replacing NaN with dot in a table

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Danielle Leblance
Danielle Leblance el 16 de En. de 2017
Respondida: Peter Perkins el 19 de En. de 2017
I have table A where many numeric variables contain NAN values. How can i replace the NaN values with a dot?

Respuestas (3)

Guillaume
Guillaume el 16 de En. de 2017
Why would you want to do that? While it is possible to do it, it's going to make using that table much harder. Matlab knows what NaN means and can be told to ignore them in calculations, it knows nothing about the meaning of '.'. Furthermore, a NaN is numeric meaning that your columns can all be basic vectors, introducing dot characters means that all columns need to change to cell arrays, complicating access for all values.
If you really want to do it, assuming all the table columns are numeric:
Aascell = table2cell(A);
Aascell{cellfun(@isnan, Aascell)} = '.';
Awithdots = cell2table(Aascell, 'VariableNames', A.Properties.VariableNames)
  5 comentarios
Walter Roberson
Walter Roberson el 17 de En. de 2017
I did hit "Expected one output from a curly brace or dot indexing expression" when I tried your code using the sample data KSSV posted.
Guillaume
Guillaume el 17 de En. de 2017
Editada: Guillaume el 17 de En. de 2017
Using R2016b with that sample data and my code, I get "Error using cellfun, Non-scalar in Uniform output, at index 1, output 1. Set 'UniformOutput' to false." I did say "assuming all columns are numeric" (and I should have added scalar)
However, I do get the expected one output error if I change the cellfun to cellfun(@(x) all(isnan, x), Aascell) to fix the above error.
Anyway, this would fix all:
Aascell = table2cell(A);
Aascell(cellfun(@(x) all(isnan(x)), Aascell)) = {'..'}; %can't be a single character due to a bug in cell2table
Awithdots = cell2table(Aascell, 'VariableNames', A.Properties.VariableNames)

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Peter Perkins
Peter Perkins el 19 de En. de 2017
As Guillaume already said, this is almost certainly a bad idea, and you should really ask yourself why you want to do it. You cannot store a dot in a numeric variable, you will have to convert each of those variables to a cell array, at which point they become useless for computation. NaN is the way to indicate missing data, I really, strongly suggest that you use NaN.
But if you really want to do this ...
1) Convert every variable in t to a cell:
>> t = array2table(randn(5));
>> t{2,1} = NaN; t{4,3} = NaN
t =
Var1 Var2 Var3 Var4 Var5
_______ ________ ________ ________ _________
-1.4557 1.3628 0.8534 -1.3116 -0.088259
NaN 0.82098 -2.0728 0.65508 1.3722
-1.1904 -0.53919 -1.2658 -0.52271 -1.8305
0.12795 2.0646 NaN -0.16537 -0.96146
-1.1664 1.1394 -0.32769 -0.54855 -0.47091
>> c = table2cell(t);
>> c(ismissing(t)) = {'.'}
c =
5×5 cell array
[-1.4557] [ 1.3628] [ 0.8534] [ -1.3116] [-0.088259]
'.' [ 0.82098] [ -2.0728] [ 0.65508] [ 1.3722]
[-1.1904] [-0.53919] [ -1.2658] [-0.52271] [ -1.8305]
[0.12795] [ 2.0646] '.' [-0.16537] [ -0.96146]
[-1.1664] [ 1.1394] [-0.32769] [-0.54855] [ -0.47091]
>> t = array2table(c,'VariableNames',t.Properties.VariableNames)
t =
Var1 Var2 Var3 Var4 Var5
_________ __________ __________ __________ ___________
[-1.4557] [ 1.3628] [ 0.8534] [ -1.3116] [-0.088259]
'.' [ 0.82098] [ -2.0728] [ 0.65508] [ 1.3722]
[-1.1904] [-0.53919] [ -1.2658] [-0.52271] [ -1.8305]
[0.12795] [ 2.0646] '.' [-0.16537] [ -0.96146]
[-1.1664] [ 1.1394] [-0.32769] [-0.54855] [ -0.47091]
2) Convert only the ones you need to:
>> t = array2table(randn(5));
>> t{2,1} = NaN; t{4,3} = NaN
t =
Var1 Var2 Var3 Var4 Var5
_______ ________ ________ _______ _________
1.3769 -0.22714 -0.72136 1.8183 0.47225
NaN 0.55456 1.2013 0.50796 0.0063762
-1.1656 0.91397 0.051885 0.42044 1.7116
-2.2186 -0.20018 NaN 0.33693 -0.23943
0.13924 0.14879 -0.45745 0.45433 -0.84344
>> t2 = varfun(@NaN2Dot,t);
>> t2.Properties.VariableNames = t.Properties.VariableNames
t2 =
Var1 Var2 Var3 Var4 Var5
_________ ________ __________ _______ _________
[ 1.3769] -0.22714 [-0.72136] 1.8183 0.47225
'.' 0.55456 [ 1.2013] 0.50796 0.0063762
[-1.1656] 0.91397 [0.051885] 0.42044 1.7116
[-2.2186] -0.20018 '.' 0.33693 -0.23943
[0.13924] 0.14879 [-0.45745] 0.45433 -0.84344
where NaN2Dot looks like this:
function x = NaN2Dot(x)
nans = isnan(x);
if any(nans)
x = num2cell(x);
x(nans) = {'.'};
end

KSSV
KSSV el 16 de En. de 2017
t = table({'smith';'jones';'doe'},[20;NaN;40],[NaN;72;66],[120;130;140],'VariableNames',{'Name' 'Age' 'Height' 'Weight'})
vars = {'Age' 'Height'};
t2 = t{:,vars};
t2(isnan(t2)) = 0;
t{:,vars} = t2
  2 comentarios
Danielle Leblance
Danielle Leblance el 16 de En. de 2017
no it is working terribly.there isn't only 1 problem to report but many.
Jan
Jan el 16 de En. de 2017
@Danielle: Then please explain the problems such that KSSV and others can fix them. We want to assist you to solve your problem. So please help us.

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