Find a specific value in a csv file

23 次查看(过去 30 天)
Alexai
Alexai 2022-7-15
评论: Voss 2022-7-19
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

回答(2 个)

Voss
Voss 2022-7-15
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 个评论
Alexai
Alexai 2022-7-19
编辑:Alexai 2022-7-19
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 2022-7-19
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'}

请先登录,再进行评论。


Abderrahim. B
Abderrahim. B 2022-7-15
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.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by