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finding a numeric pattern in a vector

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Saikrishna
Saikrishna 2023 年 4 月 19 日
コメント済み: Saikrishna 2023 年 4 月 21 日
Hi all,
I have a numeric vector and I am trying to find a pattern (with one missing number) in the vector.
Example:
my numeric vector vec = [5 6 1 2 3 3 4 5 6 1 2 6 3 5 4 2 3 11 2 31 3 4 5 1 2 6 31 11 2 5]
pattern :pat = [3 4 * 1 2]
I know the solution if there can be one or multiple missing numbers for example: [start end] = regexp(char(vec),char( [3 4 *? 1 2]),'start','end') gives start and endpoints of patterns (3 4 5 6 1 2) and (3 4 5 1 2) from the vector. But I am searching for only (3 4 5 1 2) with one missing number.

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Stephen23
Stephen23 2023 年 4 月 20 日
編集済み: Stephen23 2023 年 4 月 20 日
Your basic concept is okay. You need to select an appropriate character match and quantifier. Note that the asterisk is actually a quantifier, as is the question mark (context dependent):
You also have not taken into account any characters that need to be escaped, e.g. char(36) == '$'
Assuming only integers between 0 and 65535, here is a robust approach (no fiddling around counting characters):
V = [5,6,1,2,3,3,4,5,6,1,2,6,3,5,4,2,3,11,2,31,3,4,5,1,2,6,31,11,2,5];
F = @(n)regexptranslate('escape',char(n));
R = sprintf('%s.%s',F([3,4]),F([1,2]));
[X,Y] = regexp(F(V),R)
X = 21
Y = 25
V(X:Y)
ans = 1×5
3 4 5 1 2
  1 件のコメント
Saikrishna
Saikrishna 2023 年 4 月 21 日
thank you @Stephen23 this is what i was looking.

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その他の回答 (1 件)

Les Beckham
Les Beckham 2023 年 4 月 19 日
編集済み: Les Beckham 2023 年 4 月 19 日
Note that I added an additional test at the end of vec to make sure this handles a multi-digit number in the middle position of the pattern ([3 4 10 1 2])
vec = [5 6 1 2 3 3 4 5 6 1 2 6 3 5 4 2 3 11 2 31 3 4 5 1 2 6 31 11 2 5 3 4 10 1 2];
str = num2str(vec);
pat = ['3\s+4\s+\d+\s+1\s+2'];
result = regexp(str, pat, 'match')
result = 1×2 cell array
{'3 4 5 1 2'} {'3 4 10 1 2'}
  2 件のコメント
Saikrishna
Saikrishna 2023 年 4 月 20 日
thank you for your reply. Do you know how to get the indices of the matching pattern?
Walter Roberson
Walter Roberson 2023 年 4 月 20 日
vec = [5 6 1 2 3 3 4 5 6 1 2 6 3 5 4 2 3 11 2 31 3 4 5 1 2 6 31 11 2 5 3 4 10 1 2];
str = num2str(vec);
pat = ['3\s+4\s+\d+\s+1\s+2'];
result = regexp(str, pat)
would return the indices of the starting points inside the character vector str . Which is a bit of a problem because you would have to convert character vector indices to array indices, operating in the face of the possibility that not all entries might have the same width (if they had the same width then the calculation becomes straight forward.)
One way to get them to all have the same width is to use something like
digits_needed = length(num2str(max(vec));
fmt = sprintf('%%%dd', digits_needed);
str = join(compose(fmt, vec), ' ');
pat = '3\s+4\s+\d+\s+1\s+2';
str_locations = regexp(str, pat);
vec_indices = (str_locations - 1) / (digits_needed + 1) + 1
or something close to that

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