Averaging columns in table using only rows where a condition is met.

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Joshua Himmelstein
Joshua Himmelstein 2022-4-21
评论： Campion Loong 2022-5-18
I would like to find the mean (average) for values in columns 3, 4, 5, 6 using only rows where OOBS.night == 1. This average will be different than the entire column average, as I only want to include a subset of the column values.
I am open to suggestions - perhaps I am going about this wrong!
Thank you.
OOBS = table([hobo_times.OOBS23],[hobo_times.water_elevation_m_NAVD88],[tu_values_23'],[tu_values_27'],[tu_values_29'],[tu_values_35']);
OOBS.Properties.VariableNames = {'times','WtrLvlm','OOBS23','OOBS27','OOBS29','OOBS35'};
OOBS.night = (hour(OOBS.times)>=20 | hour(OOBS.times)<=6);
find(OOBS.night ==1 & OOBS.WtrLvlm>=0.33);
  1 个评论
Campion Loong
Campion Loong 2022-5-18
Ran in:
This is really just 1-line of code using groupsummary. The first few lines are just making up fake data, since there is no example data attached:
% Make up some data
Times = (datetime(2021,10,1):minutes(10):datetime(2021,10,31))';
WtrLvlm = rand(length(Times),1);
OOBS23 = rand(length(Times),1);
OOBS27 = rand(length(Times),1);
OOBS29 = rand(length(Times),1);
OOBS35 = rand(length(Times),1);
night = timeofday(Times) > hours(18); % assume 'night' means later than 6PM
tt = timetable(Times, WtrLvlm, OOBS23, OOBS27, OOBS29, OOBS35, night)
tt = 4321×6 timetable
Times WtrLvlm OOBS23 OOBS27 OOBS29 OOBS35 night ____________________ _______ _________ ________ ________ _______ _____ 01-Oct-2021 00:00:00 0.38644 0.89286 0.20368 0.93066 0.49296 false 01-Oct-2021 00:10:00 0.65379 0.79538 0.58104 0.63038 0.50282 false 01-Oct-2021 00:20:00 0.30412 0.46776 0.51774 0.9148 0.1635 false 01-Oct-2021 00:30:00 0.23873 0.88916 0.35675 0.69045 0.46307 false 01-Oct-2021 00:40:00 0.21866 0.76901 0.97638 0.64259 0.728 false 01-Oct-2021 00:50:00 0.21378 0.66606 0.51613 0.84456 0.10005 false 01-Oct-2021 01:00:00 0.25439 0.54039 0.049845 0.41393 0.63819 false 01-Oct-2021 01:10:00 0.16109 0.33963 0.97461 0.3907 0.80241 false 01-Oct-2021 01:20:00 0.80712 0.74499 0.81297 0.86231 0.92233 false 01-Oct-2021 01:30:00 0.78628 0.0003465 0.50948 0.4305 0.22823 false 01-Oct-2021 01:40:00 0.28834 0.55194 0.73841 0.29849 0.57742 false 01-Oct-2021 01:50:00 0.69068 0.77565 0.95863 0.61898 0.38631 false 01-Oct-2021 02:00:00 0.22047 0.10725 0.88691 0.5075 0.41322 false 01-Oct-2021 02:10:00 0.57748 0.55464 0.56385 0.18775 0.40718 false 01-Oct-2021 02:20:00 0.4007 0.48923 0.85602 0.56587 0.25958 false 01-Oct-2021 02:30:00 0.61546 0.46406 0.7246 0.009929 0.79512 false
% This 1-liner is what you are actually after
NightAvg = groupsummary(tt,'night','mean')
NightAvg = 2×7 table
night GroupCount mean_WtrLvlm mean_OOBS23 mean_OOBS27 mean_OOBS29 mean_OOBS35 _____ __________ ____________ ___________ ___________ ___________ ___________ false 3271 0.50808 0.49655 0.49466 0.49572 0.49936 true 1050 0.49481 0.50501 0.51259 0.50066 0.51605
% Now it's a slightly different 1-liner if you want to group
% by both 'night' and a 'WtrLvlm' threshold
% (like in your code example, but unlike your descriptions)
groupsummary(tt,["night" "WtrLvlm"],{'none', [0 0.33 Inf]}, "mean")
ans = 4×7 table
night disc_WtrLvlm GroupCount mean_OOBS23 mean_OOBS27 mean_OOBS29 mean_OOBS35 _____ ____________ __________ ___________ ___________ ___________ ___________ false [0, 0.33) 1033 0.49008 0.4985 0.50889 0.49185 false [0.33, Inf] 2238 0.49954 0.49289 0.48964 0.50283 true [0, 0.33) 348 0.52147 0.50503 0.49881 0.50274 true [0.33, Inf] 702 0.49685 0.51634 0.50158 0.52264

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采纳的回答

David Hill
David Hill 2022-4-21
n_idx=hour(hobo_times.OOBS23)>=20 | hour(hobo_times.OOBS23)<=6;
m=mean([tu_values_23'(n_idx);tu_values_27'(n_idx);tu_values_29'(n_idx);tu_values_35'(n_idx)]);
  1 个评论
Joshua Himmelstein
Joshua Himmelstein 2022-4-21
Thanks for the help! Adjusted it slightly as it wasn't working right off the bat!
n_idx = (find(OOBS.night ==1 & OOBS.WtrLvlm>=0.33))';
mean_turbidity = mean([OOBS.OOBS23(n_idx),OOBS.OOBS27(n_idx),OOBS.OOBS29(n_idx),OOBS.OOBS35(n_idx)]);

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