multiple digit number in to individual digits

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Raza
Raza 2014-7-22
编辑： John D'Errico 2023-2-27
i want to change the number 1123 in 1 1 2 3, want to split combine number into into individual numbers

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Azzi Abdelmalek
Azzi Abdelmalek 2014-7-22
a=1234
b=str2double(regexp(num2str(a),'\d','match'))
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Walter Roberson
Walter Roberson 2017-2-7
a = '123*+'
for K = 1 : length(a)
fprintf('character #%d of "%s" is "%c"\n', K, a, a(K));
end
Adam Danz
Adam Danz 2020-4-29
For large values such as a=11122333345555566 this will not work since num2str will convert the value to '1.112233334555557e+16'. Otherwise nice solution.

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更多回答（3 个）

John D'Errico
John D'Errico 2014-7-22
N = 1123;
Ndigits = dec2base(N,10) - '0'
Ndigits =
1 1 2 3
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John D'Errico
John D'Errico 2023-2-27
编辑：John D'Errico 2023-2-27
Ran in:
@Michael Ramage
Not difficult with a floating point number, but remember that a float is NOT an exact decimal representation of that number. But...
X = 1.2345;
dec2base(X*10000,10)
ans = '12345'
or
dec2base(X*10000,10) - '0'
ans = 1×5
1 2 3 4 5
You can even fuss around and get the decimal point in there if you want, but if you want that, then sprintf is arguably a better choice.

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Jan
Jan 2017-2-7
编辑：Jan 2017-2-14
For getting the digits, a conversion to a string is an indirection. Staying at numerical values is usually faster:
N = 1123;
m = floor(log10(N)); % [EDITED] Thanks Stephen
D = mod(floor(N ./ 10 .^ (m:-1:0)), 10);
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Jay Doshi
Jay Doshi 2022-3-25
If the number is 0006, and i want all four numbers. What can I do? Because this method just gives me d = 6.
Stephen23
Stephen23 2022-3-25
Ran in:
@Jay Doshi:
N = 6;
m = 3; % order
d = mod(floor(N ./ 10 .^ (m:-1:0)), 10)
d = 1×4
0 0 0 6

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Ramon Villamangca
Ramon Villamangca 2018-11-20
编辑：Ramon Villamangca 2018-11-20
a simple single line solution:
>> num = 12345042117;
>> arrayfun(@(x) mod(floor(num/10^x),10),floor(log10(num)):-1:0)
ans =
1 2 3 4 5 0 4 2 1 1 7
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Ramon Villamangca
Ramon Villamangca 2023-2-27
Ran in:
@Jyahway Dong: If the digits are that long, you'll probably input it as a char string, anyway. That means the solution would even be much simpler:
num = '62229893423380308135336276614282806444486645238749';
num - '0'
ans = 1×50
6 2 2 2 9 8 9 3 4 2 3 3 8 0 3 0 8 1 3 5 3 3 6 2 7 6 6 1 4 2

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