Displaying 1 result from an estimated 1 matches for "655613".
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255613
2010 Aug 03
2
subset based on column names and then subset based on the inverse (grep?, or...)
...;, "06/18/2010", "06/21/2010", "06/22/2010",
"06/22/2010", "06/21/2010", "06/19/2010", "06/19/2010"), elevation_m = c(101,
81, 59, 75, 73, 55, 55, 88, 77, 87), x = c(652159, 651646, 674147,
635466, 665726, 675295, 673098, 658917, 655613, 651748), y = c(3887647,
3886986, 3893724, 3876272, 3893886, 3895529, 3895076, 3882474,
3881587, 3884249), station = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1),
notes_ = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), .Names = c("site",
"base", "creek", "date", "el...