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
Related
As simple as in title. I have nx1 sized vector p. I'm interested in the maximum value of r = p/foo - floor(p/foo), with foo being a scalar, so I just call:
max_value = max(p/foo-floor(p/foo))
How can I get which value of p gave out max_value?
I thought about calling:
[max_value, max_index] = max(p/foo-floor(p/foo))
but soon I realised that max_index is pretty useless. I'm sorry asking this, real beginner here.
Having dropped the issue to pieces, I realized there's no unique corrispondence between values p and values in my related vector p/foo-floor(p/foo), so there's a logical issue rather than a language one.
However, given my input data, I know that the solution is unique. How can I fix this?
I ended up doing:
result = p(p/foo-floor(p/foo) == max(p/foo-floor(p/foo)))
Looks terrible, so if you know any other way...
Once you have the index, use it:
result = p(max_index)
You can create a new vector with your lets say "transformed" values:
p2 = (p/foo-floor(p/foo))
and then just use find to find the max values on p2:
max_index = find(p2 == max(p2))
that will return the index or indices of p2 with the max value of that operation, and finally just lookup the original value in p
p(max_index)
in 1 line, this is:
p(find((p/foo-floor(p/foo) == max((p/foo-floor(p/foo))))))
which is basically the same thing you did in the end :)
i have a cell array as below, which are dates. I am wondering how can i extract the year at the last 4 digits? Could anyone teach me how to locate the year in the string? Thank you!
'31.12.2001'
'31.12.2000'
'31.12.2004'
'31.12.2003'
'31.12.2002'
'31.12.2000'
'31.12.1999'
'31.12.1998'
'31.12.1997'
'31.12.2005'
'31.12.2004'
'31.12.2003'
'31.12.2002'
'31.12.2001'
'31.12.2000'
'31.12.1999'
'31.12.1998'
'31.12.2005'
'31.12.2004'
'31.12.2003'
'31.12.2002'
'31.12.2005'
Example cell array:
A = {'31.12.2001'; '31.12.2002'; '31.12.2003'};
Apply some regular expressions:
B = regexp(A, '\d\d\d\d', 'match')
B = [B{:}];
EDIT: I never realized that matlab will "nest" an extra layer of cells until I tested this. I don't like this solution as much now that I know the second line is necessary. Here is an alternative approach that gets you the years in numeric form:
C = datevec(A, 'dd.mm.yyyy');
C = C(:, 1);
SECOND EDIT: Suprisingly, if your cell array has less than 10000 elements, the regexp approach is faster on my machine. But the output of it is another cell array (which takes up much more memory than a numeric matrix). You can use B = cell2mat(B) to get a character array instead, but this brings the two approaches to approximately equal efficiency.
Just to add a fun answer, designed to take the OP to the stranger regions of Matlab:
C = char(C);
y = (D(:,7:end)-'0') * 10.^(3:-1:0).'
which is an order of magnitude faster than anything posted in the other answers :)
Or, to stay a bit closer to home,
y = cellfun(#(x)str2double(x(7:end)),C);
or, yet another regexp variation:
y = str2num(char(regexprep(C, '\d+\.\d+\.','')));
Assuming your matrix with dates is M or a cell array C:
In case your data is in a cell array start with
M = cell2mat(C)
Then get the relevant part
Y=M(:,end-4:end)
If required you can even make the year a number
Year = str2num(Y)
Using regexp this will works also with dates with slightly different formats, like 1.1.2000, which can mess with you offsets
res = regexp(dates, '(?<=\d+\.\d+\.)\d+', 'match')
I am new to Mathematica and I am having difficulties with one thing. I have this Table that generates 10 000 times 13 numbers (12 numbers + 1 that is a starting number). I need to create a Histogram from all 10 000 13th numbers. I hope It's quite clear, quite tricky to explain.
This is the table:
F = Table[(Xi = RandomVariate[NormalDistribution[], 12];
Mu = -0.00644131;
Sigma = 0.0562005;
t = 1/12; s = 0.6416;
FoldList[(#1*Exp[(Mu - Sigma^2/2)*t + Sigma*Sqrt[t]*#2]) &, s,
Xi]), {SeedRandom[2]; 10000}]
The result for the following histogram could be a table that will take all the 13th numbers to one table - than It would be quite easy to create an histogram. Maybe with "select"? Or maybe you know other ways to solve this.
You can access different parts of a list using Part or (depending on what parts you need) some of the more specialised commands, such as First, Rest, Most and (the one you need) Last. As noted in comments, Histogram[Last/#F] or Histogram[F[[All,-1]]] will work fine.
Although it wasn't part of your question, I would like to note some things you could do for your specific problem that will speed it up enormously. You are defining Mu, Sigma etc 10,000 times, because they are inside the Table command. You are also recalculating Mu - Sigma^2/2)*t + Sigma*Sqrt[t] 120,000 times, even though it is a constant, because you have it inside the FoldList inside the Table.
On my machine:
F = Table[(Xi = RandomVariate[NormalDistribution[], 12];
Mu = -0.00644131;
Sigma = 0.0562005;
t = 1/12; s = 0.6416;
FoldList[(#1*Exp[(Mu - Sigma^2/2)*t + Sigma*Sqrt[t]*#2]) &, s,
Xi]), {SeedRandom[2]; 10000}]; // Timing
{4.19049, Null}
This alternative is ten times faster:
F = Module[{Xi, beta}, With[{Mu = -0.00644131, Sigma = 0.0562005,
t = 1/12, s = 0.6416},
beta = (Mu - Sigma^2/2)*t + Sigma*Sqrt[t];
Table[(Xi = RandomVariate[NormalDistribution[], 12];
FoldList[(#1*Exp[beta*#2]) &, s, Xi]), {SeedRandom[2];
10000}] ]]; // Timing
{0.403365, Null}
I use With for the local constants and Module for the things that are other redefined within the Table (Xi) or are calculations based on the local constants (beta). This question on the Mathematica StackExchange will help explain when to use Module versus Block versus With. (I encourage you to explore the Mathematica StackExchange further, as this is where most of the Mathematica experts are hanging out now.)
For your specific code, the use of Part isn't really required. Instead of using FoldList, just use Fold. It only retains the final number in the folding, which is identical to the last number in the output of FoldList. So you could try:
FF = Module[{Xi, beta}, With[{Mu = -0.00644131, Sigma = 0.0562005,
t = 1/12, s = 0.6416},
beta = (Mu - Sigma^2/2)*t + Sigma*Sqrt[t];
Table[(Xi = RandomVariate[NormalDistribution[], 12];
Fold[(#1*Exp[beta*#2]) &, s, Xi]), {SeedRandom[2];
10000}] ]];
Histogram[FF]
Calculating FF in this way is even a little faster than the previous version. On my system Timing reports 0.377 seconds - but such a difference from 0.4 seconds is hardly worth worrying about.
Because you are setting the seed with SeedRandom, it is easy to verify that all three code examples produce exactly the same results.
Making my comment an answer:
Histogram[Last /# F]
I am studying for my algorithms class. I have a question in context to the Masters theorem:
How is n.log2(n) polynomially larger than n^(log4(3))
(log2(x) = log to the base 2 of x
log4(x) = log to the base 4 of x)
(Note: This is a solved problem on page.95 of 'Introduction to Algorithms' by Cormen et.al.)
log4(3) is less than 1, and so n^(log4(3)) is less than n^1 = n, which is less than n*log2(n).
hey guys i want your suggestion that how can change value of two variables without 3rd one. in objective cc.
is there any way so please inform me,
it can be done in any language. x and y are 2 variables and we want to swap them
{
//lets say x , y are 1 ,2
x = x + y; // 1+2 =3
y = x - y; // 3 -2 = 1
x = x -y; // 3-1 = 2;
}
you can use these equation in any language to achieve this
Do you mean exchange the value of two variables, as in the XOR swap algorithm? Unless you're trying to answer a pointless interview question, programming in assembly language, or competing in the IOCCC, don't bother. A good optimizing compiler will probably handle the standard tmp = a; a = b; b = tmp; better than whatever trick you might come up with.
If you are doing one of those things (or are just curious), see the Wikipedia article for more info.
As far as number is concerned you can swap numbers in any language without using the third one whether it's java, objective-C OR C/C++,
For more info
Potential Problem in "Swapping values of two variables without using a third variable"
Since this is explicitly for iPhone, you can use the ARM instruction SWP, but it's almost inconceivable why you'd want to. The complier is much, much better at this kind of optimization. If you just want to avoid the temporary variable in code, write an inline function to handle it. The compiler will optimize it away if it can be done more efficiently.
NSString * first = #"bharath";
NSString * second = #"raj";
first = [NSString stringWithFormat:#"%#%#",first,second];
NSRange needleRange = NSMakeRange(0,
first.length - second.length);
second = [first substringWithRange:needleRange];
first = [first substringFromIndex:second.length];
NSLog(#"first---> %#, Second---> %#",first,second);