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Matrix Indexing 的更多信息

标签

产品


版本

R2021a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by