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Interpolation Error using interp1 for interval 30 minutes

1 vue (au cours des 30 derniers jours)
Trung Hieu Le
Trung Hieu Le le 15 Juin 2016
Hi everyone,
Could you help me check my below code? I'm trying with many ways, however, It always met the error
% %%reset
clear all;
close all;
clc;
%delete NaN, continuously duplicated value and keep the last one
f=fopen('CLF1997.txt');
c=textscan(f , '%s%s%s%f' , 'Headerlines' , 1 , 'delimiter' , ' ');
fclose(f);
t =[diff(c{end})~=0;true];
C = [c{1:3}];
data = [C(t,:),num2cell(c{end}(t))];
clearvars -except data
%combine column date and time
day = data(1:end,2);
time = data(1:end,3);
ns = datenum(day, 'MM/dd/yyyy') + datenum(time, 'hh:mm:ss') - datenum('00:00:00','hh:mm:ss');
data=[data num2cell(ns)];
data(:,2:3)=[];
%data = cell2table(data,'VariableNames',{'Symbol','Price','DateTime'});
DTn = data(:,2);
ti = 1/(60/30 * 24); % Time Interval
DTiv = transpose(DTn{1}:ti:DTn{end}); % Interpolation Vector
Price = data(:,2); % Vector: Column #2 Of Table1
DT30 = interp1({DTn}, Price, {DTiv}); % Interpolated Column #2
NewTable1 = {datestr(DTiv, 'MM/dd/yyyy hh:mm:ss') DT30};
Result = [NewTable1{1} repmat(' ', size(NewTable1{2})) num2str(NewTable1{2}, '%.2f')];
Result5 = Result(1:5,:);
The error:
% Error using interp1 (line 109)
X must be a vector of numeric coordinates.
Error in Interval30minute (line 24) DT30 = interp1({DTn}, Price, {DTiv}); % Interpolated Column #2
Attached is file using this code.

Réponse acceptée

KSSV
KSSV le 15 Juin 2016
%%reset
clear all;
close all;
clc;
%delete NaN, continuously duplicated value and keep the last one
f=fopen('CLF1997.txt');
c=textscan(f , '%s%s%s%f' , 'Headerlines' , 1 , 'delimiter' , ' ');
fclose(f);
t =[diff(c{end})~=0;true];
C = [c{1:3}];
data = [C(t,:),num2cell(c{end}(t))];
clearvars -except data
%combine column date and time
day = data(1:end,2);
time = data(1:end,3);
ns = datenum(day, 'MM/dd/yyyy') + datenum(time, 'hh:mm:ss') - datenum('00:00:00','hh:mm:ss');
data=[data num2cell(ns)];
data(:,2:3)=[];
%data = cell2table(data,'VariableNames',{'Symbol','Price','DateTime'});
DTn = data(:,2);
ti = 1/(60/30 * 24); % Time Interval
DTiv = transpose(DTn{1}:ti:DTn{end}); % Interpolation Vector
Price = data(:,2); % Vector: Column #2 Of Table1
% Convert cell to matrix
DTn = cell2mat(DTn) ;
Price = cell2mat(Price) ;
% Arrange the matrix in order
[DTn,idx] = sort(DTn) ;
Price = Price(idx) ;
% Remove doubles
[DTn1,idx] = unique(DTn) ;
DTn = DTn1 ;
Price = Price(idx) ;
DT30 = interp1(DTn, Price, DTiv); % Interpolated Column #2
NewTable1 = {datestr(DTiv, 'MM/dd/yyyy hh:mm:ss') DT30};
Result = [NewTable1{1} repmat(' ', size(NewTable1{2})) num2str(NewTable1{2}, '%.2f')];
Result5 = Result(1:5,:);
  2 commentaires
Trung Hieu Le
Trung Hieu Le le 17 Juin 2016
Hi Dr. Siva Srinivas Kolukula,
After edit my data, it runs well at that time. Thanks a lots for your help
Trung Hieu Le
Trung Hieu Le le 17 Juin 2016
Hi!
Do you know how I can save the result to .txt or table? I need to divide it to 3 column for next steps. Thanks for your help.

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Plus de réponses (1)

aastha gupta
aastha gupta le 25 Oct 2019
please help me with the code.I am getting an error in linear interpolation.
Error using interp1 (line 109)
X must be a vector of numeric coordinates.
V(1)= f(0)/b(0);
V(N+1) = f(1)/b(1);
W(1) = U(1) - V(1);
W(N+1) = U(N+1)-V(N+1);
for i = 2:N
V(i) = f(x(i))./b(x(i));
W(i) = U(i) - V(i);
end
backdW(2)= (W(2)-W(1))/h;
fordW(2)= (W(3)-W(2))/h;
backdW(N)= (W(N)-W(N-1))/h;
fordW(N)= (W(N+1)-W(N))/h;
secdW(2)= (fordW(2) - backdW(2))/h;
secdW(N)= (fordW(N) - backdW(N))/h;
for i = 3:N-1
backdW(i)= (W(i)-W(i-1))/h ;
fordW(i)= (W(i+1)-W(i))/h ;
secdW(i)= (fordW(i)-backdW(i))/h ;
end
avgsecdW(2)=secdW(2);
avgsecdW(N+1)=secdW(N);
for i=3:N
avgsecdW(i)= (secdW(i)+ secdW(i-1))/2 ;
end
for i=1:N
K = cumsum(h*sqrt(abs(avgsecdW(i+1))));
end
Q(2) = K + sqrt(abs(avgsecdW(2)));
Q(N+1) = K + sqrt(abs(avgsecdW(N+1)));
for i=3:N
Q(i)= K + sqrt(abs(avgsecdW(i)));
end
M(2) = h*Q(2);
for j = 3:N+1
for i = 2:j
M(j)= cumsum(h* Q(i));
end
end
Iteration = 0;
while( max(h*Q(i))/M(N+1)) > (2/N)
Iteration = Iteration + 1;
Y = linspace(0, M(N+1), N+1)';
x(1)=0; x(N+1)=1; % avoid issues with round-off
x = interp1(M, x, Y);
M(1)=0;
end

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