How to count similar elements and store this value separately?

3 次查看（过去 30 天）
Graduate Student
Graduate Student 2018-1-29
评论： Graduate Student 2018-1-29
I have a data set that is one column and looks like:
[0000000011111022222033333330444]
I am trying to count the number of elements for a specific number. The list goes up to 170, increasing by 1 integer. And a 0 breaks up the previous number and the next number. I am attempting to count the occurrence of the numbers. For example, for 1 it would be 5 since shows up 5 times before moving onto 2.
Ideally, I am hoping to store this information separately to look like [5,2], something to show that 2 shows up 5 times.
loop_counter = 0;
m = 0;
y = zeros(length(x))
for i = 1:length(x)
if x(i,1) == 1
loop_counter = loop_counter + 1;
else
y(i, 1) = loop_counter;
m = m + 1;
end
end

请先登录，再回答此问题。

采纳的回答

Image Analyst
Image Analyst 2018-1-29
But 2 shows up 4 times, not 5. Why don't you simply use histcounts:
data = [0,0,0,0,0,0,0,0,1,1,1,1,1,0,2,2,2,2,0,3,3,3,3,3,3,3,0,4,4,4]
[counts, edges] = histcounts(data)
result = [(1:length(counts)-1)', counts(2:end)']
You'll see
result =
1 5
2 4
3 7
4 3
  3 个评论
显示 1更早的评论
Image Analyst
Image Analyst 2018-1-29
Attaching your data in advance is always a good idea. Prevents wasting time. But at least better late than never. So, this works:
s = load('sweep_ind.mat')
sweep_ind = s.sweep_ind
edges = .5 : 1 : max(sweep_ind)+ 0.5;
[counts, edges] = histcounts(sweep_ind, edges)
result = [(1:length(counts))', counts']

请先登录，再进行评论。

更多回答（1 个）

KSSV
KSSV 2018-1-29
A = '000000001111102222033333330444' ;
num = 2 ;
iwant = [length(strfind(A,num2str(num))),num]
  1 个评论
Graduate Student
Graduate Student 2018-1-29
Is there something similar to strfind for columns? I ran the code and the error I received was: Input string must have one row.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Creating and Concatenating Matrices 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