Extracting the daily time series from a text file to a single column

1 Ansicht (letzte 30 Tage)
Dear Matlab users,
I have a text file (attached here) containing daily time series from 2013 to 2019 and I want to take out the daily data points and save it in a column vector in chronological order. I would be really grateful of your help in this regard.
Best,
Shahram
  2 Kommentare
Guillaume
Guillaume am 23 Jul. 2019
It would be a lot more useful if you attached a representative text file.
Shahram Sahraei
Shahram Sahraei am 23 Jul. 2019
Dear Guillaume,
Thanks for you suggestion. I just modified my question and attached the text file to it.

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

David Wilson
David Wilson am 24 Jul. 2019
A quick hack is:
fname = 'D59.txt'
Flow = [];
fid=fopen(fname);
Yr = 2012;
R = [];
while 1
tline = fgetl(fid);
if ~ischar(tline), break, end
s = strtrim(tline);
disp(s)
if strncmp(s,'Day',3)
disp(s)
Yr = Yr+1;
X = [];
i=1;
while i < 32
s = deblank(fgetl(fid));
if isempty(s)
s = deblank(fgetl(fid));
end
%disp(s);
x = str2num(s); disp(x)
n = length(x);
if n == 8
x = [x(1:2), NaN, x(3), NaN, x(4), NaN, x(5:6), NaN, x(7), NaN x(8)];
elseif n == 12
x = [x(1:2), NaN, x(3:end)];
end
X = [X;x];
i=i+1;
end % for
Rdat = X(:,[2:end]);
r = Rdat(:);
r(find(isnan(r)))=[]; % drop bad days (leap year?)
R = [R; r];
end
end
fclose(fid);
plot(R)

Weitere Antworten (0)

Kategorien

Mehr zu Characters and Strings finden Sie in Help Center und File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by