Question

如何将以下datetime拆分为year-month-day-hour-minute-second？日期是使用：

创建的

datetime = seq.POSIXt(as.POSIXct("2015-04-01 0:00:00", tz = 'GMT'), 
                                           as.POSIXct("2015-11-30 23:59:59", tz = 'GMT'), 
                                           by="hour",tz="GMT"))

最终目标是将x分辨率的hourly汇总到6-hourly分辨率。可能aggregate datetime可以不分裂它吗？

             datetime     x 
1  2015-04-01 00:00:00    0.0     
2  2015-04-01 01:00:00    0.0     
3  2015-04-01 02:00:00    0.0     
4  2015-04-01 03:00:00    0.0     
5  2015-04-01 04:00:00    0.0     
6  2015-04-01 05:00:00    0.0     
7  2015-04-01 06:00:00    0.0     
8  2015-04-01 07:00:00    0.0     
9  2015-04-01 08:00:00    0.0     
10 2015-04-01 09:00:00    0.0     
11 2015-04-01 10:00:00    0.0     
12 2015-04-01 11:00:00    0.0     
13 2015-04-01 12:00:00    0.0     
14 2015-04-01 13:00:00    0.0     
15 2015-04-01 14:00:00    0.0     
16 2015-04-01 15:00:00    0.0     
17 2015-04-01 16:00:00    0.0     
18 2015-04-01 17:00:00    0.0     
19 2015-04-01 18:00:00    0.0     
20 2015-04-01 19:00:00    0.0     
21 2015-04-01 20:00:00    0.0     
22 2015-04-01 21:00:00    0.0     
23 2015-04-01 22:00:00    1.6     
24 2015-04-01 23:00:00    0.2     
25 2015-04-02 00:00:00    1.5     
26 2015-04-02 01:00:00    1.5     
27 2015-04-02 02:00:00    0.5     
28 2015-04-02 03:00:00    0.0     
29 2015-04-02 04:00:00    0.0     
30 2015-04-02 05:00:00    0.0     
31 2015-04-02 06:00:00    0.0     
32 2015-04-02 07:00:00    0.5     
33 2015-04-02 08:00:00    0.3     
34 2015-04-02 09:00:00    0.0     
35 2015-04-02 10:00:00    0.0     
36 2015-04-02 11:00:00    0.0     
37 2015-04-02 12:00:00    0.0     
38 2015-04-02 13:00:00    0.0     
39 2015-04-02 14:00:00    0.0     
40 2015-04-02 15:00:00    0.0     
41 2015-04-02 16:00:00    0.0     
42 2015-04-02 17:00:00    0.0     
43 2015-04-02 18:00:00    0.0     
44 2015-04-02 19:00:00    0.0     
45 2015-04-02 20:00:00    0.0     
46 2015-04-02 21:00:00    0.0     
47 2015-04-02 22:00:00    0.0     
48 2015-04-02 23:00:00    0.0 
....

输出应非常接近：

YYYY-MM-DD hh:mm:ss  YYYY-MM-DD hh:mm:ss  YYYY-MM-DD hh:mm:ss  YYYY-MM-DD hh:mm:ss
2015-04-01 00:00:00  2015-04-01 06:00:00  2015-04-01 12:00:00  2015-04-01 18:00:00
2015-04-02 00:00:00  2015-04-02 06:00:00  2015-04-02 12:00:00  2015-04-02 18:00:00 
.....

感谢您对此的看法。

修改

如何在列表对象上实现@ r2evans答案，例如：

 x = runif(5856)
    flst1=list(x,x,x,x)

    flst1=lapply(flst1, function(x){x$datetime <- as.POSIXct(x$datetime, tz = "GMT"); x})

    sixhours1=lapply(flst1, function(x) {x$bin <- cut(x$datetime,sixhours);x})

    head(sixhours1[[1]],n=7)

    ret=lapply(sixhours1, function(x) aggregate(x$precip, list(x$bin), sum,na.rm=T))

    head(ret[[1]],n=20)

Answer 1

您的最小数据不完整，因此我会随机生成一些内容：

dat <- data.frame(datetime = seq.POSIXt(as.POSIXct("2015-04-01 0:00:00", tz = "GMT"),
                                        as.POSIXct("2015-11-30 23:59:59", tz = "GMT"), 
                                        by = "hour",tz = "GMT"),
                  x = runif(5856))
# the "1+" ensures we extend at least to the end of the datetimes;
# without it, the last several rows in "bin" would be NA
sixhours <- seq.POSIXt(as.POSIXct("2015-04-01 0:00:00", tz = "GMT"),
                       1 + as.POSIXct("2015-11-30 23:59:59", tz = "GMT"), 
                       by = "6 hours",tz = "GMT")

# this doesn't have to go into the data.frame (could be a separate
# vector), but I'm including it for easy row-wise comparison
dat$bin <- cut(dat$datetime, sixhours)

head(dat, n=7)
#              datetime          x                 bin
# 1 2015-04-01 00:00:00 0.91022534 2015-04-01 00:00:00
# 2 2015-04-01 01:00:00 0.02638850 2015-04-01 00:00:00
# 3 2015-04-01 02:00:00 0.42486354 2015-04-01 00:00:00
# 4 2015-04-01 03:00:00 0.90722845 2015-04-01 00:00:00
# 5 2015-04-01 04:00:00 0.24540085 2015-04-01 00:00:00
# 6 2015-04-01 05:00:00 0.60360906 2015-04-01 00:00:00
# 7 2015-04-01 06:00:00 0.01843313 2015-04-01 06:00:00
tail(dat)
#                 datetime         x                 bin
# 5851 2015-11-30 18:00:00 0.5963204 2015-11-30 18:00:00
# 5852 2015-11-30 19:00:00 0.2503440 2015-11-30 18:00:00
# 5853 2015-11-30 20:00:00 0.9600476 2015-11-30 18:00:00
# 5854 2015-11-30 21:00:00 0.6837394 2015-11-30 18:00:00
# 5855 2015-11-30 22:00:00 0.9093506 2015-11-30 18:00:00
# 5856 2015-11-30 23:00:00 0.9197769 2015-11-30 18:00:00
nrow(dat)
# [1] 5856

工作：

ret <- aggregate(dat$x, list(dat$bin), mean)
nrow(ret)
# [1] 976
head(ret)
#               Group.1         x
# 1 2015-04-01 00:00:00 0.5196193
# 2 2015-04-01 06:00:00 0.4770019
# 3 2015-04-01 12:00:00 0.5359483
# 4 2015-04-01 18:00:00 0.8140603
# 5 2015-04-02 00:00:00 0.4874332
# 6 2015-04-02 06:00:00 0.6139554
tail(ret)
#                 Group.1         x
# 971 2015-11-29 12:00:00 0.6881228
# 972 2015-11-29 18:00:00 0.4791925
# 973 2015-11-30 00:00:00 0.5793872
# 974 2015-11-30 06:00:00 0.4809868
# 975 2015-11-30 12:00:00 0.5157432
# 976 2015-11-30 18:00:00 0.7199298

Answer 2

我使用以下方法获得了解决方案：

library(xts)
flst<- list.files(pattern=".csv")
flst1<- lapply(flst,function(x) read.csv(x,header = TRUE,stringsAsFactors=FALSE,sep = ",",fill=TRUE, 
                                        dec = ".",quote = "\"",colClasses=c('factor', 'numeric', 'NULL'))) # read files ignoring 3 column
head(flst1[[1]])
dat.xts=lapply(flst1, function(x) xts(x$precip,as.POSIXct(x$datetime)))
head(dat.xts[[1]])
ep.xts=lapply(dat.xts, function(x) endpoints(x, on="hours", k=6))#k=by .... see endpoints for "on"
head(ep.xts[[1]])
stations6hrly<-lapply(dat.xts, function(x) period.apply(x, FUN=sum,INDEX=ep))

head(stations6hrly[[703]])
                    [,1]
2015-04-01 05:00:00  0.3
2015-04-01 11:00:00  1.2
2015-04-01 17:00:00  0.0
2015-04-01 23:00:00  0.2
2015-04-02 05:00:00  0.0
2015-04-02 11:00:00  1.4

日期不是我想要的，但价值是正确的。我怀疑R中是否存在-shifttime函数，就像在CDO

中一样

将日期分为YYYY-MM-DD-HH-MM-SS和汇总日期（R）

2 个答案: