Counting the number of digits

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joseph Frank
joseph Frank 2011 年 7 月 3 日
コメント済み: Steven Lord 2023 年 6 月 14 日
Hi,
how can I compute the number of digits A=[12875] how can I get 5 as the number of digits in A?
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Walter Roberson
Walter Roberson 2017 年 9 月 22 日
Fails on negative numbers because they are already lower than 1
Steven Lord
Steven Lord 2021 年 1 月 29 日
How many digits does A have in the code below?
A = Inf;
How many digits does B have?
B = NaN;
How about C?
C = 3+4i;

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採用された回答

bym
bym 2011 年 7 月 3 日
One way:
numel(num2str(A))
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Antonino Visalli
Antonino Visalli 2022 年 8 月 9 日
numel(char(vpa(floor(pi*1e25))))
Walter Roberson
Walter Roberson 2022 年 8 月 9 日
Ran in:
S1 = num2str(floor(pi*1e25))
S1 = '3.141592653589793e+25'
numel(S1)
ans = 21
S2 = char(vpa(floor(pi*1e25)))
S2 = '31415926535897934939029504.0'
numel(S2)
ans = 28
S3 = char(vpa(floor(sym(pi)*1e25)))
S3 = '31415926535897932384626433.0'
numel(S3)
ans = 28
Actual answer should be 26

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その他の回答 (8 件)

Oleg Komarov
Oleg Komarov 2011 年 7 月 3 日
To count integer part
ceil(log10(abs(A)))
Edit
floor(log10(abs(A)+1)) + 1
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Walter Roberson
Walter Roberson 2011 年 7 月 15 日
Fails on 0, Jan.
Over integral values:
ceil(log10(max(1,abs(A)+1)))
Over real numbers,
ceil(log10(max(1,abs(A)*(1+eps))))
I think.
DGM
DGM 2023 年 1 月 28 日
編集済み: DGM 2023 年 1 月 28 日
Ran in:
It's also necessary to handle cases where A is integer-class.
For sake of clarifying the limitations and assumptions in each suggested method:
A = [-2456 -1.45 0 0.05 0.16 1.5 10 100 35465].';
y1 = ceil(log10(abs(A))); % fails on 0, subunity fractions, powers of 10
y2 = floor(log10(abs(A)+1)) + 1; % 0 has 1 digit? yes/no?
y3 = ceil(log10(max(1,abs(A)+1))); % 0 has 0 digits, but 0.05 and 0.16 both have 1 digit?
y4 = ceil(log10(max(1,abs(A)*(1+eps)))); % fails on powers of 10
y5 = numdigits(A); % where does this one fail?
table(A,y1,y2,y3,y4,y5)
ans = 9×6 table
A y1 y2 y3 y4 y5 _____ ____ __ __ __ __ -2456 4 4 4 4 4 -1.45 1 1 1 1 1 0 -Inf 1 0 0 0 0.05 -1 1 1 0 0 0.16 0 1 1 0 0 1.5 1 1 1 1 1 10 1 2 2 1 2 100 2 3 3 2 3 35465 5 5 5 5 5
function ndigits = numdigits(x)
% NDIGITS = NUMDIGITS(X)
% A simple convenience tool to count the number of digits in
% the integer part of a number. For complex inputs, the result
% is the number of digits in the integer part of its magnitude.
%
% X is a numeric scalar or array of any class or sign.
%
% Note that the integer 0 is considered to have zero digits.
% Consequently, for numbers on the interval -1<X<1, NDIGITS is 0.
ndigits = ceil(log10(abs(double(fix(x)))+1));
end

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Turner
Turner 2013 年 8 月 16 日
編集済み: Turner 2013 年 8 月 19 日
Will do the trick for all nonzero integers:
fix(abs(log10(abs(A))))+1
For a 10,000 iteration benchmark with some above answers:
Jaymin= 1.423717 seconds; Stephanie= 0.476135 seconds; Mine= 0.000878 seconds
If you don't expect 0s to appear, this is the fastest and most accurate method. Only works for decimals that satisfy -1<A<1.
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DGM
DGM 2023 年 1 月 28 日
編集済み: DGM 2023 年 1 月 28 日
Ran in:
This doesn't actually provide meaningful results on the specified interval.
x = (-0.15:0.025:0.15).';
y1 = floor(log10(abs(x)+1))+1; % works for all integers (assumes 0 is 1 digit)
y2 = fix(abs(log10(abs(x))))+1; % works for nonzero integers
table(x,y1,y2)
ans = 13×3 table
x y1 y2 ______ __ ___ -0.15 1 1 -0.125 1 1 -0.1 1 2 -0.075 1 2 -0.05 1 2 -0.025 1 2 0 1 Inf 0.025 1 2 0.05 1 2 0.075 1 2 0.1 1 2 0.125 1 1 0.15 1 1
Consider the case of abs(x) = 0.125. In what interpretation does this value have one digit?
Consider the case of abs(x) = 0.10. The naive interpretation of this value might suggest that it has either 1 or 2 digits (depending if you consider the leading integer 0). In reality, 0.1 is not exactly represented in floating point, so instead it's actually 0.999999999999999916 ... etc, or some similar approximation depending on how it was calculated. In base-2, 0.1 has as many digits as your floating-point approximation allows. If 0.125 has 1 digit, then no consistent interpretation would suggest that 0.10 has 2 digits.
As the utility for the fractional part of the number can now be ignored, the only meaningful difference between the two methods presented is the initial offset of 1 to avoid the singularity in log(x). At that point, the distinction between floor() and fix() can be ignored, since all its inputs will be positive.
Bear in mind that for practical use, we should be considering double(x) or something equivalent, otherwise both methods will fail if x is integer-class.

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Jaymin
Jaymin 2012 年 12 月 13 日
編集済み: Walter Roberson 2020 年 5 月 30 日
Long, but it gets the job done.
numel(num2str(A))-numel(strfind(num2str(A),'-'))-numel(strfind(num2str(A),'.'))
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Walter Roberson
Walter Roberson 2020 年 5 月 30 日
Fails on floor(pi*1e25)

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fugue
fugue 2013 年 5 月 26 日
numel(num2str(fix(abs(A))))
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Walter Roberson
Walter Roberson 2020 年 5 月 30 日
Fails on floor(pi*1e25)

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ARVIND KUMAR SINGH
ARVIND KUMAR SINGH 2020 年 5 月 30 日
no_of_digits = numel(num2str(abs(A)));
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Walter Roberson
Walter Roberson 2020 年 5 月 30 日
Fails on floor(pi*1e25)

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Mayur Lad
Mayur Lad 2020 年 10 月 8 日
編集済み: Walter Roberson 2020 年 10 月 8 日
x=input('Enter number: ');
disp(x)
sum= 0;
while x > 0
t = mod(x,10);
sum= sum+1;
x = (x-t)/10;
end
fprintf('no of digit is:\t %d',sum)
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manindra
manindra 2023 年 1 月 27 日
If number is 0 or negative, the control doesn't even enter into the loop. Thats just a dumb question.
Walter Roberson
Walter Roberson 2023 年 1 月 27 日
@manindra
I am not clear as to what you are saying is "dumb" ?

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Daniel Dalskov
Daniel Dalskov 2021 年 1 月 29 日
I am using a simplified version of the below to determine whether I should try to represent a number as x*pi, x*sqrt(2), x*exp(1) or a fraction. In my case I only needed to check if there are more than are 14 digits, so no for-loop needed.
It basically finds the difference between the first and last non-zero number. Sign, decimal point and exponent are not included in the count.
a = [0,12e-17,1,10,21003, round(pi,3)*1e6, round(pi,5), round(pi,10), round(pi,10)*1e-3, round(pi*1e-5,10), pi, pi*1e307, pi*1e-314];
a = [a, -a];
for i=1:length(a)
b=abs(a(i)); % sign is not important for the number of digits, but feel free to add 0>sign(a(i)) to sd
if b < 3e-323 % depending on how precise you want to be
sd = 1; % could be changed to zero depending on your usecase
else
msd=floor(log10(b)); % most significant digit
lsd=-inf;
for dp=msd:-1:-323 % again depending on how precise you want to be, numbers really close to zero gets a little ify
if mod(b,10.^dp)==0
lsd=dp-1; % least significant digit
break
end
end
sd = msd-lsd; % number of digits
end
fprintf('%d significant digits\tin\t%s\n', sd, num2str(a(i), 16))
end
Aziz ur Rehman
Aziz ur Rehman 2023 年 6 月 14 日
The approach that i followed is in which i used a recursive function to compute the sum of digits of the integer provided such that the sum of A=[12345] is 15.
function x=digit_sum(input)
if input==0
x=0;
else
digit = rem(input,10);
input=(input-digit)/10;
x=digit+digit_sum(input);
end
end
You can see that for getting the 10th of the integer, i just divided the number by 10, and the remainder gave us the last digit, which in turn was added using a recursive function.
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Steven Lord
Steven Lord 2023 年 6 月 14 日
Ran in:
Is this what you expected? I added one additional input to your function so it shows you the intermediary steps.
digit_sum(pi, true)
Digit is 3.14159, input is 0
ans = 3.1416
You can see the effect of the additional input on a problem where this function works:
digit_sum(12345, true)
Digit is 5, input is 1234 Digit is 4, input is 123 Digit is 3, input is 12 Digit is 2, input is 1 Digit is 1, input is 0
ans = 15
The following behavior is what I expected for a non-finite input, since you are recursively calling digit_sum with Inf as the input each time. The same holds if you call digit_sum with NaN as input. I'm not showing the intermediate results, as I suspect the displaying of those steps would cause this code to time out in MATLAB Answers.
digit_sum(Inf, false)
Out of memory. The likely cause is an infinite recursion within the program.

Error in solution>digit_sum (line 13)
x=digit+digit_sum(input, displaySteps);
Your code would also fail if input was a complex number.
function x=digit_sum(input, displaySteps)
if input==0
x=0;
else
digit = rem(input,10);
input=(input-digit)/10;
if displaySteps
fprintf("Digit is %g, input is %g\n", digit, input);
end
x=digit+digit_sum(input, displaySteps);
end
end

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