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POR: Extracting proper time-date out of csv
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I have a csv with some time series data with date-time formatted like this: <pre><code>mydata <- read.csv("mydata.csv") > mode(mydata$t_5min[1]) [1] "numeric" </code></pre> It looks like R is interpreting it as a factor, since it can't understand the format: <pre><code>mydata$t_5min[1] [1] 1/3/2012 16:00 27698 Levels: 10/10/2012 10:00 10/10/2012 10:05 10/10/2012 10:10 10/10/2012 10:15 ... 9/6/2012 9:55 </code></pre> I tried using strptime, which seems to work okay on single entries: <pre><code>> strptime(x=mydata$t_5min[2],format="%d/%m/%Y %H:%M", tz="") [1] "2012-04-01 06:10:00" > mode(strptime(x=mydata$t_5min[2],format="%d/%m/%Y %H:%M", tz="")) [1] "list" </code></pre> But if I try this with sapply, I get the following error: <pre><code>mydata$t_5min <- sapply(mydata$t_5min, strptime, format="%d/%m/%Y %H:%M", tz="") Error in `$<-.data.frame`(`*tmp*`, "t_5min", value = list(sec = 0, min = 0L, : replacement has 9000 rows, data has 1000 </code></pre> I tried the timeDate library with slightly better results: <pre><code>> as.timeDate(mydata$t_5min[1]) GMT [1] [2012-01-03] </code></pre> However, I need minute precision. However, the example code in the timeDate function doesn't seem to work (or I'm using it wrong, but it was kind of brief): <pre><code>as.timeDate(mydata$t_5min[2], units=c("min")) Error in as.timeDate(mydata$t_5min[2], units = c("min")) : unused argument(s) (units = c("min")) </code></pre> What's the right way to do convert this time data into something R can work with? Here is some data to duplicate these results: <pre><code>t_5min,n,value 1/3/2012 16:00,16,48.125 1/3/2012 16:05,28,44.39285714 1/3/2012 16:10,29,37.44827586 1/3/2012 16:15,28,30.39285714 1/3/2012 16:20,28,23.67857143 1/3/2012 16:25,29,19.10344828 1/3/2012 16:30,28,16.35714286 1/3/2012 16:35,29,14.34482759 1/3/2012 16:40,28,11.71428571 </code></pre>
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<r><csv><time-series><text-parsing>
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