Rich,
Did I miss something? The summarise() command is telling you that you had not
implicitly grouped the data and it made a guess. The canonical way is:
... %>% group_by(year, month, day, hour) %>% summarise(...)
You decide which fields to group by, sometimes including others so they are in
the output.
Avi
-----Original Message-----
From: R-help <r-help-boun...@r-project.org> On Behalf Of Rich Shepard
Sent: Monday, September 13, 2021 4:53 PM
To: r-help@r-project.org
Subject: [R] tidyverse: grouped summaries (with summerize)
I changed the data files so the date-times are in five separate columns:
year, month, day, hour, and minute; for example, year,month,day,hour,min,cfs
2016,03,03,12,00,149000
2016,03,03,12,10,150000
2016,03,03,12,20,151000
2016,03,03,12,30,156000
2016,03,03,12,40,154000
2016,03,03,12,50,150000
2016,03,03,13,00,153000
2016,03,03,13,10,156000
2016,03,03,13,20,154000
The script is based on the example (on page 59 of 'R for Data Science'):
library('tidyverse')
disc <- read.csv('../data/water/disc.dat', header = TRUE, sep = ',',
stringsAsFactors = FALSE) disc$year <- as.integer(disc$year) disc$month <-
as.integer(disc$month) disc$day <- as.integer(disc$day) disc$hour <-
as.integer(disc$hour) disc$min <- as.integer(disc$min) disc$cfs <-
as.double(disc$cfs, length = 6)
# use dplyr to filter() by year, month, day; summarize() to get monthly #
means, sds disc_by_month <- group_by(disc, year, month)
summarize(disc_by_month, vol = mean(cfs, na.rm = TRUE))
but my syntax is off because the results are:
> source('disc.R')
`summarise()` has grouped output by 'year'. You can override using the
`.groups` argument.
Warning messages:
1: In eval(ei, envir) : NAs introduced by coercion
2: In eval(ei, envir) : NAs introduced by coercion
> ls()
[1] "disc" "disc_by_month"
> disc_by_month
# A tibble: 590,940 × 6
# Groups: year, month [66]
year month day hour min cfs
<int> <int> <int> <int> <int> <dbl>
1 2016 3 3 12 0 149000
2 2016 3 3 12 10 150000
3 2016 3 3 12 20 151000
4 2016 3 3 12 30 156000
5 2016 3 3 12 40 154000
6 2016 3 3 12 50 150000
7 2016 3 3 13 0 153000
8 2016 3 3 13 10 156000
9 2016 3 3 13 20 154000
10 2016 3 3 13 30 155000
# … with 590,930 more rows
I have the same results if I use as.numeric rather than as.integer and
as.double. What am I doing incorrectly?
TIA,
Rich
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______________________________________________
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
