Find a specific value in a csv file

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Alexai
Alexai am 15 Jul. 2022
Kommentiert: Voss am 19 Jul. 2022
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

Antworten (2)

Voss
Voss am 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{:})
years = 5×1 cell array
{'2022'} {'2021'} {'2020'} {'2019'} {'2018'}
d = regexp(C(:,end),'D:"(\d+)"','tokens','once');
d = vertcat(d{:})
d = 5×1 cell array
{'1'} {'2'} {'3'} {'4'} {'5'}
% result as a cell array of character vectors:
result = [years d]
result = 5×2 cell array
{'2022'} {'1'} {'2021'} {'2'} {'2020'} {'3'} {'2019'} {'4'} {'2018'} {'5'}
% result as a numeric matrix:
result = str2double([years d])
result = 5×2
2022 1 2021 2 2020 3 2019 4 2018 5
  9 Kommentare
Alexai
Alexai am 19 Jul. 2022
Bearbeitet: Alexai am 19 Jul. 2022
um. sorry I have a mistake I want to read next slash ex)A/B/C0001/123456/abc0011
Thanksfully you give me a code
years = regexp(C(:,1),'/(\d{4})','tokens','once'); but I want to read next slash 123456->abc0011 How can I revise this code?
Voss
Voss am 19 Jul. 2022
file_name = 'table_data.csv';
C = readcell(file_name,'Delimiter','\t','NumHeaderLines',1)
C = 5×2 cell array
{'A/1/2/AB0000/2022abc'} {'"B":"1", "C":"2", "D":"1"'} {'A/1/2/AB0000/2021abc'} {'"B":"1", "C":"2", "D":"2"'} {'A/1/2/AB0000/2020abc'} {'"B":"1", "C":"2", "D":"3"'} {'A/1/2/AB0000/2019abc'} {'"B":"1", "C":"2", "D":"4"'} {'A/1/2/AB0000/2018abc'} {'"B":"1", "C":"2", "D":"5"'}
years = regexp(C(:,1),'/([^/]*/[^/]*)$','tokens','once');
years = vertcat(years{:})
years = 5×1 cell array
{'AB0000/2022abc'} {'AB0000/2021abc'} {'AB0000/2020abc'} {'AB0000/2019abc'} {'AB0000/2018abc'}

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Abderrahim. B
Abderrahim. B am 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 )
result = 5×1 string array
"2022,1" "2021,2" "2020,3" "2019,4" "2018,5"
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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