I'm new in MATLAB, i cannot get the answer in the format that i want.
I have a basic function call, but every execution of the program gives the result in the following format :
357341279027200000/23794118819840001
It's supposed to be in decimal, for example for same execution : 15.0181.
I could not figure out why this is happening ? Can you help me, thank you !!
Type format long on the command prompt or in your script.
If that doesnt work because the value is too large, try using vpa
Note that it's just visual, internally the value computed is precise.
>d = 357341279027200000/23794118819840001
d =
15.0181
>> d * 23794118819840001 == 357341279027200000
ans =
1
>> 15.0181 * 23794118819840001 == 357341279027200000
ans =
0
Are you sure that you are not using format rat (rational). This is the reason why you may be having fractional values. If you want decimals, try format long or format long g (Long g provides the optimal length and accuracy as a decimal, up to 10 places.)
Related
I have two questions:
In the following MATLAB code x is a date time value of the format "datetime(Y,M,D,H,MI,S,MS)". The display(x) command displays '00:00:00'. However the 'if condition' displays 'Well received!' which means the real value of x is greater than 0 as opposed to '00:00:00' displayed by the display(x) command. Please suggest how I could display the full value of x up to milliseconds or microseconds.
How can I save '0000,00,00,00,00,00,200' as a date time value?
send = datetime(2016,08,31,06,01,00,00);
receive=datetime(2016,08,31,06,01,00,100);
x=receive-send;
display(x);
if (x>0)
disp('Well received!')
else
disp('Late!')
end
The solution of your first question is, that you might convert your datetime-variable to a formatted string:
disp(datestr(x,'HH:MM:SS:FFF'));
This gives you the output 00:00:00:100, because F is the symbolic identifier for milliseconds.
Furthermore it seems, datetime doesn't support milliseconds. In this case you should use the MATLAB serial date number:
http://de.mathworks.com/help/matlab/ref/datenum.html
The variable x created in your example is a duration object. You can specify the display of milliseconds (as well as smaller decimal fractions of seconds) by setting the Format property.
>> x.Format = 'hh:mm:ss.SSS';
>> display(x);
x = 00:00:00.100
This is presumably also what you want when you ask about saving '0000,00,00,00,00,00,200' as a date time value. It's not really a date and time but a duration, and can be created using the duration constructor.
>> duration(00,00,00,200,'Format','hh:mm:ss.SSS')
ans =
00:00:00.200
Most operations acting on these duration objects will work as expected, such as comparison with the > operator:
>> x > duration(00,00,00,200)
ans =
0
I need to use the equation k=k0*e^(-Q/RT) where T is 10 variables between 90 and 500, to generate the 10 variables I used T = linspace(90,500,10), but when I try to generate the equation it wont let me, k0=1200,Q=8000,R=2 so when I type in k=k0*exp(-Q./(R.*T) the numbers are kind of funky with some 0.0000, am I doing something wrong? Thanks
T = linspace(90,500,10);
k0=1200;
Q=8000;
R=2;
k=k0*exp(-Q./(R.*T));
format longG
k =
Columns 1 through 3
5.98693127135402e-17 1.83626027549851e-10 3.07185900119097e-07
Columns 4 through 6
2.60112352637493e-05 0.000498552970093129 0.00409767412987074
Columns 7 through 9
0.0198590098887835 0.067709291730596 0.180526236997812
Column 10
0.402555153483014
results = [T';k'];
Nothing wrong there, just the format that went wrong I reckon. For formatting options, see either the documentation on format, or this question.
Stability (Numerical analysis)
Trying to apply the answer I saw in this question, a+x=a worked just fine with a+eps(a)/2. Suppose we have x^b+a=a, where b is a small integer, say 3 and a=2000. Then a+(eps(a))^3 or a+(eps(a)/2)^3 will always return number a. Can someone help with the measurement of x? Any way, even different from eps will do just fine.
p.s. 1938+(eps(1938)/0.00000000469)^3 is the last number that returns ans = 1.9380e+003.
1938+(eps(1938)/0.0000000047)^3 returns a=1938. Does that have to do with anything?
x = (eps(a)/2).^(1/(b-eps(a)/2))
if b = 3,
(eps(1938)/2).^(1/(3-eps(1938)/2)) > eps(1938)/0.0000000047
I simply want to generate a series of dates 1 year apart from today.
I tried this
CurveLength=30;
t=zeros(CurveLength);
t(1)=datestr(today);
x=2:CurveLength-1;
t=addtodate(t(1),x,'year');
I am getting two errors so far?
??? In an assignment A(I) = B, the number of elements in B and
Which I am guessing is related to the fact that the date is a string, but when I modified the string to be the same length as the date dd-mmm-yyyy i.e. 11 letters I still get the same error.
Lsstly I get the error
??? Error using ==> addtodate at 45
Quantity must be a numeric scalar.
Which seems to suggest that the function can't be vectorised? If this is true is there anyway to tell in advance which functions can be vectorised and which can not?
To add n years to a date x, you do this:
y = addtodate(x, n, 'year');
However, addtodate requires the following:
x must be a scalar number, not a string.
n must be a scalar number, not a vector.
Hence the errors you get.
I suggest you use a loop to do this:
CurveLength = 30;
t = zeros(CurveLength, 1);
t(1) = today; % # Whatever today equals to...
for ii = 2:CurveLength
t(ii) = addtodate(t(1), ii - 1, 'year');
end
Now that you have all your date values, you can convert it to strings with:
datestr(t);
And here's a neat one-liner using arrayfun;
datestr(arrayfun(#(n)addtodate(today, n, 'year'), 0:CurveLength))
If you're sequence has a constant known start, you can use datenum in the following way:
t = datenum( startYear:endYear, 1, 1)
This works fine also with months, days, hours etc. as long as the sequence doesn't run into negative numbers (like 1:-1:-10). Then months and days behave in a non-standard way.
Here a solution without a loop (possibly faster):
CurveLength=30;
t=datevec(repmat(now(),CurveLength,1));
x=[0:CurveLength-1]';
t(:,1)=t(:,1)+x;
t=datestr(t)
datevec splits the date into six columns [year, month, day, hour, min, sec]. So if you want to change e.g. the year you can just add or subtract from it.
If you want to change the month just add to t(:,2). You can even add numbers > 12 to the month and it will increase the year and month correctly if you transfer it back to a datenum or datestr.
I'm trying to write a program that cleans data, using Matlab. This program takes in the max and min that the data can be, and throws out data that is less than the min or greater than the max. There looks like a small issue with the cleaning part. This case ONLY happens when the minimum range of the variable being checked is 0. If this is the case, for one reason or another, the program won't throw away data points that are between 0 and -1. I've been trying to fix this for some time now, and noticed that this is the only case where this happens, and if you try to run a SQL query selecting data that is < 0, it will leave out data between 0 and -1, so effectively the same error as what's happening to me. Wondering if anyone might recognize this and know what it could be.
I would write such a function as:
function data = cleanseData(data, limits)
limits = sort(limits);
data = data( limits(1) <= data & data <= limits(2) );
end
an example usage:
a = rand(100,1)*10;
b = cleanseData(a, [-2 5]);
c = cleanseData(a, [0 -1]);
-1 is less than 0, so 0 should be the max value. And if this is the case it will keep points between -1 and 0 by your definition of the cleaning operation:
and throws out data that is less than the min or greater than the max.
If you want to throw away (using the above definition)
data points that are between 0 and -1
then you need to set 0 as the min value and -1 as the max value --- which does not make sense.
Also, I think you mean
and throws out data that is less than the min AND greater than the max.
It may be that the floats are getting casted to ints before the comparison. I don't know matlab, but in python int(-0.5)==0, which could explain the extra data points getting in. You can test this by setting the min to -1, if you then also get values from -1 to -2 then you'll need to make sure casting isn't being done.
If I try to mimic your situation with SQL, and run the following query against a datatable that has 1.00, 0.00, -0.20, -0.80. -1.00, -1.20 and -2.00 in the column SomeVal, it correctly returns -0.20 and -0.80, which is as expected.
SELECT SomeVal
FROM SomeTable
WHERE (SomeVal < 0) AND (SomeVal > - 1)
The same is true for MatLab. Perhaps there's an error in your code. Dheck the above statement with your own SELECT statement to see if something's amiss.
I can imagine such a bug if you do something like
minimum = 0
if minimum and value < minimum