identifying and isolating consecutive numbers

37 次查看（过去 30 天）
Claudia
Claudia 2012-6-14
评论： Eric Homer 2024-2-16
I have a vector, for example, A= [1 2 3 4 14 15 23 24 25 ]
and I want a code that will identify regions of consecutive numbers and separate them into their own array. ie, a code that will split A into
B = [1 2 3 4] C = [14 15] D = [23 24 25]
I would like this code to be able to work on a matrix A of variable length. Any suggestions?
Thank you!

请先登录，再回答此问题。

采纳的回答

Maziyar
Maziyar 2015-10-8
编辑：Maziyar 2015-10-8
A(end+1) = 2 % Add a new isolating point end
I_1 = find(D ~= 1); % Find indexes of isolating points
[m,n] = size(I_1);
Start_Idx = 1 ; % Set start index
for i = 1:n
End_Idx = I_1(i); % Set end index
Sequ = A(Start_Idx:End_Idx) % Find consecuative sequences
Start_Idx = End_Idx + 1;
% update start index for the next consecuitive sequence
end
  4 个评论
显示 2更早的评论
Samantha Plesce
Samantha Plesce 2021-10-27
I was trying to use this snipet for the same application. I have been testing it with various arrays that contain consecutive sequences. I am having an issue with actually finding the correct set of sequences.
When A = ...
A= [2 4 5 7 8 9 10 -3 -2 -1 0 20] % your array of values
The desired seq cell array should contian:
{[4 5], [7 8 9 10], [-3 -2 -1 0]}
While these sets do appear in the seq cell array, it is also accounting for the first and last value even though they are not apart of a sequence of consecutive values.
seq =
1×5 cell array
{[2]} {1×2 double} {1×4 double} {1×4 double} {[20]}

请先登录，再进行评论。

更多回答（4 个）

Walter Roberson
Walter Roberson 2012-6-14
The splits should occur at places where diff(A) has 1's . (You can find the runs of 1's by looking at diff(diff(A)).
Once you know the length of each piece, you can use mat2cell() to break up the vector into cell arrays. (Writing to individual variables is not a good practice for something like this.)
Diego Tasso
Diego Tasso 2012-6-14
Use regexp to do this, something like ( but not exactly):
[B] = regexp(A,'[1-4]','match')
repeat for C and D replacing [1-4] with the number ranges you want to isolate...this might work...not sure.
  2 个评论
Diego Tasso
Diego Tasso 2012-6-14
By the way this is the cheap inefficient way to get this done...I am sure you can create a loop to do so for you.
Guillaume
Guillaume 2015-10-8
regexp is a cheap and efficient way of locating patterns in strings. It does not apply to numbers.
A loop is the most inefficient way of dealing with the problem.

请先登录，再进行评论。

Frank Uhlig
Frank Uhlig 2020-5-5
编辑：Frank Uhlig 2020-5-5
Here is a simple sequence that gives the adjacent integers without the non-adjacent ones:
Strat with k = [ 1 2 3 4 7 8 9 12 13 140],
>> k = [ 1 2 3 4 7 8 9 12 13 140],
k =
1 2 3 4 7 8 9 12 13 140
>> i = find(diff(k) == 1),
i =
1 2 3 5 6 8
>> all = unique(sort([k(i),k(i)+1]))
all =
1 2 3 4 7 8 9 12 13
And you have all adjacent integer groups united in ascending order.
Sorting the last output into individual adjacent integer groups now is another problem.
Eric
Eric 2023-12-2
编辑：Eric 2023-12-2
I know this is a very late answer but wasn't sure how to implement the answer by Maziyar because it uses a variable 'D' that is not defined anywhere. I ended up writing my own and it turned out well so I thought I'd share it.
A= [1 2 3 4 14 15 23 24 25 ]
assert(size(A,1)==1 && isa(A,'double'));
p=find(diff(A)>1);
ind=[A(1),A(p+1);A(p),A(end)];
% ind =
% 1 14 23
% 4 15 25
It takes in a vector A that would have a series of sequential numbers and returns a matrix where the top rop is the start of each sequence and the bottom row is the end of each sequence.
  2 个评论
Katerina F
Katerina F 2024-2-16
I am getting this error:
Operands to the logical AND (&&) and OR (||) operators must be convertible to
logical scalar values. Use the ANY or ALL functions to reduce operands to logical
scalar values.
Any suggestions please?
Eric Homer
Eric Homer 2024-2-16
What's the data you are passing in. Can you please share what is causing the issue?

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Matrix Indexing 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by