# Hi guys. I need help splitting a number into its individual parts and then add them. E.g. the number would be 1994 = 1 + 9 + 9 + 4 = 23

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tyler brecht on 20 Sep 2014
Commented: John D'Errico on 9 Oct 2020
I don't want the number to turn a scalar into an array eg x = 1994 = 1 9 9 4 I want it the scalar x = 1994 to split into multiple scalars. I hope this makes sense and your help will be much appreciated

John on 20 Sep 2014
Edited: John on 20 Sep 2014
When you say you do not want the number to turn to a scalar, I think what you are saying is you don't want to convert the scalar number x = 1994 into a vector v = [1 9 9 4].
If you are allowed to use string functions and are loose on your restriction, consider this:
x = 1994
%Convert x = 1994 to a vector of characters
a = num2str(x); %a now holds ['1' '9' '9' '4']
%Go through each of the elements in the vector, a, convert them to numbers and add them up
sum = 0;
for i = 1:size(a, 2)
sum = sum + str2num(a(i));
end
%sum now contains the sum of the individual digits
But this seems like a typical undergraduate homework problem :-) for a comp. sci./electrical enginering course :-) and the string approach above is unlikely to be the proper solution. I hope no one else provides the mathematical solution to this :-) as this is not a MATLAB question.
As a hint, consider the significance of the position of the individual digits in the number. We call them the one's, ten's, hundred's, and thousand's position for a reason :-)

tyler brecht on 20 Sep 2014
Wow u are wise :), thanks a lot John this is exactly what I was looking for:). I'm actually doing a mechanical engineering course, but this solution will help me a lot in completing my assignment, which contains a if function and while loop
John on 20 Sep 2014
A glimpse from the mathematical vantage point:
x = 1994;
a = 1994 / 1000;
a = floor(a); %a = 1, ..cough.. the thousand's position
b = x - a * 1000; % b = 994
b = b / 100;
b = floor(b); % b = 9, ..cough.. the hundred's position
None of the variables above are vectors.

Guillaume on 20 Sep 2014
n = 1994;
num2str(n) - '0'

tyler brecht on 22 Sep 2014
Thank you :)
CHAIYUT CHAROENPHON on 16 Jul 2016
work perfectly.
HARSHA HN on 11 Jul 2019

per isakson on 20 Sep 2014
Edited: per isakson on 20 Sep 2014
A one-liner with a lot of Matlab
>> sum( arrayfun( @(a) str2double(a), num2str( 1994 ) ) )
ans =
23
or even more matlab-ish
>> sum( arrayfun( @str2double, num2str( 1994 ) ) )
ans =
23
This is essential the same as John's solution with the for-loop replaced by the function, arrayfun

tyler brecht on 20 Sep 2014
Yes I see your solution is more efficient and more advanced, I would use it but I need to get to understand the arrayfun function first before using it
Star Strider on 20 Sep 2014
per isakson on 20 Sep 2014
Fewer lines of code is good, but one should only use code that one understands and is able to take responsibility for.

Jan on 11 Jul 2019
Edited: Jan on 11 Jul 2019
N = 1994;
m = floor(log10(N));
D = mod(floor(N ./ 10 .^ (m:-1:0)), 10);
>> D = [1, 9, 9, 4]

Tejas Mahajan on 9 Oct 2020
Hi Jan,
John D'Errico on 9 Oct 2020
@Tejas Mahajan - easy enough. But you should make the effort yourself. What, for example does this do?
10 .^ (m:-1:0)
Now, what would happen when you do this?
N ./ 10 .^ (m:-1:0)
Now add one more layer around that?
floor(N ./ 10 .^ (m:-1:0))
Try that part for 1994. Now, look at the last step in his code.
mod(floor(N ./ 10 .^ (m:-1:0)), 10);
What did that do?
When you don't understand a piece of code, break it apart. Start in the middle, then work outwards, one part at a time until you do see what it does. You won't learn otherwise.

Joachim Posselt on 12 Apr 2017
Edited: Joachim Posselt on 12 Apr 2017
yyyy = 1994;
% dig = digits extract // Ziffern extrahieren - mit Mathe, nicht mit Strings!!
dig_t = floor (yyyy / 1000); % tausender // thousands
dig_h = yyyy - dig_t * 1000;
dig_h = floor (dig_h / 100); % hunderter // hundreds
dig_z = yyyy - dig_t * 1000 - dig_h*100;
dig_z = floor (dig_z / 10); % zehner // ten
dig_e = yyyy - dig_t * 1000 - dig_h*100 -dig_z*10;
dig_e = floor (dig_e / 1); % einer // ones