I would like merge three arrays in one array.
Dim arr() = 1 4 7 10
Dim arr2() = 2 5 8 11
Dim arr3() = 3 6 9 12
Dim arr4()
for i=0 to ubond.(arr1)+2
??????????????
Next
The result should be arr4()= {1 2 3 4 5 6 7 8 9 10 11 12}
Best Regards
I'd just concatenate and sort them with LINQ:
Dim arr4 = arr.Concat(arr2).Concat(arr3).OrderBy(Function(x) x)
Whether or not you call .ToArray() is optional and depends what you do with arr4. If you will enumerate it with a foreach loop you can leave it like above. If you will random access it or pass it round as an array of int then call ToArray on it
Make sure you Imports System.Linq
If you don't want to use LINQ, a simple loop:
Dim arr4() as New Integer(arr.Length * 3 - 1)
For x = 0 to arr.Length - 1 Step 1
arr4(3*x) = arr(x)
arr4(3*x+1) = arr2(x)
arr4(3*x+2) = arr3(x)
Next x
This works completely differently - the LINQ version would have made an array of 1 4 7 2 5 8 3 6 9 and then sorted it, this one puts the bits in the right order naturally, but if your data didn't sort this way the two approaches would produce different outputs!
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.
This question already has an answer here:
Matlab - sort cell array of objects by property
(1 answer)
Closed 8 years ago.
I have an array of objects, each with various properties, and I would like to figure out how to run "sort" on the array, specifying a specific property of each object to sort by.
For example, say my objects are "chairs" and I want to sort by the property of numlegs (number of legs), then I would be able to run a sort function on an array of chairs, and it would sort them by the number of legs they have like "sort(chairs,numlegs)". Is there any way to do this?
Thanks!
Sorting object array
define a class
_
classdef SimpleClass
properties
id
numlegs
end
methods
function obj = SimpleClass(id,numlegs)
if nargin > 0
obj.id = id;
obj.numlegs = numlegs;
end
end
end
end
sort
_
chairs = SimpleClass.empty(20,0);
for ii = 1:20
chairs(ii) = SimpleClass(ii, randi(4,1));
end
[~, ind] = sort([chairs.numlegs]);
chairs_sorted = chairs(ind);
Output
_
>> [chairs_sorted.numlegs]
ans =
Columns 1 through 10
1 1 1 1 1 1 1 1 2 3
Columns 11 through 20
3 3 3 3 3 3 3 4 4 4
>> [chairs_sorted.id]
ans =
Columns 1 through 10
3 5 8 9 10 11 17 19 12 1
Columns 11 through 20
2 4 6 7 14 15 20 13 16 18
Sorting structure array
chairs = struct('id',num2cell(1:20), 'numlegs',num2cell(randi(4, 1, 20)));
[~, ind] = sort([chairs.numlegs]);
chairs_sorted = chairs(ind);
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.
This question already has answers here:
Repeat copies of array elements: Run-length decoding in MATLAB
(5 answers)
Closed 8 years ago.
My question is similar to this one, but I would like to replicate each element according to a count specified in a second array of the same size.
An example of this, say I had an array v = [3 1 9 4], I want to use rep = [2 3 1 5] to replicate the first element 2 times, the second three times, and so on to get [3 3 1 1 1 9 4 4 4 4 4].
So far I'm using a simple loop to get the job done. This is what I started with:
vv = [];
for i=1:numel(v)
vv = [vv repmat(v(i),1,rep(i))];
end
I managed to improve by preallocating space:
vv = zeros(1,sum(rep));
c = cumsum([1 rep]);
for i=1:numel(v)
vv(c(i):c(i)+rep(i)-1) = repmat(v(i),1,rep(i));
end
However I still feel there has to be a more clever way to do this... Thanks
Here's one way I like to accomplish this:
>> index = zeros(1,sum(rep));
>> index(cumsum([1 rep(1:end-1)])) = 1;
index =
1 0 1 0 0 1 1 0 0 0 0
>> index = cumsum(index)
index =
1 1 2 2 2 3 4 4 4 4 4
>> vv = v(index)
vv =
3 3 1 1 1 9 4 4 4 4 4
This works by first creating an index vector of zeroes the same length as the final count of all the values. By performing a cumulative sum of the rep vector with the last element removed and a 1 placed at the start, I get a vector of indices into index showing where the groups of replicated values will begin. These points are marked with ones. When a cumulative sum is performed on index, I get a final index vector that I can use to index into v to create the vector of heterogeneously-replicated values.
To add to the list of possible solutions, consider this one:
vv = cellfun(#(a,b)repmat(a,1,b), num2cell(v), num2cell(rep), 'UniformOutput',0);
vv = [vv{:}];
This is much slower than the one by gnovice..
What you are trying to do is to run-length decode. A high level reliable/vectorized utility is the FEX submission rude():
% example inputs
counts = [2, 3, 1];
values = [24,3,30];
the result
rude(counts, values)
ans =
24 24 3 3 3 30
Note that this function performs the opposite operation as well, i.e. run-length encodes a vector or in other words returns values and the corresponding counts.
accumarray function can be used to make the code work if zeros exit in rep array
function vv = repeatElements(v, rep)
index = accumarray(cumsum(rep)'+1, 1);
vv = v(cumsum(index(1:end-1))+1);
end
This works similar to solution of gnovice, except that indices are accumulated instead being assigned to 1. This allows to skip some indices (3 and 6 in the example below) and remove corresponding elements from the output.
>> v = [3 1 42 9 4 42];
>> rep = [2 3 0 1 5 0];
>> index = accumarray(cumsum(rep)'+1, 1)'
index =
0 0 1 0 0 2 1 0 0 0 0 2
>> cumsum(index(1:end-1))+1
ans =
1 1 2 2 2 4 5 5 5 5 5
>> vv = v(cumsum(index(1:end-1))+1)
vv =
3 3 1 1 1 9 4 4 4 4 4