Interpolation Error using interp1 for interval 30 minutes
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Trung Hieu Le
am 15 Jun. 2016
Beantwortet: aastha gupta
am 25 Okt. 2019
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.
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Akzeptierte Antwort
KSSV
am 15 Jun. 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,:);
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aastha gupta
am 25 Okt. 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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