multiple digit number in to individual digits
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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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采纳的回答
Azzi Abdelmalek
2014-7-22
a=1234
b=str2double(regexp(num2str(a),'\d','match'))
3 个评论
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
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.
更多回答(3 个)
John D'Errico
2014-7-22
N = 1123;
Ndigits = dec2base(N,10) - '0'
Ndigits =
1 1 2 3
3 个评论
John D'Errico
2023-2-27
编辑:John D'Errico
2023-2-27
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)
or
dec2base(X*10000,10) - '0'
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.
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);
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
3 个评论
Ramon Villamangca
2023-2-27
@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'
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