Operator '>' is not supported for operands of type 'cell'.

1 次查看（过去 30 天）
Sanley Guerrier
Sanley Guerrier 2023-12-8
评论： Sanley Guerrier 2023-12-10
Can someone help with the logical operator " extract rows +/- 1 where columns T And V are not zero"? Thank you!
clear
close all
clc
T1= readtable("Performance history + distress ( d-3) reduced.xls",VariableNamingRule="preserve");
T2 = readtable("D-3 Road sections (HPMA) Tran red v2.xlsx",VariableNamingRule="preserve");
% Join tables
joinedData = innerjoin(T2,T1,"LeftKeys",["FROM_REFP","TO_REFP"],"RightKeys",...
["begin_refp","end_refp"]);
T = joinedData;
T.Properties.VariableNames = num2cell(char(64+(1:size(T,2))));
%Set the empty cell "iri" ==0
if isempty(T{:,22})
T(:,22) =0;
end
% set H = G where H == 0
idx = T.H == 0;
T.H(idx) = T.G(idx);
% extract rows +/- 1 where columns T And V are not zero
idx = T.T>0 | T.V >0;
idx = any([idx [false; idx(1:end-1)] [idx(2:end); false]],2);
TT= T(idx,:); %extracted_rows
writetable(TT, "Merged&CleanData.csv")
  6 个评论
显示 4更早的评论
Sanley Guerrier
Sanley Guerrier 2023-12-8
I want to extract non empty rows in Column R that correspond to nonzero row in column T and add +/- 1 rows.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Walter Roberson
Walter Roberson 2023-12-8
移动：Walter Roberson 2023-12-8
Ran in:
filename = 'T.xlsx';
opt = detectImportOptions(filename);
opt = setvaropts(opt, [2:6,13:15,18:19], 'Prefixes', "'", 'Suffixes', "'");
T = readtable(filename, opt);
T.Properties.VariableNames = num2cell(char(64+(1:size(T,2))));
mask = T.T ~= 0 & ~strcmp(T.R, "");
mask = conv(mask, [1 1 1], 'same') > 0;
selected_rows = T(mask,:);
selected_rows
selected_rows = 0×23 empty table
  5 个评论
显示 3更早的评论
Sanley Guerrier
Sanley Guerrier 2023-12-8
移动：Voss 2023-12-9
It is working perfectly fine.
Appreciate you, Voss.

请先登录，再进行评论。

更多回答（2 个）

VBBV
VBBV 2023-12-8
T is a table array/datatype, and you try to access variable inside the table with same name , this is not valid operation
idx = T.T>0 | T.V >0; %T.T is not valid operation
  7 个评论
显示 5更早的评论
Walter Roberson
Walter Roberson 2023-12-8
Ran in:
@VBBV
Nearly the only variable name that you cannot use in a table is Properties .
T = table((1:5).', 'VariableNames', {'T'})
T = 5×1 table
T _ 1 2 3 4 5
T.T
ans = 5×1
1 2 3 4 5
Stephen23
Stephen23 2023-12-8
Ran in:
"T.T is not valid operation"
Why not?
T = array2table(pi, "VariableNames","T")
T = table
T ______ 3.1416
T.T % why do you think that this is "not valid operation" ?
ans = 3.1416

请先登录，再进行评论。

Voss
Voss 2023-12-8
Ran in:
T = readtable('T.xlsx');
T.Properties.VariableNames = num2cell(char(64+(1:size(T,2))))
T = 16×23 table
A B C D E F G H I J K L M N O P Q R S T U V W ___ _______ _____________ _____________ ____________________ _________ ____ ____ ____ ___ ___ ___ _____________ _______ _________ ____ ____ ______ _________ _____ __ __ __ 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 1977 {''''} {''BAB''} 0 20 0 0 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 1981 {''''} {''BAB''} 0 20 0 0 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 1984 {''''} {''BAB''} 0 2 0 26 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 1985 {''''} {''BAB''} 0 4 0 30 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 1988 {''''} {''BAB''} 0 10 0 30 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 1989 {''''} {''BAB''} 0 2 0 40 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 1990 {''''} {''BAB''} 223 4 0 38 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 1991 {''''} {''BAB''} 165 10 0 30 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 1992 {''''} {''BAB''} 222 12 0 24 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 1994 {''''} {''BAB''} 73 46 0 2 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 1996 {''''} {''BAB''} 79 40 0 10 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 1998 {''''} {''BAB''} 65 44 0 10 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2000 {''''} {''BAB''} 78 12 32 6 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2001 {''''} {''BAB''} 99.5 0 0 0 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2002 {''''} {''BAB''} 119.5 18 26 6 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2003 {''''} {''BAB''} 142 0 0 0
"extract rows +/- 1 where columns T And V are not zero"
idx = T.T ~= 0 & T.V ~= 0;
idx = any([idx [false; idx(1:end-1)] [idx(2:end); false]],2);
result = T(idx,:)
result = 5×23 table
A B C D E F G H I J K L M N O P Q R S T U V W ___ _______ _____________ _____________ ____________________ _________ ____ ____ ____ ___ ___ ___ _____________ _______ _________ ____ ____ ______ _________ _____ __ __ __ 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 1998 {''''} {''BAB''} 65 44 0 10 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2000 {''''} {''BAB''} 78 12 32 6 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2001 {''''} {''BAB''} 99.5 0 0 0 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2002 {''''} {''BAB''} 119.5 18 26 6 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2003 {''''} {''BAB''} 142 0 0 0
  3 个评论
显示 1更早的评论
Voss
Voss 2023-12-8
编辑：Voss 2023-12-8
Ran in:
T = readtable('T.xlsx');
T.Properties.VariableNames = num2cell(char(64+(1:size(T,2))));
"extract non empty rows in Column R that correspond to nonzero row in column T and add +/- 1 rows"
idx = T.T ~= 0 & ~strcmp(T.R,char([39 39]));
% idx = T.T ~= 0 & ~strcmp(T.R,''''''); % alternate, equivalent to the above
idx = any([idx [false; idx(1:end-1)] [idx(2:end); false]],2);
result = T(idx,:)
result = 12×23 table
A B C D E F G H I J K L M N O P Q R S T U V W ___ _______ _____________ _____________ ____________________ _________ ____ ____ ____ ___ ____ ___ _____________ _______ _________ ____ ____ _______________ _________ _____ __ __ __ 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2004 {'''' } {''BAB''} 138.5 6 24 26 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2005 {''BAB Rural''} {''BAB''} 55 0 0 0 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2006 {'''' } {''BAB''} 47 0 0 0 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2009 {'''' } {''BAB''} 54.5 0 0 0 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2010 {''Chip Seal''} {''BAB''} 60 26 0 0 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2011 {'''' } {''BAB''} 75.5 0 0 0 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2012 {'''' } {''BAB''} 69.5 38 0 0 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2013 {''Patching'' } {''BAB''} 72 0 0 0 107 {''U''} {''0+0.000''} {''0+0.151''} {''2010 RECON BIT''} {''BAB''} 2005 2013 4500 2.9 166 107 {''MN107-U''} {''U''} {''BAB''} 4500 2014 {'''' } {''BAB''} 88.5 38 4 0 238 {''U''} {''0+0.000''} {''0+0.302''} {''2014 BAB'' } {''BAB''} 2014 2014 1400 3.4 82.5 238 {''MN238-U''} {''U''} {''BAB''} 1400 2013 {'''' } {''BAB''} 293.5 10 20 12 238 {''U''} {''0+0.000''} {''0+0.302''} {''2014 BAB'' } {''BAB''} 2014 2014 1400 3.4 82.5 238 {''MN238-U''} {''U''} {''BAB''} 1400 2014 {''BAB Urban''} {''BAB''} 70.5 0 0 0 238 {''U''} {''0+0.000''} {''0+0.302''} {''2014 BAB'' } {''BAB''} 2014 2014 1400 3.4 82.5 238 {''MN238-U''} {''U''} {''BAB''} 1400 2015 {'''' } {''BAB''} 69.5 0 0 0
For this purpose, I considered the 1x2 character vector where each element is the single quote character (i.e., the character vector char([39 39])) to be the "empty" entries in column R you want to avoid. No doubt using setvaropts as in Walter's approach results in a table whose contents are easier to deal with.
Sanley Guerrier
Sanley Guerrier 2023-12-10
Thanks a lot, Voss. Both approaches work perfectly great. Your comment is very helpful.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Data Type Identification 的更多信息

标签

产品

Community Treasure Hunt

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

Start Hunting!

Translated by