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 commentaires

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 le 27 Juil 2015
I get this: "Undefined function 'isnan' for input arguments of type 'table'.
hager fouda
hager fouda le 1 Sep 2023
Thank you so much. it works.

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 Réponse acceptée

Peter Perkins
Peter Perkins le 26 Juil 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 commentaire

Chris Hooper
Chris Hooper le 23 Août 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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Plus de réponses (9)

Akira Agata
Akira Agata le 17 Fév 2017

10 votes

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 le 29 Juil 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 commentaire

Dooyoung Kim
Dooyoung Kim le 25 Juin 2018
This works for me too! Thanks for the suggestion.

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Steven Lord
Steven Lord le 23 Juil 2018

7 votes

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 le 8 Fév 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 le 19 Mar 2020

1 vote

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.
Andy
Andy le 23 Juil 2018

0 votes

so, in other words:
mainTTable{:,:}(ismissing(mainTTable)) = 0;
carolina franco
carolina franco le 28 Jan 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 commentaires

Stephen23
Stephen23 le 28 Jan 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 le 28 Jan 2020
Right! Thanks

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

1 commentaire

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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