Change 15 minute data to daily mean in R - r

Question

I have 15 minute data that I want to change into daily mean. I just listed the Columbia data below, but there are other sites (CR1 and CR2) where I didn't list that data. I put my code at the bottom. I get an error at
x <- xts(d[,-1], as.POSIXct(d[,1], format="%Y-%m-%d %H:%M", tz = "EST"))
Error in as.POSIXct.default(d[, 1], format = "%Y-%m-%d %H:%M", tz = "EST") :
do not know how to convert 'd[, 1]' to class “POSIXct”"
I'm pretty new to R so I'm sorry if the answer is something incredibly simple and I should have caught it.
datetime Discharge Columbia
2014-01-19 22:00 6030 4.3
2014-01-19 22:15 5970 4.28
2014-01-19 22:30 5880 4.25
2014-01-19 22:45 5830 4.23
2014-01-19 23:00 5710 4.19
2014-01-19 23:15 5620 4.16
2014-01-19 23:30 5510 4.12
2014-01-19 23:45 5400 4.08
2014-01-20 00:00 5340 4.06
2014-01-20 00:15 5290 4.04
2014-01-20 00:30 5260 4.03
2014-01-20 00:45 5210 4.01
2014-01-20 01:00 5180 4
2014-01-20 01:15 4990 3.93
2014-01-20 01:30 4830 3.87
2014-01-20 01:45 4810 3.86
2014-01-20 02:00 4780 3.85
2014-01-20 02:15 4780 3.85
2014-01-20 02:30 4760 3.84
2014-01-20 02:45 4760 3.84
2014-01-20 03:00 4760 3.84
2014-01-20 03:15 4760 3.84
USGS_Columbia_Data <- read.csv("~/Desktop/R/USGS_Columbia_Data.csv",header=TRUE)
## daily averages of the data
library(xts)
d <- structure(list(datetime = (USGS_Columbia_Data[1]),
Columbia = (USGS_Columbia_Data[3]),
CR1 = (USGS_Columbia_Data[5]),
CR2 = (USGS_Columbia_Data[7])),
.Names = c("datetime", "Columbia", "CR1", "CR2"),
row.names = c(NA, -3L), class = "data.frame")
x <- xts(d[,-1], as.POSIXct(d[,1], format="%Y-%m-%d %H:%M", tz = "EST"))
apply.daily(x, colMeans)

Answer 1

The other answer works, apparently, but you can (and probably should) use xts for something like this. The problem is with your use of structure(...) to create the data frame. USGS_Columbia_Data is already a data frame. If you want to extract columns 1,3,5, and 7, do this:
d <- USGS_Columbia_Data[,c(1,3,5,7)]
colnames(d) <- c("datetime","Columbia","CR1","CR2"")
You may not need the second line if USGS_Columbia_Data already has those column names. Having done that, you can create a date-indexed xts object as follows:
x <- xts(d[,-1], as.Date(d[,1], format="%Y-%m-%d"))
Then either of the following will work: (note I'm using the d from your example here).
apply.daily(x,mean)
# Discharge Columbia
# 2014-01-19 5743.75 4.201250
# 2014-01-20 4965.00 3.918571
aggregate(x,as.Date,mean)
# Discharge Columbia
# 2014-01-19 5743.75 4.201250
# 2014-01-20 4965.00 3.918571
will work.
If you want to leave the index as POSIXct, use this:
x <- xts(d[,-1], as.POSIXct(d[,1], format="%Y-%m-%d %H:%M"))
apply.daily(x,mean)
# Discharge Columbia
# 2014-01-19 23:45:00 5743.75 4.201250
# 2014-01-20 03:15:00 4965.00 3.918571
But note the index is the last time on each date, not the date itself.

Answer 2

You could use cut and aggregate
# make certain datetime is class POSIXct
d$datetime <- as.POSIXct(d$datetime, tz='EST')
aggregate(list(Discharge = d$Discharge, Columbia = d$Columbia), list(time = cut(d$datetime, "1 day")), mean)
> aggregate(list(Discharge = d$Discharge, Columbia = d$Columbia), list(datetime = cut(t$datetime, "1 day")), mean)
time Discharge Columbia
1 2014-01-19 5743.75 4.201250
2 2014-01-20 4965.00 3.918571