Find a specific value in a csv file
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How can I make it?
In csv file, I want to find the year in Part 1 and D value ("D":"1") in Part2
Finally I want to make :
2022, 1
2021, 2
2020, 3
2019, 4
2018, 5
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Answers (2)
Voss
on 15 Jul 2022
Maybe this will work on your file (if not, upload the file here)
file_name = 'table_data.csv';
C = readcell(file_name,'Delimiter',',','NumHeaderLines',1);
years = regexp(C(:,1),'(\d{4})','tokens','once');
years = vertcat(years{:})
d = regexp(C(:,end),'D:"(\d+)"','tokens','once');
d = vertcat(d{:})
% result as a cell array of character vectors:
result = [years d]
% result as a numeric matrix:
result = str2double([years d])
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Voss
on 19 Jul 2022
file_name = 'table_data.csv';
C = readcell(file_name,'Delimiter','\t','NumHeaderLines',1)
years = regexp(C(:,1),'/([^/]*/[^/]*)$','tokens','once');
years = vertcat(years{:})
Abderrahim. B
on 15 Jul 2022
Hi!
Try this:
% Create, split, and extract from part 1
part1 = "A/1/2/" + string(2022:-1:2018) ;
part1 = split(part1, '/') ;
yrs = part1(:,:,end) ;
% Create, split, and extract from part 2
part2 = " ""B"":""1"", ""C"":""2"", ""D"":""" + string(1:5) + '"' ;
part2 = split(part2, '"') ;
dig = str2double(part2(:,:,end-1)) ;
% Result
result = transpose(yrs + "," + dig )
Use datetime if you want to convert this string array to date time array.
Please keep in mind this a way from many to solve it, you can also use regular experssion or patterns to do this.
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