Extracting column and specific sorted row data from a large text file

Hello everybody,
I have a large text file of 4Gb. I wanted to import the file into MATLAB and extract certain column and row.
It is required to extract the data corresponding to Time=86400. For e.g., Line 5 should be skipped and Line 17 should be included.
In Line 17, the cell number (1) and slope angle (14.6) along with Z corresponding to the lowest positive value of P (in several cases, a set of both positive and negative values of P are seen) should be selected.
I have attached a sample file.
The final output will look like:
Cell_number Slope_Angle Z P
1 14.6 0.10000E-02 0.93631E-03
2 14.4 0.10000E-02 0.93631E-03
Thanks a lot !!

 Réponse acceptée

Andrei Bobrov
Andrei Bobrov le 11 Déc 2017
Modifié(e) : Andrei Bobrov le 12 Déc 2017
[EDIT]
f = fopen('FILE_MATLAB.txt');
c = textscan(f,'%s','delimiter','\n');
fclose(f);
c = c{:};
cw = c(find(~cellfun(@isempty,regexp(c,'86400.000'))) + (0:11))';
d = regexp(cw,'-?\d+(\.\d+)?(E[+-]\d+)?','match');
N = cell2mat(cellfun(@(x)str2double(x(1:2)),d,'un',0));
[m,n] = size(N);
D = N(2:end,2:2:end);
D(D < 0) = nan;
[~,ii] = min(D);
out = reshape(permute(reshape(N([ones(1,n);kron(ii+1,[1 1])] + m*(0:n-1)),2,2,[]),[2,1,3]),4,[])';
OTHER variant (for data from your file FILE_MATLAB.txt) 
f = fopen('FILE_MATLAB.txt');
c0 = textscan(f,repmat('%16.6f ',1,6),'Headerlines',4,'CollectOutput',1);
fclose(f);
c = c0{:};
t = c(:,4)==86400;
ii = (find(t) + (1:11))';
c2 = c(ii + size(c,1));
c2(c2<0) = nan;
[~,jj] = min(c2);
out = [c(t,1:2),c(ii(1,:) + jj - 1,1:2)];

3 commentaires

NIKIL
NIKIL le 12 Déc 2017
Modifié(e) : NIKIL le 12 Déc 2017
Thanks a lot Andrei!
I removed the previous comment since the error occurred while running in MacOS. In windows its running fine.
The code ran extremely fast on the 4 gb file.
However, the code couldnt extract the lowest positive value for P. Instead it is extracting only the uppermost value.
For e.g.: when the P values are as shown below (new file attached), it will extract 0.93631 instead of 8.20E-05
p
-0.93631
0.28183
8.20E-05
0.84361
1.1245
1.4054
-1.6863
-1.9672
2.2481
2.529
2.8099
Is there a way to solve this?
Thank you Andrei. The application of code to a 4 gb file worked within 34.69 mins.

Connectez-vous pour commenter.

Plus de réponses (1)

fid = fopen('FILE_MATLAB.txt');
Cell_number = [];
Slope_Angle = [];
Z = [];
P = [];
for i = 1:4 % Skip header lines
tline = fgetl(fid);
end
while ~feof(fid)
s = textscan(fid,'%f',4);
d = textscan(fid,'%f %f %f %f %f %f',11,'CollectOutput', 1);
time = s{1}(4);
if abs(time-86400) < 1e-5
Cell_number = [Cell_number;s{1}(1)];
data = d{1};
[~,rowno] = min(data(:,2));
Slope_Angle = [Slope_Angle;s{1}(2)];
Z = [Z;data(rowno,1)];
P = [P;data(rowno,5)];
end
end
fclose(fid);
fprintf('%12s %12s %12s %12s\n','Cell_number','Slope_Angle','Z','P');
for i = 1:length(Z)
fprintf('%12d %12g %12.2e %12.4e\n',Cell_number(i),Slope_Angle(i),...
Z(i),P(i));
end

1 commentaire

NIKIL
NIKIL le 11 Déc 2017
Modifié(e) : NIKIL le 12 Déc 2017
Thank you very much for the help. Currently i am running the code on 4 gb file. However, its been running for more than 14hrs now. :(

Connectez-vous pour commenter.

Catégories

En savoir plus sur Large Files and Big Data dans Centre d'aide et File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by