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

조회 수: 107 (최근 30일)
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

채택된 답변

John
John 2014년 9월 20일
편집: John 2014년 9월 20일
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 :-)
  댓글 수: 2
tyler brecht
tyler brecht 2014년 9월 20일
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
John 2014년 9월 20일
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.

댓글을 달려면 로그인하십시오.

추가 답변 (4개)

Guillaume
Guillaume 2014년 9월 20일
n = 1994;
num2str(n) - '0'
  댓글 수: 6
Jan
Jan 2021년 6월 16일
@andrea: num2str converts a number to the corresponding CHAR vector:
n = 1994;
num2str(n) % '1994'
If you subtract another CHAR from this vector, this is done elementwise in Matlab:
'1994' - '0' % Is the same as:
['1', '9', '9', '4'] - '0'
% This is calculated elementwise:
['1' - '0', '9' - '0', '9' - '0', '4' - '0']
If CHARs appear as input to a numerical operation, their ASCII values are used. So this is converted to:
[49 - 48, 57 - 48, 57 - 48, 52 - 48]
which is:
[1, 9, 9, 4]

댓글을 달려면 로그인하십시오.


Jan
Jan 2019년 7월 11일
편집: Jan 2019년 7월 11일
N = 1994;
m = floor(log10(N));
D = mod(floor(N ./ 10 .^ (m:-1:0)), 10);
>> D = [1, 9, 9, 4]
  댓글 수: 2
John D'Errico
John D'Errico 2020년 10월 9일
@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.

댓글을 달려면 로그인하십시오.


per isakson
per isakson 2014년 9월 20일
편집: per isakson 2014년 9월 20일
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
  댓글 수: 10
per isakson
per isakson 2014년 9월 20일
편집: per isakson 2016년 7월 17일
Fewer lines of code is good, but one should only use code that one understands and is able to take responsibility for.

댓글을 달려면 로그인하십시오.


Joachim Posselt
Joachim Posselt 2017년 4월 12일
편집: Joachim Posselt 2017년 4월 12일
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

카테고리

Help CenterFile Exchange에서 Numeric Types에 대해 자세히 알아보기

Community Treasure Hunt

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

Start Hunting!

Translated by