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I have a 12-bit binary that I need to convert to a decimal. For example:
A = [0,1,1,0,0,0,0,0,1,1,0,0];
Bit 1 is the most significant bit, Bit 12 is the least significant bit.
Note: This answer applies primarily to unsigned data types. For converting to signed types, a few extra steps are necessary, discussed here.
The bin2dec function is one option, but requires you to change the vector to a string first. bin2dec can also be slow compared to computing the number yourself. Here's a solution that's about 75 times faster:
>> A = [0,1,1,0,0,0,0,0,1,1,0,0];
>> B = sum(A.*2.^(numel(A)-1:-1:0))
B =
1548
To explain, A is multiplied element-wise by a vector of powers of 2, with the exponents ranging from numel(A)-1 down to 0. The resulting vector is then summed to give the integer represented by the binary pattern of zeroes and ones, with the first element in the array being considered the most significant bit. If you want the first element to be considered the least significant bit, you can do the following:
>> B = sum(A.*2.^(0:numel(A)-1))
B =
774
Update: You may be able to squeeze even a little more speed out of MATLAB by using find to get the indices of the ones (avoiding the element-wise multiplication and potentially reducing the number of exponent calculations needed) and using the pow2 function instead of 2.^...:
B = sum(pow2(find(flip(A))-1)); % Most significant bit first
B = sum(pow2(find(A)-1)); % Least significant bit first
Extending the solution to matrices...
If you have a lot of binary vectors you want to convert to integers, the above solution can easily be modified to convert all the values with one matrix operation. Suppose A is an N-by-12 matrix, with one binary vector per row. The following will convert them all to an N-by-1 vector of integer values:
B = A*(2.^(size(A, 2)-1:-1:0)).'; % Most significant bit first
B = A*(2.^(0:size(A, 2)-1)).'; % Least significant bit first
Also note that all of the above solutions automatically determine the number of bits in your vector by looking at the number of columns in A.
Dominic's answer assumes you have access to the Data Acquisition toolbox. If not use bin2dec:
A = [0,1,1,0,0,0,0,0,1,1,0,0];
bin2dec( sprintf('%d',A) )
or (in reverse)
A = [0,1,1,0,0,0,0,0,1,1,0,0];
bin2dec( sprintf('%d',A(end:-1:1)) )
depending on what you intend to be bit 1 and 12!
If the MSB is right-most (I'm not sure what you mean by Bit 1, sorry if that seems stupid):
Try:
binvec2dec(A)
Output should be:
ans =
774
If the MSB is left-most, use fliplr(A) first.
The eps routine in MATLAB essentially returns the positive distance between floating point numbers. It can take an optional argument, too.
My question: How does MATLAB calculate this value? (Does it use a lookup table, or does it use some algorithm to calculate it at runtime, or something else...?)
Related: how could it be calculated in any language providing bit access, given a floating point number?
WIkipedia has quite the page on it
Specifically for MATLAB it's 2^(-53), as MATLAB uses double precision by default. Here's the graph:
It's one bit for the sign, 11 for the exponent and the rest for the fraction.
The MATLAB documentation on floating point numbers also show this.
d = eps(x), where x has data type single or double, returns the positive distance from abs(x) to the next larger floating-point number of the same precision as x.
As not all fractions are equally closely spaced on the number line, different fractions will show different distances to the next floating-point within the same precision. Their bit representations are:
1.0 = 0 01111111111 0000000000000000000000000000000000000000000000000000
0.9 = 0 01111111110 1100110011001100110011001100110011001100110011001101
the sign for both is positive (0), the exponent is not equal and of course their fraction is vastly different. This means that the next floating point numbers would be:
dec2bin(typecast(eps(1.0), 'uint64'), 64) = 0 01111001011 0000000000000000000000000000000000000000000000000000
dec2bin(typecast(eps(0.9), 'uint64'), 64) = 0 01111001010 0000000000000000000000000000000000000000000000000000
which are not the same, hence eps(0.9)~=eps(1.0).
Here is some insight into eps which will help you to write an algorithm.
See that eps(1) = 2^(-52). Now, say you want to compute the eps of 17179869183.9. Note that, I have chosen a number which is 0.1 less than 2^34 (in other words, something like 2^(33.9999...)). To compute eps of this, you can compute log2 of the number, which would be ~ 33.99999... as mentioned before. Take a floor() of this number and add it to -52, since eps(1) = 2^(-52) and the given number 2^(33.999...). Therefore, eps(17179869183.9) = -52+33 = -19.
If you take a number which is fractionally more than 2^34, e.g., 17179869184.1, then the log2(eps(17179869184.1)) = -18. This also shows that the eps value will change for the numbers that are integer powers of your base (or radix), in this case 2. Since eps value only changes at those numbers which are integer powers of 2, we take floor of the power. You will be able to get the perfect value of eps for any number using this. I hope it is clear.
MATLAB uses (along with other languages) the IEEE754 standard for representing real floating point numbers.
In this format the bits allocated for approximating the actual1 real number, usually 32 - for single or 64 - for double precision, are grouped into: 3 groups
1 bit for determining the sign, s.
8 (or 11) bits for exponent, e.
23 (or 52) bits for the fraction, f.
Then a real number, n, is approximated by the following three - term - relation:
n = (-1)s * 2(e - bias) * (1 + fraction)
where the bias offsets negatively2 the values of the exponent so that they describe numbers between 0 and 1 / (1 and 2) .
Now, the gap reflects the fact that real numbers does not map perfectly to their finite, 32 - or 64 - bit, representations, moreover, a range of real numbers that differ by abs value < eps maps to a single value in computer memory, i.e: if you assign a values val to a variable var_i
var_1 = val - offset
...
var_i = val;
...
val_n = val + offset
where
offset < eps(val) / 2
Then:
var_1 = var_2 = ... = var_i = ... = var_n.
The gap is determined from the second term containing the exponent (or characteristic):
2(e - bias)
in the above relation3, which determines the "scale" of the "line" on which the approximated numbers are located, the larger the numbers, the larger the distance between them, the less precise they are and vice versa: the smaller the numbers, the more densely located their representations are, consequently, more accurate.
In practice, to determine the gap of a specific number, eps(number), you can start by adding / subtracting a gradually increasing small number until the initial value of the number of interest changes - this will give you the gap in that (positive or negative) direction, i.e. eps(number) / 2.
To check possible implementations of MATLAB's eps (or ULP - unit of last place , as it is called in other languages), you could search for ULP implementations either in C, C++ or Java, which are the languages MATLAB is written in.
1. Real numbers are infinitely preciser i.e. they could be written with arbitrary precision, i.e. with any number of digits after the decimal point.
2. Usually around the half: in single precision 8 bits mean decimal values from 1 to 2^8 = 256, around the half in our case is: 127, i.e. 2(e - 127)
2. It can be thought that: 2(e - bias), is representing the most significant digits of the number, i.e. the digits that contribute to describe how big the number is, as opposed to the least significant digits that contribute to describe its precise location. Then the larger the term containing the exponent, the smaller the significance of the 23 bits of the fraction.
Didn't know how to paraphrase the question well.
Function for example:
Data:https://www.dropbox.com/s/wr61qyhhf6ujvny/data.mat?dl=0
In this case how do I calculate that the rest point of this function is ~1? I have access to the vector that makes the plot.
I guess the mean is an approximation but in some cases it can be pretty bad.
Under the assumption that the "rest" point is the steady-state value in your data and the fact that the steady-state value happens the majority of the times in your data, you can simply bin all of the points and use each unique value as a separate bin. The bin with the highest count should correspond to the steady-state value.
You can do this by a combination of histc and unique. Assuming your data is stored in y, do this:
%// Find all unique values in your data
bins = unique(y);
%// Find the total number of occurrences per unique value
counts = histc(y, bins);
%// Figure out which bin has the largest count
[~,max_bin] = max(counts);
%// Figure out the corresponding y value
ss_value = bins(max_bin);
ss_value contains the steady-state value of your data, corresponding to the most occurring output point with the assumptions I laid out above.
A minor caveat with the above approach is that this is not friendly to floating point data whose unique values are generated by floating point values whose decimal values beyond the first few significant digits are different.
Here's an example of your data from point 2300 to 2320:
>> format long g;
>> y(2300:2320)
ans =
0.99995724232555
0.999957488454868
0.999957733165346
0.999957976465197
0.999958218362579
0.999958458865564
0.999958697982251
0.999958935720613
0.999959172088623
0.999959407094224
0.999959640745246
0.999959873049548
0.999960104014889
0.999960333649014
0.999960561959611
0.999960788954326
0.99996101464076
0.999961239026462
0.999961462118947
0.999961683925704
0.999961904454139
Therefore, what I'd recommend is to perhaps round so that the first 5 or so significant digits are maintained.
You can do this to your dataset before you continue:
num_digits = 5;
y_round = round(y*(10^num_digits))/(10^num_digits);
This will first multiply by 10^n where n is the number of digits you desire so that the decimal point is shifted over by n positions. We round this result, then divide by 10^n to bring it back to the scale that it was before. If you do this, for those points that were 0.9999... where there are n decimal places, these will get rounded to 1, and it may help in the above calculations.
However, more recent versions of MATLAB have this functionality already built-in to round, and you can just do this:
num_digits = 5;
y_round = round(y,num_digits);
Minor Note
More recent versions of MATLAB discourage the use of histc and recommend you use histcounts instead. Same function definition and expected inputs and outputs... so just replace histc with histcounts if your MATLAB version can handle it.
Using the above logic, you could also use the median too. If the majority of data is fluctuating around 1, then the median would have a high probability that the steady-state value is chosen... so try this too:
ss_value = median(y_round);
I am trying to compare an array of doubles to a scalar double for equality, but equality is never recognized under certain circumstances. I suspect that this has to do with the way the double is represented (e.g., 1.0 vs 1.00), but I can't figure it out.
For instance, I have generated an array composed of thousands of double values, the last few of which at some instant in time are given by
10.6000
-11.0000
10.2000
22.6000
3.4000
If I test for equality to 10.2 (or 10.2000) by the command array==10.2 (or array=10.2000), I return an array of 0s. If I place the values shown into an array manually (e.g., array=[10.6000 -11.0000 10.2000 22.6000 3.4000]), then the command is successful (i.e., array==10.2 returns 0 0 1 0 0). Could someone please explain why the equality succeeds if I input the values manually, but fails if the array is generated in the context of a program? I am able to rectify the comparison failure by using an approximate rather than an exact comparison (e.g., (array<10.20001) & (array>10.19999)), but this seems unsatisfying.
Edit: The values in the array are generated by iterative addition or subtraction of a constant double (e.g., 0.2). The modulus of this array by 0.2 should therefore be everywhere equal to 0. In fact, the modulus of each element is equal to either 0 or 0.2, as shown below for the above sequence of numbers in the array:
mod(array,0.2)
...
0.2000
0
0.2000
0.2000
0
Again, if the values are placed in an array manually and the modulus is taken, the expected value of all 0s is obtained.
The reason is MATLAB truncated the numbers in the array to preserve only 4 digits after the decimal point when displaying,. That is, the real value of your array may be [10.600000000001, -10.99999999999, ...]. You are right, this is due to the internal representation of floating-point numbers in computer, which may cause tiny errors in the computations.
Personally I think there are two solutions, one is approximate matching like you did, while the other is to round the array up first (say, with this tool from FileExchange), and then do exact matching.
Something is probably in single precision somewhere, and double elsewhere. The binary representation of e.g. 10.2 in each is different as they terminate after a different number of bits. Thus they are different:
>> if (single(10.2) == 10.2) disp('honk'); end
>> if (single(10.2) == single(10.2)) disp('honk'); end
honk
You'll need to check for equality within some small difference:
eps = 0.001;
result = abs(array-10.2) < eps;
You can find the precision used in an array using whos:
>> whos A
Name Size Bytes Class Attributes
A 1x2 8 single
Create a MATLAB function file that will accept a modulo values (from 3 to 9; that is Z3 to Z9) and
will output the least possible value describe by the modulo conditions.
Sample Simulation:
Z = [ 3 4 5 ]; % Modulo Z3, Z4, and Z5
r = [ 2 1 4 ]; % Remainder Values
Least Possible Value is 29.
The Z inputs must be an array matrix ... where in you can type any number from 3 to 9 .... and you can type 3,4,5,6,7,8,9 in any order, in any pairings or groupings ...
The r inputs should be equal to the number of z inputs too...
the output should yield the least possible value though modulo conditions...
I have a 12-bit binary that I need to convert to a decimal. For example:
A = [0,1,1,0,0,0,0,0,1,1,0,0];
Bit 1 is the most significant bit, Bit 12 is the least significant bit.
Note: This answer applies primarily to unsigned data types. For converting to signed types, a few extra steps are necessary, discussed here.
The bin2dec function is one option, but requires you to change the vector to a string first. bin2dec can also be slow compared to computing the number yourself. Here's a solution that's about 75 times faster:
>> A = [0,1,1,0,0,0,0,0,1,1,0,0];
>> B = sum(A.*2.^(numel(A)-1:-1:0))
B =
1548
To explain, A is multiplied element-wise by a vector of powers of 2, with the exponents ranging from numel(A)-1 down to 0. The resulting vector is then summed to give the integer represented by the binary pattern of zeroes and ones, with the first element in the array being considered the most significant bit. If you want the first element to be considered the least significant bit, you can do the following:
>> B = sum(A.*2.^(0:numel(A)-1))
B =
774
Update: You may be able to squeeze even a little more speed out of MATLAB by using find to get the indices of the ones (avoiding the element-wise multiplication and potentially reducing the number of exponent calculations needed) and using the pow2 function instead of 2.^...:
B = sum(pow2(find(flip(A))-1)); % Most significant bit first
B = sum(pow2(find(A)-1)); % Least significant bit first
Extending the solution to matrices...
If you have a lot of binary vectors you want to convert to integers, the above solution can easily be modified to convert all the values with one matrix operation. Suppose A is an N-by-12 matrix, with one binary vector per row. The following will convert them all to an N-by-1 vector of integer values:
B = A*(2.^(size(A, 2)-1:-1:0)).'; % Most significant bit first
B = A*(2.^(0:size(A, 2)-1)).'; % Least significant bit first
Also note that all of the above solutions automatically determine the number of bits in your vector by looking at the number of columns in A.
Dominic's answer assumes you have access to the Data Acquisition toolbox. If not use bin2dec:
A = [0,1,1,0,0,0,0,0,1,1,0,0];
bin2dec( sprintf('%d',A) )
or (in reverse)
A = [0,1,1,0,0,0,0,0,1,1,0,0];
bin2dec( sprintf('%d',A(end:-1:1)) )
depending on what you intend to be bit 1 and 12!
If the MSB is right-most (I'm not sure what you mean by Bit 1, sorry if that seems stupid):
Try:
binvec2dec(A)
Output should be:
ans =
774
If the MSB is left-most, use fliplr(A) first.