search for: hourlydata

...of the hourly data based on the hours within a day... and we wanted to do Anova analysis... We have two choices: Please see below: Why are these two approaches giving very different p-values? And which one shall I use? Thanks a lot! 1. treating the hours as double/floating numbers: anova(lm(hourlydata~as.double(hours_factors))) Df Sum Sq Mean Sq F value Pr(>F) as.double(hours_factors) 1 0.0002 0.00019876 1.3425 0.2466 Residuals 14868 2.2013 0.00014806 2. treating the hours as factors: anova(lm(hourlydata~hours_factors)) Df Sum Sq Mean Sq F value Pr(>F) hours_factors 9 0.00077 8.59...

....167 9999 9999 1.208 9999 9999 1.25 9999 9999.....and so on (the time column is in fractions of a day) I want to be able to group the data by day. I managed to do this using: Day1H = hourlydata[c(1:24),] but I'd like to be able to create groups for each day without doing this manually for each set of 24 rows. Any suggestions greatly appreciated Thanks -- View this message in context: http://n4.nabble.com/Grouping-rows-of-data-by-day-tp2016063p2016063.html Sent from the R help m...