# I want to count the number of times each event (a, b, c... up until j) occurs, a(t<1 hour) b(1 hour<t<2 hours) c(2 hours<t<3 hours)... up to j( t>9 hours) based on the following sample in order to find the probability of any individual event occuring

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GS25 on 9 Jan 2020
Commented: GS25 on 21 Jan 2020
'02:47:09'
'03:28:26'
'13:57:18'
'08:58:46'
'10:31:50'
'11:03:35'
'09:11:01'
'01:37:30'
'04:38:33'
'03:02:50'
'05:41:13'
'01:21:29'
'04:44:26'
'01:06:18'
'03:47:08'
'03:33:12'
'05:30:32'
'08:48:40'
'00:00:02'
'08:52:09'
'04:57:18'
'03:47:12'
'03:39:37'
'03:28:25'
'00:43:12'
'03:41:52'
'03:47:26'
'08:00:04'
'01:16:49'
'01:00:12'
'00:01:34'
'04:28:35'
'14:28:52'
'05:46:56'
'02:38:23'
'00:01:19'
'06:21:24'
'05:38:55'
'01:05:57'
'13:36:44'

David Hill on 9 Jan 2020
If your data is not in a cell array, convert to a cell array. For example:
A={'02:47:09';'03:28:26';'13:57:18';'08:58:46';'10:31:50'};
A=hours(duration(A));
a=nnz(A<1);
b=nnz(A>=1&A<2);
c=nnz(A>=2&A<3);
d=nnz(A>=3&A<4);
e=nnz(A>=4&A<5);
f=nnz(A>=5&A<6);
g=nnz(A>=6&A<7);
h=nnz(A>=7&A<8);
i=nnz(A>=8&A<9);
j=nnz(A>=9);

Stephen Cobeldick on 9 Jan 2020
Repeated code is not very generalized, but could easily be replaced by a histogram function.
GS25 on 21 Jan 2020
Thank you guys!