如何按小时数平均一个时间段? [英] How to averaging over a time period by hours?
问题描述
我是R的新手，并且体验了我的第一个困难.我有一个大约10000 obs的数据集.捕获事件发生的365天之内仅在每个月的前14天标记出这种情况.我想通过对相应月份的以前出现次数(按小时)进行平均来补充另外的16天.
Im new to R and expirience my first difficulties. I have a data set of ca.10000 obs. of 365 days where I capture occurences of an event. This occurrences are marked out only for the first 14 days of each month. I would like to complement the additional 16 days by averaging over the previous occurrences of the corresponding month(by hour).
结构如下:
day hours occurrence
2000-01-01 1 5
2000-01-01 2 6
2000-01-01 3 7
... ... ...
2000-01-01 23 3
2000-01-01 24 2
... ... ...
2000-01-02 1 4
2000-01-02 2 2
2000-01-02 3 5
... ... ...
2000-01-02 23 2
2000-01-02 24 1
...
...
2000-01-15 1 average of the previous 1 hours((5+4+n)/2*k))
2000-01-15 2 average of the previous 2 hours ((6+2+n)/2*k))
2000-01-15 3 average of the previous 3 hours((7+5+n)/2*k))
... ... ...
2000-01-15 23 average of the previous 23 hours
2000-01-15 24 average of the previous 24 hours
... ... ...
... ... ...
2000-01-30
2000-01-30
2000-01-30
2000-01-30
... ... ...
... ... ...
2000-02-01
2000-02-01
2000-02-01
2000-02-01
... ... ...
...
... ... ...
2000-12-24
我尝试了
aggregate( occurences ~ hours, mean)
但是结果没有意义，我尝试了
but the results were pointless and I tried
tapply( X = occurences, INDEX = list(hours), FUN = Mean )
不幸的是，两个都没有像我想象的那样工作.我认为有必要在功能中包含相应的月份.但是我的能力似乎有限.
Unfortunately both didnt work as I imagined. I think its necessary to include the corresponding month into the function. However my means seems to be limited.
推荐答案
您可以尝试一下.请注意，为了使示例更小，我仅选择每月1-4天和0-1小时的数据.第一天每个月的第2天都有发生的数据，第2天和第2天3个缺少发生的数据.
You may try this. Please note that in order to make the example smaller, I select data only for day 1-4 and hour 0-1 each month. Day 1 & 2 in each month have data on occurrence, and day 2 & 3 are missing data for occurrence.
library(dplyr)
# create dummy data
set.seed(123) # for reproducibility of sample
d1 <- data.frame(time = seq(from = as.POSIXct("2000-01-01"),
to = as.POSIXct("2000-02-28"),
by = "hour"))
d1 <- d1 %>%
mutate(hour = as.integer(format(time, "%H")),
day = as.integer(format(time, "%d")), # <~~ only needed to generate sample data
month = as.integer(format(time, "%m")),
occurence = sample(1:10, length(time), replace = TRUE),
occurence = ifelse(day %in% 1:2, occurence, NA)) %>% # <~~~ data only for day 1-2
filter(hour %in% 0:1 & day %in% 1:4) %>% # <~~~ smaller example: select hour 0-1, day 1-4
select(-day)
# calculate mean occurrence per month and hour
d2 <- d1 %>%
group_by(month, hour) %>%
summarise(mean_occ = round(mean(occurence, na.rm = TRUE), 1))
d2
# month hour mean_occ
# 1 1 0 5.0
# 2 1 1 8.0
# 3 2 0 5.5
# 4 2 1 6.5
# replace missing occurrence with mean_occ
d3 <- d1 %>%
left_join(d2, by = c("hour", "month")) %>%
mutate(occurence2 = ifelse(is.na(occurence), mean_occ, occurence)) %>%
select(-month, -mean_occ)
d3
# hour time occurence occurence2
# 1 0 2000-01-01 00:00:00 3 3.0
# 2 1 2000-01-01 01:00:00 8 8.0
# 3 0 2000-01-02 00:00:00 7 7.0
# 4 1 2000-01-02 01:00:00 8 8.0
# 5 0 2000-01-03 00:00:00 NA 5.0
# 6 1 2000-01-03 01:00:00 NA 8.0
# 7 0 2000-01-04 00:00:00 NA 5.0
# 8 1 2000-01-04 01:00:00 NA 8.0
# 9 0 2000-02-01 00:00:00 4 4.0
# 10 1 2000-02-01 01:00:00 6 6.0
# 11 0 2000-02-02 00:00:00 7 7.0
# 12 1 2000-02-02 01:00:00 7 7.0
# 13 0 2000-02-03 00:00:00 NA 5.5
# 14 1 2000-02-03 01:00:00 NA 6.5
# 15 0 2000-02-04 00:00:00 NA 5.5
# 16 1 2000-02-04 01:00:00 NA 6.5
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