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使用dplyr从特定小时中选择数据并使用该数据计算

lubridate dplyr r

Lmm · 技术社区 · 6 年前

我在加拿大有一个时间段/山地时间。数据是每小时。每天我想取0:29:05(fn)的f值(fn=当地时间午夜后的f值),每天我想计算z=(fn-f)/fn

但是,需要在Cadadian/Mountain中选择0:29:05的fn,每天每小时z需要使用当天的fn计算。

虚拟数据:

 datetime <- seq(
 from=as.POSIXct("2012-1-1 0:29:05", tz="Canada/Mountain"),
 to=as.POSIXct("2012-2-1 0:29:05", tz="Canada/Mountain"),
 by="hour")

 #variable F
 F <- runif(745, min = 0, max =2)

df <- as.data.frame(cbind(datetime,F))
library(lubridate)
#make sure its in  "POSIXct" "POSIXt" format
df$datetime <- as_datetime(df$datetime)

现在,在使用分钟数据集时,我已经在dplyr中获得了一些帮助,但显然我的理解仍然很差,因为我无法将其转换为小时1数据示例。下面是我的尝试…我想在这种情况下变异是正确的选择?

df2 <- df %>%
group_by(Date = as.Date(datetime)) %>%
mutate(Fn = F[hour(datetime) == 0]), 
z = (Fn - F)/Fn) %>%
ungroup() %>%
select(-Date)

谢谢。

1 回复 | 直到 6 年前

parkerchad81 6 年前

  library(lubridate)
  library(tidyverse)

datetime <- seq(
   from = as.POSIXct("2012-1-1 0:29:05", tz = "Canada/Mountain"),
   to = as.POSIXct("2012-2-1 0:29:05", tz = "Canada/Mountain"),
   by = "hour"
   )

f <- runif(745, min = 0, max =2) #variable f 
df <- data.frame(datetime, f)

# method using fill function from tidyr package
df2 <- df %>%
   mutate(Date = as.Date(datetime, tz = "Canada/Mountain")) %>% 
   left_join( #this will grab the f value at 0:29:05 of each day
     df %>% filter(hour(datetime) == 0) %>% select(datetime, Fn = f),
     by = 'datetime'
   ) %>% 
   group_by(Date) %>% 
   fill(Fn, Fn, .direction = 'down') %>% #fills in NA values with f values of the following day
   mutate(
     Z = ( Fn - f ) / Fn
   ) %>% 
   ungroup() %>% 
   select(-Date)

# method not using fill
df3 <- df %>%
   mutate(Date = as.Date(datetime, tz = "Canada/Mountain")) %>% 
   left_join( #this will grab the f value at 0:29:05 of each day
     df %>% filter(hour(datetime) == 0) %>% select(datetime, Fn = f),
     by = 'datetime'
   ) %>% 
   group_by(Date) %>% 
   mutate(
     Fn = na.omit(Fn),
     Z = ( Fn - f ) / Fn
   ) %>% 
   ungroup() %>% 
   select(-Date)

# both methods result in the same result, as shown below
# A tibble: 745 x 4
   datetime                f    Fn       Z
   <dttm>              <dbl> <dbl>   <dbl>
 1 2012-01-01 00:29:05 0.590 0.590  0     
 2 2012-01-01 01:29:05 1.57  0.590 -1.66  
 3 2012-01-01 02:29:05 0.537 0.590  0.0900
 4 2012-01-01 03:29:05 0.691 0.590 -0.171 
 5 2012-01-01 04:29:05 0.719 0.590 -0.218 
 6 2012-01-01 05:29:05 0.811 0.590 -0.374 
 7 2012-01-01 06:29:05 0.248 0.590  0.581 
 8 2012-01-01 07:29:05 1.98  0.590 -2.35  
 9 2012-01-01 08:29:05 0.839 0.590 -0.422 
10 2012-01-01 09:29:05 0.733 0.590 -0.242 
# ... with 735 more rows

只是一个想法,但你不应该说出R对象的名字 F 因为它通常是为 FALSE 价值观。