Question

我有两个数据帧（df1，df2），包含大致相同时间段但不同时间戳的一些测量值。 df1具有每小时数据，df2具有每小时2-3次测量的数据。我想：

将df2的小时平均值与df中的每小时值进行比较，即每个数据框每小时一个值
在df2（df2 $ hrly）中创建一个新元素，其值等于df2中每个时间戳的df1的每小时值，即每小时2-3个值（取决于时间的数量 - df2中的那个小时的邮票）

subset，filter并不适用于这种情况 - 我不想使用循环。我正在考虑使用strftime和aggregate - 有更好的方法吗？我正在学习data.table包 - 也许，有更快/更方便的方法？

这是df1和df2的样子：

> glimpse(df1)
Observations: 7,770
Variables: 7
$ lat      <dbl> 30.46198, 30.46198, 30.46198, 30.46198, 30.46198, 30....
$ lon      <dbl> -91.17922, -91.17922, -91.17922, -91.17922, -91.17922...
$ date_gmt <chr> "2016-01-01", "2016-01-01", "2016-01-01", "2016-01-01...
$ time_gmt <chr> "06:00", "07:00", "08:00", "09:00", "10:00", "11:00",...
$ dust     <dbl> 10.7, 8.0, 8.3, 11.1, 9.1, 10.5, 9.7, 13.5, 10.5, 10....
$ state    <chr> "Louisiana", "Louisiana", "Louisiana", "Louisiana", "...
$ tme      <dttm> 2016-01-01 06:00:00, 2016-01-01 07:00:00, 2016-01-01...

df1$tme是POSIxct个对象（tz = "GMT"）

> glimpse(df2)
Observations: 5,000
Variables: 9
$ dp1        <dbl> 0.96, 0.97, 0.98, 0.99, 0.99, 0.99, 0.99, 0.99, 0.9...
$ dp2        <dbl> 1.51, 1.53, 1.55, 1.56, 1.56, 1.56, 1.56, 1.56, 1.5...
$ hz         <dbl> 54.13, 54.55, 54.91, 55.03, 54.98, 55.00, 55.13, 55...
$ rh         <dbl> 68.15, 68.56, 69.84, 68.32, 69.62, 71.14, 70.42, 70...
$ degc       <dbl> 82.88, 82.33, 82.26, 82.62, 82.20, 81.60, 82.05, 81...
$ cfm        <dbl> 3993, 3990, 3989, 3928, 3967, 4045, 4002, 3979, 403...
$ dust       <dbl> 0.02, 0.02, 0.02, 0.02, 0.02, 0.02, 0.02, 0.02, 0.0...
$ time_stamp <dttm> 2016-06-01 17:48:10, 2016-06-01 18:08:12, 2016-06-...
$ dur        <dbl> 0.0000000, 0.3338889, 0.6677778, 1.0013889, 1.33555...

df2$time_stamp是POSIxct对象（tz = "EST"）

Answer 1

由于我没有测试数据，这是我能做的最好的。希望它有效。

我假设您要比较尘埃变量（数据帧中只有常见变量）。我也假设比较意味着你只想看三角洲。

<强>步骤：

确保您的时区相同
将时间戳转换为每小时数据
按小时计算变量/ s的平均值
根据时间戳合并
计算比较的增量

<强> TESTDATA：

library(data.table)
df1<-data.table(tme=seq.POSIXt(as.POSIXct("2016-01-01 00:00",tz="GMT"),by=3600, length.out = 100),dust=rnorm(100))
df2<-data.table(matrix(rnorm(1000*8),1000,8))
setnames(df2, c("dp1","dp2", "hz","rh","degc", "cfm", "dust","dur"))
df2[,time_stamp:=seq.POSIXt(as.POSIXct("2016-01-01 00:00",tz="EST"),by=360, length.out = 1000)]

dplyr::glimpse(df1)
dplyr::glimpse(df2)

<强>代码：

#first snippet
attr(df2$time_stamp,"tzone")<-"GMT" #make same timezone
df2[, tme:=lubridate::round_date(time_stamp, unit = "hours")] #make hourly timestamps
df3<-df2[, mean(dust), by=c("tme")] #group by tme I am assuming you want to compare the only common variable dust
setnames(df3, c("tme","dustmean"))
df_compare<-merge(df1, df3, by="tme", all=T) #this will include all observations from both data.tables
df_compare[,delta_dust:=dust-dustmean] #is that what you want as comparison?
plot(df_compare$delta_dust)

<强>代码2：对于所有带有EST时间和round_date的变量（列）。

#second snippet
attr(df1$tme,"tzone")<-"EST" #make same timezone
df2[, tme:=lubridate::round_date(time_stamp, unit="hours")] #make hourly timestamps
cols2mean<-colnames(df2)
cols2mean<-cols2mean[!(cols2mean %in% c("tme", "time_stamp"))]
df3<-df2[, lapply(.SD, mean), by=c("tme"), .SDcols=cols2mean] #all variables except tme and time_stamp
df_compare<-merge(df1, df3, by="tme", all=T) #this will include all observations from both data.tables
df_compare[,delta_dust:=dust.x-dust.y] #one example
plot(df_compare$delta_dust)

通过R中的时间戳进行算术运算

1 个答案: