R: isolation of initial 10% - r

Question

My data frame has a 10 columns and 100,000 rows, each row is an observation and the columns are data pertaining to each observation. One of the columns has the date of an observation in the julian day(ie feb 4= day 34). I want to reduce my data set so I'd have the first 10% observations PER year PER species. Ie, for species 1 in the year 1901 I want the average day of appearance based on the first 10% of observations.
Example of what I have: note id= species but as a number. ie blue=1
date=c(3,84,98,100,34,76,86...)
species=c(blue,purple,grey,purple,green,pink,pink,white...)
id=c(1,2,3,2,4,5,5,6...)
year=c(1901,2000,1901,1996,1901,2000,1986...)
habitat=c(forest,plain,mountain...)
ect
What i want:
date=c(3,84,76,86...)
species=c(purple,pink,pink, white...)
id=c(2,5,5,6...)
year=c(1901,2000,2000,1986...)
habitat=c(forest,plain,mountain...)
new=c(3,84,79,86...)

Answer 1

Assuming the data set dd defined below
set.seed(123)
n <- 100000
dd <- data.frame(year = sample(1901:2000, n, replace = TRUE),
date = sample(0:364, n, replace = TRUE),
species = sample(1:5, n, replace = TRUE))
1) base Aggregate dd with the indicated function. No packages are used:
avg10 <- function(date) {
ok <- seq_along(date) <= length(date) / 10
if (any(ok)) mean(date[ok]) else NA
}
aggregate(date ~ species + year, dd, avg10)
2) data.table Here is a data.table solution:
data.table(dd)[,
{ok <- .I <= .10 * .N; if (any(ok)) mean(date[ok]) else NA}, by = "species,year"]
Note: If you don't want NA's then use this instead of either of the if statements above to get the first point in that case:
if (any(ok)) mean(date[ok]) else date[1]

Answer 2

Just as for your last question, dplyr may work well for you:
Some data:
library(dplyr)
set.seed(42)
n <- 500
dat <- data.frame(date = sample(365, size=n, replace=TRUE),
species = sample(5, size=n, replace=TRUE),
year = 1980 + sample(20, size=n, replace=TRUE))
How it looks without filtering:
dat %>% group_by(year, species) %>% arrange(year, date)
## Source: local data frame [500 x 3]
## Groups: year, species
## date species year
## 1 50 1 1981
## 2 138 1 1981
## 3 174 1 1981
## 4 179 1 1981
## 5 200 1 1981
## 6 332 1 1981
## 7 31 2 1981
## 8 52 2 1981
## 9 196 2 1981
## 10 226 2 1981
## .. ... ... ...
How it looks with the first 10% by date within each year:
dat %>%
group_by(year, species) %>%
filter(ntile(date, 10) == 1) %>%
arrange(year, date)
## Source: local data frame [100 x 3]
## Groups: year, species
## date species year
## 1 50 1 1981
## 2 31 2 1981
## 3 63 3 1981
## 4 112 4 1981
## 5 1 5 1981
## 6 40 1 1982
## 7 103 2 1982
## 8 40 3 1982
## 9 86 4 1982
## 10 48 5 1982
## .. ... ... ...
I think the ntile trick is doing what you want: it breaks the data into roughly equal-sized bins, so it should be giving you the lowest 10% of your dates.
EDIT
Sorry, I missed the mean in there:
dat %>% group_by(year, species) %>%
filter(ntile(date, 10) == 1) %>%
summarise(date = mean(date)) %>%
arrange(year, date)
## Source: local data frame [99 x 3]
## Groups: year
## year species date
## 1 1981 5 1
## 2 1981 2 31
## 3 1981 1 50
## 4 1981 3 63
## 5 1981 4 112
## 6 1982 1 40
## 7 1982 3 40
## 8 1982 5 48
## 9 1982 4 86
## 10 1982 2 103
## .. ... ... ...