How can I make a function with a vector as input and a matrix as an output?
I have to write a function that will convert cubic meters to liters and English gallons. The input should be a vector containing volume values in m ^ 3 to be converted. The result should be a matrix in which the first column contains the result in m ^ 3, the second liter, the third English gallon.
I tried this:
function [liter, gallon] = function1 (x=[a, b, c, d]);
liter= a-10+d-c;
gallon= b+15+c;
endfunction
You're almost there.
The x=[a,b,c,d] part is superfluous, your argument should be just x.
function [liter, gallon] = function1 (x);
a = x(1); b = x(2); c = x(3); d = x(4);
liter = a - 10 + d - c;
gallon = b + 15 + c;
endfunction
If you want your code to be safe and guard against improper inputs, you can perform such checks manually inside the function, e.g.
assert( nargin < 1 || nargin > 4, "Wrong number of inputs supplied");
The syntax x=[a,b,c,d] does not apply to octave; this is reserved for setting up default arguments, in which case a, b, c, and d should be given specific values that you'd want as the defaults. if you had said something like x = [1,2,3,4], then this would be fine, and it would mean that if you called the function without an argument, it would set x up to this default value.
Related
Currently, I am working on a program that integrates x + x^2 + e^x + 2cos(x/2) - 1 with three input variables, a, b, and n. What I need returned is the numerical integral from a to b with n increments. The function also has to return trapezoids for each n as a column vector. Thus, the integral value as a scalar, and a vector of values.
I've gotten to a point where the function int_f_1 is undefined for some reason, and I have no idea why. I thought by nesting that function under the test function, it would help. But it does not, and I don't know why that is. Any suggestions?
function [y] = test_function_1(x);
y = x + x.^2 + exp(x) + 2*cos(x/2) - 1
end
function [int_f, increment] = int_f_1 (a, b, n);
f = #test_function_1;
h = a + b ./ n
increments = h
int_f = integral(h, f)
end
I have 2 different very simple functions with the same input-output structure (Both return a count(*) when avg of 3 notes is >= 4 (function1) and the other a count(*) when avg of 3 notes is < 4 (function2)), They both work properly in separate but now i need to join both into just one function with 2 outputs, I now maybe is a very easy question but i am only getting started with Haskell:
function1::[(String, Int,Int,Int)]->Int
function1 ((name,note1,note2,note3):xs) =
if (note1+note2+note3) `div` 3 >=4 then length xs else length xs
function2::[(String, Int,Int,Int)]->Int
function2 ((name,note1,note2,note3):xs) =
if (note1+note2+note3) `div` 3 <4 then length xs else length xs
Thanks!
You can use &&& from Control.Arrow.
combineFunctions f1 f2 = f1 &&& f2
Then use it like this :
combinedFunc = combineFunctions function1 function2
(res1,res2) = combinedFunc sharedArg
You already use tuples (name,note1,note2,note3) in your input data, so you must be familiar with the concept.
The simplest way to produce two outputs simultaneously is to put the two into one tuple:
combinedFunction f1 f2 input = (out1, out2)
where
out1 = f1 input
out2 = f2 input
It so happens that this can be written shorter as combinedFunction f1 f2 = f1 &&& f2 and even combinedFunction = (&&&), but that's less important for now.
A more interesting way to produce two outputs simultaneously is to redefine what it means to produce an output:
combinedFunWith k f1 f2 input = k out1 out2
where
out1 = f1 input
out2 = f2 input
Here instead of just returning them in a tuple, we pass them as arguments to some other user-specified function k. Let it decide what to do with the two outputs!
As can also be readily seen, our first version can be expressed with the second, as combinedFunction = combinedFunWith (,), so the second one seems to be more general ((,) is just a shorter way of writing a function foo x y = (x,y), without giving it a name).
Is there a way to supply arguments using varargin in MATLAB in the following manner?
Function
func myFunc(varargin)
if a not given as argument
a = 2;
if b not given as argument
b = 2;
if c not given as argument
c = a+b;
d = 2*c;
end
I want to call the above function once with b = 3 and another time while the previous one is running in the same command window with a = 3 and c = 3 and letting b take the default value in the function this time. How can it be done using varargin?
Here's the latest and greatest way to write the function (using arguments blocks from R2019b)
function out = someFcn(options)
arguments
options.A = 3;
options.B = 7;
options.C = [];
end
if isempty(options.C)
options.C = options.A + options.B;
end
out = options.A + options.B + options.C;
end
Note that this syntax does not allow you to say options.C = options.A + options.B directly in the arguments block.
In MATLAB < R2021a, you call this like so
someFcn('A', 3)
In MATLAB >= R2021a, you can use the new name=value syntax
someFcn(B = 7)
Here are two ways to do this which have been available since 2007a (i.e. a long time!). For a much newer approach, see Edric's answer.
Use nargin and ensure your inputs are always in order
Use name-value pairs and an input parser
nargin: slightly simpler but relies on consistent input order
function myFunc( a, b, c )
if nargin < 1 || isempty(a)
a = 2;
end
if nargin < 2 || isempty(b)
b = 2;
end
if nargin < 3 || isempty(c)
c = a + b;
end
end
Using the isempty check you can optionally provide just later arguments, for example myFunc( [], 4 ) would just set b=4 and use the defaults otherwise.
inputParser: more flexible but can't directly handle the c=a+b default
function myFunc( varargin )
p = inputParser;
p.addOptional( 'a', 2 );
p.addOptional( 'b', 2 );
p.addOptional( 'c', NaN ); % Can't default to a+b, default to NaN
p.parse( varargin{:} );
a = p.Results.a;
b = p.Results.b;
c = p.Results.c;
if isnan(c) % Handle the defaulted case
c = a + b;
end
end
This would get used like myFunc( 'b', 4 );. This approach is also agnostic to the input order because of the name-value pairs, so you can also do something like myFunc( 'c', 3, 'a', 1 );
I wrote this mips code to find the gcf but I am confused on getting the number of instructions executed for this code. I need to find a linear function as a function of number of times the remainder must be calculated before an answer. i tried running this code using Single step with Qtspim but not sure on how to proceed.
gcf:
addiu $sp,$sp,-4 # adjust the stack for an item
sw $ra,0($sp) # save return address
rem $t4,$a0,$a1 # r = a % b
beq $t4,$zero,L1 # if(r==0) go to L1
add $a0,$zero,$a1 # a = b
add $a1,$zero,$t4 # b = r
jr gcf
L1:
add $v0,$zero,$a1 # return b
addiu $sp,$sp,4 # pop 2 items
jr $ra # return to caller
There is absolutely nothing new to show here, the algorithm you just implemented is the Euclidean algorithm and it is well known in the literature1.
I will nonetheless write an informal analysis here as link only questions are evil.
First lets rewrite the code in an high level formulation:
unsigned int gcd(unsigned int a, unsigned int b)
{
if (a % b == 0)
return b;
return gcd(b, a % b);
}
The choice of unsigned int vs int was dicated by the MIPS ISA that makes rem undefined for negative operands.
Out goal is to find a function T(a, b) that gives the number of step the algorithm requires to compute the GDC of a and b.
Since a direct approach leads to nothing, we try by inverting the problem.
What pairs (a, b) makes T(a, b) = 1, in other words what pairs make gcd(a, b) terminates in one step?
We clearly must have that a % b = 0, which means that a must be a multiple of b.
There are actually an (countable) infinite number of pairs, we can limit our selves to pairs with the smallest, a and b2.
To recap, to have T(a, b) = 1 we need a = nb and we pick the pair (a, b) = (1, 1).
Now, given a pair (c, d) that requires N steps, how do we find a new pair (a, b) such that T(a, b) = T(c, d) + 1?
Since gcd(a, b) must take one step further then gcd(c, d) and since starting from gcd(a, b) the next step is gcd(b, a % b) we must have:
c = b => b = c
d = a % b => d = a % c => a = c + d
The step d = a % c => a = c + d comes from the minimality of a, we need the smallest a that when divided by c gives d, so we can take a = c + d since (c + d) % c = c % c d % c = 0 + d = d.
For d % c = d to be true we need that d < c.
Our base pair was (1, 1) which doesn't satisfy this hypothesis, luckily we can take (2, 1) as the base pair (convince your self that T(2, 1) = 1).
Then we have:
gcd(3, 2) = gcd(2, 1) = 1
T(3, 2) = 1 + T(2, 1) = 1 + 1 = 2
gcd(5, 3) = gcd(3, 2) = 1
T(5, 3) = 1 + T(3, 2) = 1 + 2 = 3
gcd(8, 5) = gcd(5, 3) = 1
T(8, 5) = 1 + T(5, 3) = 1 + 3 = 4
...
If we look at the pair (2, 1), (3, 2), (5, 3), (8, 5), ... we see that the n-th pair (starting from 1) is made by the number (Fn+1, Fn).
Where Fn is the n-th Fibonacci number.
We than have:
T(Fn+1, Fn) = n
Regarding Fibonacci number we know that Fn ∝ φn.
We are now going to use all the trickery of asymptotic analysis, particularly in the limit of the big-O notation considering φn or φn + 1 is the same.
Also we won't use the big-O symbol explicitly, we rather assume that each equality is true in the limit. This is an abuse, but makes the analysis more compact.
We can assume without loss of generality that N is an upper bound for both number in the pair and that it is proportional to φn.
We have N ∝ φn that gives logφ N = n, this ca be rewritten as log(N)/log(φ) = n (where logs are in base 10 and log(φ) can be taken to be 1/5).
Thus we finally have 5logN = n or written in reverse order
n = 5 logN
Where n is the number of step taken by gcd(a, b) where 0 < b < a < N.
We can further show that if a = ng and b = mg with n, m coprimes, than T(a, b) = T(n, m) thus the restriction of taking the minimal pairs is not bounding.
1 In the eventuality that you rediscovered such algorithm, I strongly advice against continue with reading this answer. You surely have a sharp mind that would benefit the most from a challenge than from an answer.
2 We'll later see that this won't give rise to a loss of generality.
Here is the Main Program:
PROGRAM integration
EXTERNAL funct
DOUBLE PRECISION funct, a , b, sum, h
INTEGER n, i
REAL s
PARAMETER (a = 0, b = 10, n = 200)
h = (b-a)/n
sum = 0.0
DO i = 1, n
sum = sum+funct(i*h+a)
END DO
sum = h*(sum-0.5*(funct(a)+funct(b)))
PRINT *,sum
CONTAINS
END
And below is the Function funct(x)
DOUBLE PRECISION FUNCTION funct(x)
IMPLICIT NONE
DOUBLE PRECISION x
INTEGER K
Do k = 1,10
funct = x ** 2 * k
End Do
PRINT *, 'Value of funct is', funct
RETURN
END
I would like the 'Sum' in the Main Program to print 10 different sums over 10 different values of k in Function funct(x).
I have tried the above program but it just compiles the last value of Funct() instead of 10 different values in sum.
Array results require an explicit interface. You would also need to adjust funct and sum to actually be arrays using the dimension statement. Using an explicit interface requires Fortran 90+ (thanks for the hints by #francescalus and #VladimirF) and is quite tedious:
PROGRAM integration
INTERFACE funct
FUNCTION funct(x) result(r)
IMPLICIT NONE
DOUBLE PRECISION r
DIMENSION r( 10 )
DOUBLE PRECISION x
END FUNCTION
END INTERFACE
DOUBLE PRECISION a , b, sum, h
DIMENSION sum( 10)
INTEGER n, i
PARAMETER (a = 0, b = 10, n = 200)
h = (b-a)/n
sum = 0.0
DO i = 1, n
sum = sum+funct(i*h+a)
END DO
sum = h*(sum-0.5*(funct(a)+funct(b)))
PRINT *,sum
END
FUNCTION funct(x)
IMPLICIT NONE
DOUBLE PRECISION funct
DIMENSION funct( 10)
DOUBLE PRECISION x
INTEGER K
Do k = 1,10
funct(k) = x ** 2 * k
End Do
PRINT *, 'Value of funct is', funct
RETURN
END
If you can, you should switch to a more modern Standard such as Fortran 90+, and use modules. These provide interfaces automatically, which makes the code much simpler.
Alternatively, you could take the loop over k out of the function, and perform the sum element-wise. This would be valid FORTRAN 77:
PROGRAM integration
c ...
DIMENSION sum( 10)
c ...
INTEGER K
c ...
DO i = 1, n
Do k = 1,10
sum(k)= sum(k)+funct(i*h+a, k)
End Do
END DO
c ...
Notice that I pass k to the function. It needs to be adjusted accordingly:
DOUBLE PRECISION FUNCTION funct(x,k)
IMPLICIT NONE
DOUBLE PRECISION x
INTEGER K
funct = x ** 2 * k
PRINT *, 'Value of funct is', funct
RETURN
END
This version just returns a scalar and fills the array in the main program.
Apart from that I'm not sure it is wise to use a variable called sum. There is an intrinsic function with the same name. This could lead to some confusion...