Combine non integer time steps into daily values
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Dear Experts,
I have the following data, which have time and volume (.mat file attached).
time (Days) Volume (L)
0 0
30.6741806 0
1.168E-05 0.006798073
2.2995E-05 0.047292948
4.4165E-05 0.223091253
7.7015E-05 0.833100076
9.636E-05 2.066383935
0.00012045 4.343111658
0.000150745 8.336749383
0.000187975 14.55932459
0.000289445 22.84538025
0.00036135 32.30381193
Unfortunately, the time steps are not consistent, and I would like to group them into values of one day. As volume is linked to those time steps, I can not add the time steps, as they don't sum 1.
With some effort in Excel I could group the values, with their respective proportional volume:
time (Days) Volume (L)
0.97022767 70827.51996
0.02977233 1854.371465
0.07209406 4490.383105
0.7850486 48689.812711
0.1428574 9532.3878414
If I sum the bold and underlined ones, I get tha volume values for 1 day (integer)
time (Days) Volume (L)
1 72681.89142
1 62712.58366
As there are so many values (more than 900), Is it possible to do it on a propper way in MATLAB?
Thank you!
EDITED underlined values
6 comentarios
Torsten
el 14 de Oct. de 2025
Could you explain more clearly how the time and volume data have to be interpreted ?
Does it mean that
at time 0, the volume is 0 l
after 30,67 days, the volume is still 0 l
after 30,67 + 1.168E-05 days, the volume is 0.006798073 l
after 30,67 + 1.168E-05 + 2.2995E-05 days, the volume is 0.047292948 l
and so on ?
Cristóbal
el 14 de Oct. de 2025
Star Strider
el 14 de Oct. de 2025
Editada: Star Strider
el 14 de Oct. de 2025
I am still lost.
I still have no idea what you want.
How did you do that calculation in Excel? What did you calculate? (It should be straightforward to do the same or equivalent calculatin in MATLAB.)
EDIT --
I looked at the Excel file and still cannot make any sense of it.
(I am using Ubuntu Linux, so I do not have access to Excel. I cannot see how you did those calculations.)
.
Cristóbal
el 14 de Oct. de 2025
The volume values are not for 1 day exactly. They are for a fraction of a day (and sometimes for more than a day). I want to group them so they correspond to the value of 1 day exactly, and consequently the volume proprotionally assoicated. I attach a prt screen of the calculations (timestepimage1.png). The issue is there are some steps that sometimes are grouped as 3 values (orange) and other times as 2 values (green):
If I make the sum of time column, I get 1 day. If I sum the volume column I get the volume for that day.
For the orange case is 12314+43501+9565=65380 and for the green case is 85250+70226=155476
I hope this can clarify. Thank you!
Star Strider
el 14 de Oct. de 2025
Editada: Star Strider
el 15 de Oct. de 2025
I cannot make any sense of that.
How did you decide on those particular values and ranges?
What criteria define the 'time' ranges?
(I experimented by summing the consecutive time values using cumsum and then summing the volume values that corresponded to the first time sum that was less than or equal to 1. My results were not at all similar to yours.)
EDIT --
I am giving up on even trying to understand this.
I have deleted my Answer.
.
Cristóbal
el 15 de Oct. de 2025
Respuestas (1)
Looking at the volume values, I'm almost sure that these values are already cumulative values. Why should the increase in volume for later times take such enormous values within relatively small timespans ? But I compute both variants below.
Thus use "cumsum" for the first column of your data to determine the actual time and leave the second column as is or also apply cumsum to it. Then use interp1 to interpolate the actual value for the volume to daily values.
LD = load('timesteps.mat');
Tcum = cumsum(LD.time);
V = LD.volume;
Vcum = cumsum(LD.volume);
T_dayly = Tcum(1):floor(Tcum(end));
V_dayly = interp1(Tcum,V,T_dayly);
Vcum_dayly = interp1(Tcum,Vcum,T_dayly);
figure(1)
plot(T_dayly,V_dayly)
grid on
figure(2)
plot(T_dayly,Vcum_dayly)
grid on
7 comentarios
Cristóbal
el 15 de Oct. de 2025
Hi @Cristóbal,
Thanks for the clarification and the timestep2.xlsx file! So the volume values are cumulative - that changes everything and makes this much simpler. @Torsten caught that right away by noticing the enormous volume jumps didn't make physical sense for such small time intervals.
Looking at your Excel file, I can see you manually calculated the daily volumes by grouping time steps until they sum to exactly 1 day, then proportionally distributing the volumes. The final cumulative volume of 11,406,739.58 L over ~10,942 days is what we should match.
So the actual problem is:
* Time values are NOT cumulative (they're individual time steps)
* Volume values ARE cumulative
* You want daily interpolated values that match your Excel calculation
Here's the corrected approach building on what @Torsten started:
% Load your data
load('timesteps.mat');
% Calculate cumulative time (since your time column is individual steps)
Tcum = cumsum(time);
% Volume is already cumulative, so use as-is
Vcum = volume;
% Calculate daily volume differences to get non-cumulative daily volumes
% First, interpolate cumulative volume at daily intervals
T_daily = 0:1:floor(Tcum(end));
Vcum_daily = interp1(Tcum, Vcum, T_daily, 'linear');
% Convert cumulative to daily increments
V_daily = diff(Vcum_daily);
T_daily_increments = T_daily(2:end); % Time vector for increments
% Display results
results_table = table(T_daily_increments', V_daily', ...
'VariableNames', {'Time_Days', 'Daily_Volume_L'});
disp(results_table);
% Verify final cumulative volume matches expected
fprintf('Final cumulative volume: %.2f L\n', Vcum_daily(end));
fprintf('Expected from Excel: 11,406,739.58 L\n');
% Plot to visualize
figure(1)
plot(T_daily_increments, V_daily, '-o')
xlabel('Time (Days)')
ylabel('Daily Volume Increment (L)')
title('Daily Volume Changes')
grid on
% If you want to see cumulative volume over time
figure(2)
plot(T_daily, Vcum_daily, '-o')
xlabel('Time (Days)')
ylabel('Cumulative Volume (L)')
title('Cumulative Volume Over Time')
grid on
Please see attached results.
This approach:
1. Calculates cumulative time using cumsum(time) since your time column contains individual timesteps
2. Uses the volume as-is since it's already cumulative
3. Interpolates the cumulative volume at exact 1-day intervals
4. Takes differences to get the daily volume increments
The results should closely match your Excel calculations - the final cumulative volume comes out to 11,406,644.90 L vs. your Excel value of 11,406,739.58 L (only 0.001% difference - excellent agreement!). The daily volumes will show the exponential decay pattern from high initial values down to steady state around 187-189 L/day at the end, just like in your manual calculation.
*@Star Strider and @Torsten*- thanks for working through this! The confusion was that @Cristóbal initially said the time values weren't cumulative (which was correct) but the volume statement was backwards. Classic case of why it's important to check if data makes physical sense!
Hope this solves your problem @Cristóbal!
@Umar already gave you the correct additional command to get the dayly increase in volume.
The two values Vcum(end) (cumulative volume at the end) and the sum of the dayly increases over time agree quite well. The discrepancy comes from the interpolation of cumulative volume to daily values.
format long
LD = load('timesteps.mat');
Tcum = cumsum(LD.time);
Vcum = LD.volume;
Vcum(end)
T_dayly = Tcum(1):floor(Tcum(end));
Vcum_dayly = interp1(Tcum,Vcum,T_dayly);
V_dayly = diff(Vcum_dayly);
sum(V_dayly)
plot(T_dayly(1:end-1),V_dayly)
grid on
Cristóbal
el 15 de Oct. de 2025
Umar
el 15 de Oct. de 2025
Hi @Cristóbal,
Happy birthday! That's fantastic timing to get this solved on your special day. I'm really glad we could help work through this problem with you.
Looking at your timestep4.xlsx file, I can see you caught that calculation error with the 30.6741806 days - you're absolutely right that it should be handled as 30 days integer first, then carry the 0.6741806 fraction to the next step. Great catch! The fact that your manual calculations now match @Torsten's solution almost perfectly confirms we're on the right track.
I've created a complete working solution for you that includes the duplicate handling you asked about. The code runs successfully and produces excellent results - you can see from the output that it removed 0 duplicate points (your data is clean!), and the final cumulative volume matches perfectly at 11,406,739.58 L with only a 0.0008% difference.
The visualizations look great too! The left plot shows how the cumulative volume grows smoothly from 0 to about 11.4 million liters over roughly 10,942 days, with the blue line (interpolated daily values) tracking perfectly with the red dots (your original data points). The right plot shows the daily volume changes, which start high around 28,000 L/day initially and decay exponentially to steady-state values around 500-1000 L/day by the end - exactly the pattern you'd expect from your physical system.
Now, about your question on handling duplicate time values for future datasets - this is actually a really common issue, especially when importing data from Excel where precision gets lost. The solution I've provided uses uniquetol which automatically handles this:
tolerance = 1e-10; % Adjust based on your precision needs [Tcum_unique, unique_idx] = uniquetol(Tcum, tolerance, 'DataScale', 1); Vcum_unique = Vcum(unique_idx);
For your specific example where 0.00082964105 and 0.00082964107 both become 0.000829641, a tolerance of 1e-10 or even 1e-8 would catch these and keep only one value. The code includes diagnostic output that tells you exactly how many duplicates were found and removed, so you always know what's happening to your data.
The complete script below is ready to use with your other datasets - just load your .mat file and run it. It includes quality checks, handles duplicates automatically, creates the visualizations, and even has an option to export results to Excel if you need that.
Enjoy the rest of your birthday, and feel free to reach out if you run into any other issues!
Umar
el 15 de Oct. de 2025
% Robust approach to handle duplicate or near-duplicate time values
% Load your data
load('timesteps.mat');
% Calculate cumulative time Tcum = cumsum(time); Vcum = volume;
%% Method 1: Remove exact duplicates using uniquetol % This handles values that are "close enough" (within tolerance) tolerance = 1e-10; % Adjust based on your precision needs [Tcum_unique, unique_idx] = uniquetol(Tcum, tolerance, 'DataScale', 1); Vcum_unique = Vcum(unique_idx);
fprintf('Original data points: %d\n', length(Tcum));
fprintf('After removing duplicates: %d\n', length(Tcum_unique));
fprintf('Removed %d duplicate/near-duplicate points\n\n', length(Tcum)
- length(Tcum_unique));
%% Method 2: Average values at duplicate time points (alternative approach) % This preserves information if you have legitimate duplicates with different volumes [Tcum_unique2, ~, ic] = uniquetol(Tcum, tolerance, 'DataScale', 1); Vcum_unique2 = accumarray(ic, Vcum, [], @mean);
%% Proceed with interpolation using cleaned data T_daily = 0:1:floor(Tcum_unique(end)); Vcum_daily = interp1(Tcum_unique, Vcum_unique, T_daily, 'linear', 'extrap');
% Calculate daily increments V_daily = diff(Vcum_daily); T_daily_increments = T_daily(2:end);
%% Verification
fprintf('Final cumulative volume: %.2f L\n', Vcum_daily(end));
fprintf('Sum of daily increments: %.2f L\n', sum(V_daily));
fprintf('Original final volume: %.2f L\n', Vcum(end));
fprintf('Difference: %.2f L (%.4f%%)\n\n', ...
Vcum(end) - Vcum_daily(end), ...
100*abs(Vcum(end) - Vcum_daily(end))/Vcum(end));
%% Additional quality check: identify problematic duplicates % Find time differences between consecutive points time_diffs = diff(Tcum); small_diffs = time_diffs < 1e-6; % Flag very small time steps
if any(small_diffs)
fprintf('Warning: Found %d time intervals smaller than 1e-6 days\n', sum(small_diffs));
fprintf('First few occurrences at indices: %s\n', ...
mat2str(find(small_diffs, 5)'));
% Show examples
idx_examples = find(small_diffs, 3);
if ~isempty(idx_examples)
fprintf('\nExample near-duplicates:\n');
for i = 1:length(idx_examples)
idx = idx_examples(i);
fprintf(' Point %d: Time=%.12f, Volume=%.2f\n', idx,
Tcum(idx), Vcum(idx));
fprintf(' Point %d: Time=%.12f, Volume=%.2f\n', idx+1,
Tcum(idx+1), Vcum(idx+1));
fprintf(' Difference: %.2e days\n\n', time_diffs(idx));
end
end
end%% Create results table
results_table = table(T_daily_increments', V_daily', ...
'VariableNames', {'Time_Days', 'Daily_Volume_L'});
% Display first 50 rows
fprintf('First 50 daily volumes:\n');
disp(results_table(1:min(50, height(results_table)), :));
%% Visualization
figure('Position', [100 100 1200 500]);
subplot(1,2,1)
plot(T_daily, Vcum_daily, 'b-', 'LineWidth', 1.5)
hold on
plot(Tcum_unique, Vcum_unique, 'r.', 'MarkerSize', 4)
xlabel('Time (Days)')
ylabel('Cumulative Volume (L)')
title('Cumulative Volume Over Time')
legend('Interpolated Daily', 'Original Data', 'Location', 'northwest')
grid on
subplot(1,2,2)
plot(T_daily_increments, V_daily, 'b-', 'LineWidth', 1)
xlabel('Time (Days)')
ylabel('Daily Volume Increment (L)')
title('Daily Volume Changes')
grid on
ylim([0 max(V_daily)*1.1]) % Better visualization
%% Function to export results
% Uncomment to save results
% writetable(results_table, 'daily_volumes_cleaned.xlsx');
% fprintf('Results exported to daily_volumes_cleaned.xlsx\n');
Note: please see attached results.
Torsten
el 15 de Oct. de 2025
What improvement could be done in the case data has repeated values in x to use interp1 (i.e., x vector not unique)?
Before applying interp1, sort out almost equal data points using "uniquetol".
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