我想使用基数R(没有任何特定的软件包)来表示以下SQL查询:
select month, day, count(*) as count, avg(dep_delay) as avg_delay
from flights
group by month, day
having count > 1000
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它选择平均起飞延迟和繁忙日(每天有超过1000个航班的天数)的每日航班数。数据集为nycflights13,其中包含2013年从纽约出发的航班信息。
注意,我可以轻松地在dplyr中将其编写为:
flights %>%
group_by(month, day) %>%
summarise(count = n(), avg_delay = mean(dep_delay, na.rm = TRUE)) %>%
filter(count > 1000)
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由于我之前曾被提醒过by(@Parfait的帽子尖)的优雅之处,因此这里是使用by以下解决方案:
res <- by(flights, list(flights$month, flights$day), function(x)
if (nrow(x) > 1000) {
c(
month = unique(x$month),
day = unique(x$day),
count = nrow(x),
avg_delay = mean(x$dep_delay, na.rm = TRUE))
})
# Store in data.frame and order by month, day
df <- do.call(rbind, res);
df <- df[order(df[, 1], df[, 2]) ,];
# month day count avg_delay
#[1,] 7 8 1004 37.296646
#[2,] 7 9 1001 30.711499
#[3,] 7 10 1004 52.860702
#[4,] 7 11 1006 23.609392
#[5,] 7 12 1002 25.096154
#[6,] 7 17 1001 13.670707
#[7,] 7 18 1003 20.626789
#[8,] 7 25 1003 19.674134
#[9,] 7 31 1001 6.280843
#[10,] 8 7 1001 8.680402
#[11,] 8 8 1001 43.349947
#[12,] 8 12 1001 8.308157
#[13,] 11 27 1014 16.697651
#[14,] 12 2 1004 9.021978
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小智 0
这不是一个特别优雅的解决方案,但这将使用 Base R 完成您想要的事情
flights_split <- split(flights, f = list(flights$month, flights$day))
result <- lapply(flights_split, function(x) {
if(nrow(x) > 1000) {
data.frame(month = unique(x$month), day = unique(x$day), avg_delay = mean(x$dep_delay, na.rm = T), count = nrow(x))
} else {
NULL
}
}
)
do.call(rbind, result)
# month day mean_delay n
# 12.2 12 2 9.021978 1004
# 8.7 8 7 8.680402 1001
# 7.8 7 8 37.296646 1004
# 8.8 8 8 43.349947 1001
# 7.9 7 9 30.711499 1001
# 7.10 7 10 52.860702 1004
# 7.11 7 11 23.609392 1006
# 7.12 7 12 25.096154 1002
# 8.12 8 12 8.308157 1001
# 7.17 7 17 13.670707 1001
# 7.18 7 18 20.626789 1003
# 7.25 7 25 19.674134 1003
# 11.27 11 27 16.697651 1014
# 7.31 7 31 6.280843 1001
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