Home/ Questions/Q 9196321
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-06-17T21:49:31+00:00

Hello I have dataframe that is organized like the example below. I have a

  • 0

Hello I have dataframe that is organized like the example below. I have a timestamp, a grouping variable, and several variables with numeric values per timestamp. 

# dput of subset of data
structure(list(TIMESTAMP = structure(1:15, .Label = c("1/1/2012 11:00", 
"1/1/2012 12:00", "1/1/2012 13:00", "1/1/2012 14:00", "1/1/2012 15:00", 
"1/2/2012 11:00", "1/2/2012 12:00", "1/2/2012 13:00", "1/2/2012 14:00", 
"1/2/2012 15:00", "4/7/2012 11:00", "4/7/2012 12:00", "4/7/2012 13:00", 
"4/7/2012 14:00", "4/7/2012 15:00"), class = "factor"), P = c(992.4, 
992.4, 992.4, 992.4, 992.4, 992.4, 992.4, 992.4, 992.4, 992.4, 
239, 239, 239, 239, 239), WS = c(4.023, 3.576, 4.023, 6.259, 
4.47, 3.576, 3.576, 2.682, 4.023, 3.576, 2.682, 3.129, 2.682, 
2.235, 2.682), WD = c(212L, 200L, 215L, 213L, 204L, 304L, 276L, 
273L, 307L, 270L, 54L, 24L, 304L, 320L, 321L), AT = c(16.11, 
18.89, 20, 20, 19.44, 10.56, 11.11, 11.67, 12.22, 11.11, 17.22, 
18.33, 19.44, 20.56, 21.11), FT = c(17.22, 22.22, 22.78, 22.78, 
20, 11.11, 15.56, 17.22, 17.78, 15.56, 24.44, 25.56, 29.44, 30.56, 
29.44), H = c(50L, 38L, 38L, 39L, 48L, 24L, 19L, 18L, 16L, 18L, 
23L, 20L, 18L, 17L, 15L), B = c(1029L, 1027L, 1026L, 1024L, 1023L, 
1026L, 1025L, 1024L, 1023L, 1023L, 1034L, 1033L, 1032L, 1031L, 
1030L), FM = c(14.9, 14.4, 14, 13.7, 13.6, 13.1, 12.8, 12.3, 
12, 11.7, 12.8, 12, 11.4, 10.9, 10.4), GD = c(204L, 220L, 227L, 
222L, 216L, 338L, 311L, 326L, 310L, 273L, 62L, 13L, 312L, 272L, 
281L), MG = c(8.047, 9.835, 10.28, 13.41, 11.18, 9.388, 8.941, 
8.494, 9.835, 10.73, 6.706, 7.153, 8.047, 8.047, 7.6), SR = c(522L, 
603L, 604L, 526L, 248L, 569L, 653L, 671L, 616L, 487L, 972L, 1053L, 
1061L, 1002L, 865L), WS2 = c(2.235, 3.576, 4.47, 4.47, 5.364, 
4.023, 2.682, 3.576, 3.576, 4.023, 3.129, 3.129, 3.576, 2.682, 
3.129), WD2 = c(200L, 201L, 206L, 210L, 211L, 319L, 315L, 311L, 
302L, 290L, 49L, 39L, 15L, 348L, 329L)), .Names = c("TIMESTAMP", 
"P", "WS", "WD", "AT", "FT", "H", "B", "FM", "GD", "MG", "SR", 
"WS2", "WD2"), class = "data.frame", row.names = c(NA, -15L))

I am trying to figure out the best way to deal with timestamps for future manipulations. I have read about lubridate (e.g. here), zoo and POSIXt. However, I feel there might be some r/timestamp trickery that I am unaware of that will make working with timestamps easier (i.e. I might not fully understand timestamps).

Ultimately, I want to do something to the effect of creating a new dataframe that consists of the average of all of these values for some range of date or time. For example, the average value of each variable between 12:00 and 16:00 daily.

Are one of the theses three packages better than the other for performing this sort of task? Could you point me to an example or solution that might do the averaging I wrote about above? Or, are these more suited to figuring out times (e.g. number of hours, days, etc between something [e.g. arrivals and departures]) or can they be used for working with timestamps for other dataframe tasks (like averaging)?

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    I think the process that may help you most is to mutate your [TIMESTAMP] data into a grouping variable. Then, I recommend using one of the many data summary packages to create the report. My personal preference is to use the ‘plyr’ package for both tasks, and I use it in this example.

    Step 1: Use the ‘as.POSIXct’ function to convert your timestamp data to POSIX datetimes for use with the various datetime functions. Use no parameters to simply convert the data without any adjustments.

    data$TIMESTAMP <- as.POSIXct(data$TIMESTAMP)

    Update: Since the time is in not in the unambiguous, decreasing-order format (i.e. YYYY/MM/DD HH:MM:SS), the ‘as.POSIXct’ function won’t be able to do a quick conversion of the data. Use ‘as.POSIXct’ only when you are using the unambiguous format. For other arrangements, use the ‘strptime’ function, specifying the current format like so:

    data$TIMESTAMP <- strptime(data$TIMESTAMP, "%m/%d/%Y %H:%M")

    This tells the ‘strptime’ function what format is currently in use, and exports a POSIX-compatible datetime. There should not be any need to use the ‘as.character’ function, unless your current data is not a string.

    Step 2: Use ‘plyr’ function ‘ddply’ (takes dataframe and returns dataframe) to create a new variable for use in the groupings. Use the ‘format’ function to extract the data you want from the TIMESTAMP values. Look at the ‘format’ documentation for available formats. In this case, here is how you would create a [MONTH] variable:

    library(plyr)
data <- ddply(data, .(TIMESTAMP), mutate, MONTH=format(TIMESTAMP, "%m")

    Step 3: Use ‘plyr’ function ‘ddply’ to summarize the data by your new variable.

    ddply(data, .(MONTH), summarize, V1_AVG=mean(V2), V2_AVG=mean(V2))

    If you wanted to also summarize by a second variable (like [GROUP]), just include that in the second function variable, like so:

    ddply(data, .(MONTH, GROUP), summarize, V1_AVG=mean(V2), V2_AVG=mean(V2))

    Technically, you could do this all in one statement, but experience has taught me caution. I recommend doing each step by itself to make sure nothing gets messed up.

    You can parse your data however you like by fiddling around like this just as long as your timestamps have been converted to POSIX datetimes. The ‘plyr’ package is extremely flexible for stuff like this.

    Update: As per the OP’s request, I am including how you would do the same calculation but using only data between the hours of 12p and 4p. You don’t actually have to use any particular package to subset your data like this, since it’s a straight data filter. Just change the data set inputted into the ‘ddply’ function like so:

    # Use one of the following lines, which both do the same thing.
# I'm just including both as different examples of logic that can be used.
data_Subset <- data[format(data$TIMESTAMP, "%H") >= 12 & format(data$TIMESTAMP, "%H") < 16,]
data_Subset <- data[format(data$TIMESTAMP, "%H") %in% 12:15,]

# Then summarize using the new data frame as an input
ddply(data_Subset, .(MONTH, GROUP), summarize, V1_AVG=mean(V2), V2_AVG=mean(V2))

    Here, we are filtering the data frame to only show rows (with all columns) where the hour (%H) is equal to 12 through 15. This effectively includes all times from 12:00 to 15:59. If you start getting into very large data sets, you may have to look for other solutions (like the ‘data.table’ package), but otherwise, this is your fastest option.

    Again, this only works, because we have transformed our datetimes into POSIX-compatible datetimes.

    • 0