I'm trying to simplify this task to a one-line command, but I just cannot figure out how it should be done. I tried everything that came to my mind without success. My code right now is:
A_temp = A;
A_temp(A_temp > 3) = A_temp(A_temp > 3) + 1;
A_final = A_temp;
A_final(:,end+1) = 1;
This will achieve the same result as your above code:
A_final = [A+(A > 3) ones(size(A, 2), 1)];
Related
So I have arr = randi([0,20],20,1). I want to show: If there are numbers less than 5, fprintf('Yes\n') only once. Im using a for loop (for i = 1 : length(arr)) and indexing it.
As your description, maybe you need if statement within for loop like below
for i = 1:length(arr)
if arr(i) < 5
fprintf('Yes\n');
break
end
end
If you want to print Yes once, you can try
if any(arr < 5)
fprintf('Yes\n')
endif
If you don't want to use break, the code below might be an option
for i = 1:min(find(arr <5))
if (arr(i) < 5)
fprintf('Yes\n');
end
end
You can use a break statement upon finding the first value under 5 and printing the Yes statement.
Using a break Statement:
arr = randi([0,20],20,1);
for i = 1: length(arr)
if arr(i) < 5
fprintf("Yes\n");
break;
end
end
Extension:
By Using any() Function:
Alternatively, if you'd like to concise it down without the need for a for-loop the any() function can be used to determine if any values within the array meet a condition in this case arr < 5.
arr = randi([0,20],20,1);
if(any(arr < 5))
fprintf("Yes\n");
end
By Using a While Loop:
Check = 0;
arr = randi([0,20],20,1);
i = 1;
while (Check == 0 && i < length(arr))
if arr(i) < 5
fprintf("Yes\n");
Check = 1;
end
i = i + 1;
end
function fnum = fib(n)
if (n == 1) || (n == 2)
fnum = 1;
else
fnum = fib(n-1) + fib(n-2);
end
Can you explain how does each step outputs for the given input. For example inputting 7 gives me 13, 5 gives me 5, but I am not able to track how. I would highly appreciate your reply.
Recursion basically means that the function calls itself.
If we follow your function for fib(3), you will see that what it does is call fib(2)+fib(1). The values of these are defined, and are 1, so it will return 2.
If you call it with fib(4), it will go and compute fib(3)+fib(2). You already know what fib(3) does (see previous paragraph), and we already mentioned that fib(2) returns 1.
If you call it with fib(5) it will go and compute fib(4)+fib(3). See previous paragraph.
This is a very useful way of programming as it is a very simple function to compute something that is arguably more complicated. The most important thing is that you make sure that any recursive function has strong stopping criteria, else it can go forever!
Do you know how Fibonacci series is defined? This function implements that recursively.
Longer answer
Fibonacci series is defined as
n(1) = 1
n(2) = 1
n(k+1) = n(k) + n(k-1)
So when you put 5 as argument, the expansion becomes
n(4+1) = n(4)+n(3)
= n(3)+n(2)+n(2)+n(1)
= n(2)+n(1)+1+1+1
= 1+1+1+1+1
= 5
A much easier back of envelop method is to start from first index and add last two terms to arrive at the next.
1, 1, 2 <- (1+1), 3 <- (2+1), 5 <- (3+2), ...
The Fibonnacci series is defined as f(1) = 1, f(2) = 1 and for all n > 2, f(n) = f(n-1) + f(n-2)
So when you call fib(1) it returns 1 same for fib(2). But when you call fib(3) it returns fib(3-1) + fib(3-2) which is fib(2) + fib(1) = 2. And then when you call fib(4)it returns fib(3) + fib(2) = (fib(2) + fib(1)) + fib(1) = 3. And recursively the fibonnaci series is equal to 1, 1, 3, 5, 8, 13, 21, ...
For the code when n is different than 1 or 2 it call the function fib recursively. And when is equals to 1 or 2 it returns 1.
Consider the following problem:
We are given two arrays A and B such that A and B are sorted
except A has B.length additional 0s appended to its end. For instance, A and B could be the following:
A = [2, 4, 6, 7, 0, 0, 0]
B = [1, 7, 9]
Our goal is to create one sorted list by inserting each entry of B
into A in place. For instance, running the algorithm on the above
example would leave
A = [1, 2, 4, 6, 7, 7, 9]
Is there a clever way to do this in better than O(n^2) time? The only way I could think of is to insert each element of B into A by scanning linearly and performing the appropriate number of shifts, but this leads to the O(n^2) solution.
Some pseudo-code (sorta C-ish), assuming array indexing is 0-based:
pA = A + len(A) - 1;
pC = pA; // last element in A
while (! *pA) --pA; // find the last non-zero entry in A
pB = B + len(B) - 1;
while (pA >= A) && (pB >= B)
if *pA > *pB
*pC = *pA; --pA;
else
*pC = *pB; --pB;
--pC
while (pB >= B) // still some bits in B to copy over
*pC = *pB; --pB; --pC;
Not really tested, and just written off the top of my head, but it should give you the idea... May not have the termination and boundary conditions exactly right.
You can do it in O(n).
Work from the end, moving the largest element towards the end of A. This way you avoid a lot of trouble to do with where to keep the elements while iterating. This is pretty easy to implement:
int indexA = A.Length - B.Length - 1;
int indexB = B.Length - 1;
int insertAt = A.Length;
while (indexA > 0 || indexB > 0)
{
insertAt--;
A[insertAt] = max(B[indexB], A[indexA]);
if (A[indexA] <= B[indexB])
indexB--;
else
indexA--;
}
I'm new to Matlab and now learning the basic grammar.
I've written the file GetBin.m:
function res = GetBin(num_bin, bin, val)
if val >= bin(num_bin - 1)
res = num_bin;
else
for i = (num_bin - 1) : 1
if val < bin(i)
res = i;
end
end
end
and I call it with:
num_bin = 5;
bin = [48.4,96.8,145.2,193.6]; % bin stands for the intermediate borders, so there are 5 bins
fea_val = GetBin(num_bin,bin,fea(1,1)) % fea is a pre-defined 280x4096 matrix
It returns error:
Error in GetBin (line 2)
if val >= bin(num_bin - 1)
Output argument "res" (and maybe others) not assigned during call to
"/Users/mac/Documents/MATLAB/GetBin.m>GetBin".
Could anybody tell me what's wrong here? Thanks.
You need to ensure that every possible path through your code assigns a value to res.
In your case, it looks like that's not the case, because you have a loop:
for i = (num_bins-1) : 1
...
end
That loop will never iterate (so it will never assign a value to res). You need to explicitly specify that it's a decrementing loop:
for i = (num_bins-1) : -1 : 1
...
end
For more info, see the documentation on the colon operator.
for i = (num_bin - 1) : -1 : 1
Is it somehow possible to concatenate two matlab structures recursively without iterating over all leaves of one of the structures.
For instance
x.a=1;
x.b.c=2;
y.b.d=3;
y.a = 4 ;
would result in the following
res = mergeStructs(x,y)
res.a=4
res.b.c=2
res.b.d=3
The following function works for your particular example. There will be things it doesn't consider, so let me know if there are other cases you want it to work for and I can update.
function res = mergeStructs(x,y)
if isstruct(x) && isstruct(y)
res = x;
names = fieldnames(y);
for fnum = 1:numel(names)
if isfield(x,names{fnum})
res.(names{fnum}) = mergeStructs(x.(names{fnum}),y.(names{fnum}));
else
res.(names{fnum}) = y.(names{fnum});
end
end
else
res = y;
end
Then res = mergeStructs(x,y); gives:
>> res.a
ans =
4
>> res.b
ans =
c: 2
d: 3
as you require.
EDIT: I added isstruct(x) && to the first line. The old version worked fine because isfield(x,n) returns 0 if ~isstruct(x), but the new version is slightly faster if y is a big struct and ~isstruct(x).