I have a dataset, and it looks like this.
1 2 3 4 5
2 3 4 5 6
3 4 5 6 7
4 5 6 7 8
5 6 7 8 9
This is the desired output: (Nothing is recorded with an X value).
X X 3 X X
X 3 4 5 X
3 4 5 6 7
X 5 6 7 X
X X 7 X X
When it writes, I should get "3, 3, 4, 5, 3, 4, 5, 6, 7, 5, 6, 7, 7" as output.
I have tried a lot of ideas so far and at this moment, nothing seems to be working. My do loops end up creating lots of output and/or it reads the wrong values. I have even tried messing around with different storage modes, but nothing seems to be applicable.
If you had the foresight to declare your array like this
integer, dimension(-2:2,-2:2) :: my-array
that is, with the centre element having index (0,0) then I think this nested loop will output the elements in the order you require:
do row = -2,2
do col = -2+abs(row), 2-abs(row)
write(*,*) my_array(row, col)
end do
end do
If you did not have the foresight to declare your array that way, the arithmetic is a little more fiddly (but no worse than that) so I'll leave it to you. And for a production code you'll want to replace all those 2s with a parameter or a call to ubound or lbound.
Related
I know that in python I can do something as follows.
for i in range(10, 0, -1):
print(i)
Which will output:
10
9
8
7
6
5
4
3
2
1
I'm very much new to julia and I know I can create normal loops as follows.
for i=1:10
println(i)
end
Intuitively, I tried something like as follows (since I thought it behaved similar to python's range([start], stop[, step]) function).
for i=10:1:-1
println(i)
end
Although it didn't fail, it didn't print anything either. What am I doing wrong?
Is there an intuitive way to loop backwards in julia?
Try this:
julia> for i=10:-1:1
println(i)
end
10
9
8
7
6
5
4
3
2
1
or this
julia> for i=reverse(1:10)
println(i)
end
10
9
8
7
6
5
4
3
2
1
As #phipsgabler noted you can also use:
julia> range(10, 1, step=-1)
10:-1:1
to get the same result again (note though that you have to use 1 as a second index).
From my practice range is usually more useful with with length keyword argument:
julia> range(10, 1, length=10)
10.0:-1.0:1.0
(notice that in this case you get a vector of Float64 not Int)
I have for example a=[1 2 3 4 5 6 7 8 9 10]; and I have to delete each 2 following numbers from 3.
like at the end it should be a=[1 4 7 10];
How to do this without a for loop.
And also if there is a way to guarantee that at the end the resulting array will have exact number of entries, like here it should be a with 4 entries at the end.
But for example we have b=[1 2 3 4 5 6 7 8 9 ]; and if I want make sure that at the end I still have 4 entries in the rest array, so that b can't be equal to [1 4 7] because I need 4 entries for sure.
You can use indexing for this:
A = 1:10;
B = A(1:3:end)
B =
[1 4 7 10]
Or, if you really want to remove elements:
A = 1:10;
A(2:3:end) = [];
A(3:3:end) = [];
For your second question regarding length checking, it's unclear what you're asking. Would an if comparison be enough ?
if numel(A) ~= 4
% ... Handle unexpected values here
end
Best,
As you mentioned in the question and in the comments that you need 4 elements at the end and if elements are less than 4 then you want to include the last element/s of b, the following should work:-
b=[1 2 3 4 5 6 7 8 9]
b_req=b(1:3:end);
temp=length(b_req);
if temp<4 b_req(end+1:4)=b(end-3+temp:end); % for including the elements of b so that total elements are 4 at the end
elseif temp>4 b_req=b_req(1:4); % for removing the extra elements
end
b_req
Output:-
b =
1 2 3 4 5 6 7 8 9
b_req =
1 4 7 9
and
if instead b=[1 2 3 4 5 6 7 8 9 10]; then the same code gives what you require, i.e. b_req = [1 4 7 10]
This code speaks for itself:
a = 1:15; % some vector
% returns every third element after the first one:
third_elemets = a(1:3:end);
% returns the missing elements for the vector to be in size 4 from the end of a
last_elements = a(end-3+length(third_elemets):end);
% take maximum 4 elements from a
logic_ind = true(min(4,length(third_elemets)),1);
% and concatanate them with last_elements (if there are any)
a = [third_elemets(logic_ind) last_elements]
and under the assumption that whenever there are less than 4 elements you simply take the last one(s) - it should always work.
Original Question
So if I have an array, s:
s= [4 5 2 5 8 4 11 6]
How would I go about replacing any number greater than 6 to be equal to the previous number. Giving s to be:
s=[4 5 2 5 5 4 4 6]
Updated Question
I want to check if the number is at least double the previous number, and if so to replace with the previous number. Using the a different example to above, s:
s= [8 9 3 6 7 2 5]
Would be replaced with, s:
s= [8 9 3 3 3 2 2]
Using the cummax function (introduced in R2014b):
s = s(cummax((s<=6).*(1:numel(s))));
Example:
>> s = [5 7 9 2 5 8 4 11 6]
s =
5 7 9 2 5 8 4 11 6
>> s = s(cummax((s<=6).*(1:numel(s))))
s =
5 5 5 2 5 5 4 4 6
All you need to do is, iterate over all the values of your array and check if current value is greater than twice the previous value (you can replace the > with >= (greater than or equal to) if that's what you need. And whenever your required condition is met, you just need to copy value from s(i-1) i.e. the previous index to s(i) i.e. current place.
Your loop starts from i=2 instead of 1 because there is no previous value to check in case of i=1. In MALTAB indexing starts from i instead of 0 as in most languages.
s = [4 2 7 5 1];
for i=2:numel(s)
if s(i) > 2*s(i-1)
s(i) = s(i-1);
end
end
Given A = [3 4 5 6 7 8 9 10 11 1 2 3 4 5 6 8]
Output B = [3 4 5 6 8]
Is there a Matlab function or command to get this result? I am new to Matlab. Just now I am doing it going through for each element and keeping a counter for it. I have very big array so this is taking too much time.
Use a combination of unique and histc:
uA = unique(A); %// find unique values
B = uA(histc(A, uA)>=2); %// select those that appear at least twice
The above code gives the values that appear at least twice. If you want values that appear exactly twice, replace >= by ==.
I have an array A, and want to reshape the last four elements of each column into a 2x2 matrix. I would like the results to be stored in a cell array B.
For example, given:
A = [1:6; 3:8; 5:10]';
I would like B to contain three 2x2 arrays, such that:
B{1} = [3, 5; 4, 6];
B{2} = [5, 7; 6, 8];
B{3} = [7, 9; 8, 10];
I can obviously do this in a for loop using something like reshape(A(end-3:end, ii), 2, 2) and looping over ii. Can anyone propose a vectorized method, perhaps using something similar to cellfun that can apply an operation repeatedly to columns of an array?
The way I do this is to look at the desired indices and then figure out a way to generate them, usually using some form of repmat. For example, if you want the last 4 items in each column, the (absolute) indices into A are going to be 3,4,5,6, then add the number of rows to that to move to the next column to get 9,10,11,12 and so on. So the problem becomes generating that matrix in terms of your number of rows, number of columns, and the number of elements you want from each column (I'll call it n, in your case n=4).
octave:1> A = [1:6; 3:8; 5:10]'
A =
1 3 5
2 4 6
3 5 7
4 6 8
5 7 9
6 8 10
octave:2> dim=size(A)
dim =
6 3
octave:3> n=4
n = 4
octave:4> x=repmat((dim(1)-n+1):dim(1),[dim(2),1])'
x =
3 3 3
4 4 4
5 5 5
6 6 6
octave:5> y=repmat((0:(dim(2)-1)),[n,1])
y =
0 1 2
0 1 2
0 1 2
0 1 2
octave:6> ii=x+dim(1)*y
ii =
3 9 15
4 10 16
5 11 17
6 12 18
octave:7> A(ii)
ans =
3 5 7
4 6 8
5 7 9
6 8 10
octave:8> B=reshape(A(ii),sqrt(n),sqrt(n),dim(2))
B =
ans(:,:,1) =
3 5
4 6
ans(:,:,2) =
5 7
6 8
ans(:,:,3) =
7 9
8 10
Depending on how you generate x and y, you can even do away with the multiplication, but I'll leave that to you. :D
IMO you don't need a cell array to store them either, a 3D matrix works just as well and you index into it the same way (but don't forget to squeeze it before you use it).
I gave a similar answer in this question.