Replace NaN's in table with zero

Hello, I have a 1501x7 table called 'x' and there appears to be NaN's in the fourth and sixth column called "Age" and "height". I would like a way to replace NaN's with zeros. Take note, that I have already tried:
k = find(isnan(x))';
x(k) = 0;
% and
x(isnan(x)) = 0;
Yet, neither work because I am using a table, not a matrix. I have also tried converting my table into a cell array, and using these same functions, but they still do not work. They return:"Undefined function 'isnan' for input arguments of type 'cell'" ALSO, please note that the table has columns full of text. So, cell2mat does not work.

3 comentarios

AFiorillo
AFiorillo el 24 de Jul. de 2015
When indexing into a table or cell, parentheses and curly braces have very different meanings. Using parenthesis result in a slice of that table/cell, so the result is also a table or a cell. Using curly braces yields the contents of those slices. See here for a more comprehensive explanation on accessing data in tables.
For your code, try
k = find(isnan(x))';
x{k} = 0;
% and
x{isnan(x)} = 0;
xander fong
xander fong el 27 de Jul. de 2015
I get this: "Undefined function 'isnan' for input arguments of type 'table'.
hager fouda
hager fouda el 1 de Sept. de 2023
Thank you so much. it works.

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 Respuesta aceptada

Peter Perkins
Peter Perkins el 26 de Jul. de 2015
There's a function called standardizeMissing that would replace a non-NaN value with NaN, but normally, replacing NaN with a constant value (as opposed to, for example, some sort estimated value) would be kind of a funny thing to do. I'll assume you have a good reason.
Either of the following should work:
>> t = table({'smith';'jones';'doe'},[20;NaN;40],[NaN;72;66],[120;130;140],'VariableNames',{'Name' 'Age' 'Height' 'Weight'})
t =
Name Age Height Weight
_______ ___ ______ ______
'smith' 20 NaN 120
'jones' NaN 72 130
'doe' 40 66 140
>> vars = {'Age' 'Height'};
>> t2 = t{:,vars};
>> t2(isnan(t2)) = 0;
>> t{:,vars} = t2
t =
Name Age Height Weight
_______ ___ ______ ______
'smith' 20 0 120
'jones' 0 72 130
'doe' 40 66 140
>> t = table({'smith';'jones';'doe'},[20;NaN;40],[NaN;72;66],[120;130;140],'VariableNames',{'Name' 'Age' 'Height' 'Weight'});
>> [~,vars] = ismember({'Age' 'Height'},t.Properties.VariableNames)
vars =
2 3
>> for i=vars, t.(i)(isnan(t.(i))) = 0; end
Hope this helps.

1 comentario

Chris Hooper
Chris Hooper el 23 de Ag. de 2024
NaN does not work as desired in unique function:
An option in the unique fuction to treat NaN as a distinct would negate one reason for replacing NaN.

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Más respuestas (9)

Akira Agata
Akira Agata el 17 de Feb. de 2017
If you have R2016a or later version, you can use ismissing function and make it much easier.
For example:
% Make a sample table 'T' and replace 'NaN' with 0
T = table({'smith';'jones';'doe'},[20;NaN;40],[NaN;72;66],[120;130;140],'VariableNames',{'Name' 'Age' 'Height' 'Weight'});
idx = ismissing(T(:,{'Age','Height'}));
T{:,{'Age','Height'}}(idx) = 0;
Yuting Mou
Yuting Mou el 29 de Jul. de 2016
I also run across the problem, but there seems to be an easier way:
x.age(isnan(x.age)) = 0;
This is OK in my case

1 comentario

Dooyoung Kim
Dooyoung Kim el 25 de Jun. de 2018
This works for me too! Thanks for the suggestion.

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Steven Lord
Steven Lord el 23 de Jul. de 2018

7 votos

I would use the fillmissing function introduced in release R2016b. See the "Table with Multiple Data Types" example on that documentation page for a demonstration of how to replace NaN values with 0.
J.M. Verduijn
J.M. Verduijn el 8 de Feb. de 2019
for i= 1: width(T)
T.(i)(isnan(T.(i))) = 0;
end
Works for me, replaces all NaN values in table T with 0
Zachary Smith
Zachary Smith el 19 de Mzo. de 2020

1 voto

If you are using readtable() to load the table from a file, then you can add the name-value pair argument 'EmptyValue',0 to do this automatically.
Aron Magesh
Aron Magesh el 7 de Mzo. de 2021
Editada: Aron Magesh el 7 de Mzo. de 2021

1 voto

Just use fillmissing function if the data is in a table or timetable.
Andy
Andy el 23 de Jul. de 2018

0 votos

so, in other words:
mainTTable{:,:}(ismissing(mainTTable)) = 0;
carolina franco
carolina franco el 28 de En. de 2020
Hi,
Another simple way to understand what's going on .
For me, it works well in R2014a. You only need to enter the matrix with NaN values without specifying the columns where NaN values are.
%Input
m_data=C{1,1}; % Matrix with NaN values
%Code
s1=size(m_data,1);
for i= 1: s1
msubs=m_data(i,1:end); % Save existing data in ith row of m_data
msubs=msubs(isnan(m_data(i,1:end))==0); %Substitute matrix/ taking only non-NaN values
m_data(i,1:end)=0; %Erase all existing values in ith row of m_data
m_data(i,1:size(msubs,2))=msubs; %Substitute values without NaN
end

2 comentarios

Stephen23
Stephen23 el 28 de En. de 2020
Note that all the original question explicitly states that "..I am using a table, not a matrix", and all of the other answers work with tables, not numeric matrices. Tables are a container array type:
carolina franco
carolina franco el 28 de En. de 2020
Right! Thanks

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Gabor
Gabor el 11 de Mzo. de 2021
T{:,2:4}(ismissing(T{:,2:4})) = 0;
2:4 are the columns which are containing NaN values.

1 comentario

Interesting, that does work.
T = table({'smith';'jones';'doe'},[20;NaN;40],[NaN;72;66],[120;130;140],'VariableNames',{'Name' 'Age' 'Height' 'Weight'})
T = 3x4 table
Name Age Height Weight _________ ___ ______ ______ {'smith'} 20 NaN 120 {'jones'} NaN 72 130 {'doe' } 40 66 140
T{:,2:4}(ismissing(T{:,2:4})) = 0
T = 3x4 table
Name Age Height Weight _________ ___ ______ ______ {'smith'} 20 0 120 {'jones'} 0 72 130 {'doe' } 40 66 140

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