Extract Numbers from Mixed string

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Jason
Jason 2024 年 2 月 9 日
コメント済み: Jason 2024 年 2 月 9 日
Hello, I have this string
'single snap detector: 1 S2L:232867 S2R:3151621 S3L:0 S3R:0'
Whats the best way to get the 4 numbers and then the S2R - S2L value?
I have tried this:
B='single snap detector: 1 S2L:232867 S2R:3151621 S3L:0 S3R:0'
pat=regexpPattern("S2L:[/d]")
extract(B,pat)
But Im getting an empty array
  1 件のコメント
Jason
Jason 2024 年 2 月 9 日
Actually, this seems to work not sure if its th e best way:
sscanf(B,'single snap detector: 1 S2L:%d S2R:%d S3L:%d S3R:%d')
ans =
232867
3151621
0
0

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採用された回答

Hassaan
Hassaan 2024 年 2 月 9 日
% Your input string
B = 'single snap detector: 1 S2L:232867 S2R:3151621 S3L:0 S3R:0';
% Regular expression pattern to match the labels and their corresponding values
pattern = 'S2L:(\d+) S2R:(\d+) S3L:(\d+) S3R:(\d+)';
% Using regexp to find the matches and extract the numbers
tokens = regexp(B, pattern, 'tokens');
% tokens{1} will contain a cell array of the matched groups if a match is found
if ~isempty(tokens)
% Converting the extracted strings to numbers
numbers = str2double(tokens{1});
S2L = numbers(1);
S2R = numbers(2);
S3L = numbers(3);
S3R = numbers(4);
% Calculate S2R - S2L
difference = S2R - S2L;
else
S2L = NaN;
S2R = NaN;
S3L = NaN;
S3R = NaN;
difference = NaN;
end
% Displaying the results
S2L, S2R, S3L, S3R, difference
S2L = 232867
S2R = 3151621
S3L = 0
S3R = 0
difference = 2918754
-----------------------------------------------------------------------------------------------------------------------------------------------------
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その他の回答 (3 件)

Vaibhav
Vaibhav 2024 年 2 月 9 日
編集済み: Vaibhav 2024 年 2 月 9 日
Hi Jason
You can consider using the below regular expression to extract the required values.
'S2L:(\d+) S2R:(\d+) S3L:(\d+) S3R:(\d+)'
Hope this helps!

Stephen23
Stephen23 2024 年 2 月 9 日
編集済み: Stephen23 2024 年 2 月 9 日
B = 'single snap detector: 1 S2L:232867 S2R:3151621 S3L:0 S3R:0';
Method one: SSCANF:
V = sscanf(B,'%*[^:]:%d') % very efficient
V = 5×1
1 232867 3151621 0 0
Method two: REGEXP (could also be EXTRACT):
T = regexp(B,'(S\w+):(\d+)','tokens');
T = vertcat(T{:}) % all relevant parts
T = 4×2 cell array
{'S2L'} {'232867' } {'S2R'} {'3151621'} {'S3L'} {'0' } {'S3R'} {'0' }
V = str2double(T(:,2))
V = 4×1
232867 3151621 0 0
  3 件のコメント
Stephen23
Stephen23 2024 年 2 月 9 日
@Jason: if you need only the numeric values and want efficient code then you won't get faster than the SSCANF call I showed you.
Jason
Jason 2024 年 2 月 9 日
Yeah thats what Im actually using :-), thankyou!

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Mathieu NOE
Mathieu NOE 2024 年 2 月 9 日
hello
i supposed you wanted to do this :
B='single snap detector: 1 S2L:232867 S2R:3151621 S3L:0 S3R:0'
B = 'single snap detector: 1 S2L:232867 S2R:3151621 S3L:0 S3R:0'
% pat=regexpPattern("S2L:[/d]")
pat=regexpPattern("S2L:[\d]")
pat = pattern
Matching: regexpPattern("S2L:[\d]")
extract(B,pat)
ans = 1×1 cell array
{'S2L:2'}
my2 cents suggestion to extract all numbers present in this char array (up to you to pick the ones you are intersted in);
B='single snap detector: 1 S2L:232867 S2R:3151621 S3L:0 S3R:0'
B = 'single snap detector: 1 S2L:232867 S2R:3151621 S3L:0 S3R:0'
C = regexp(B,'\d+(\.)?(\d+)?','match');
% convert string to num array
for ci = 1:numel(C)
out(ci,:) = str2double(C{ci});
end
out
out = 9×1
1 2 232867 2 3151621 3 0 3 0
  1 件のコメント
Jason
Jason 2024 年 2 月 9 日
Thnakyou for pointing out my error!

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