Question

我在加拿大/山区有一个时间序列。数据是每小时。每天我想取0:29:05（Fn）的F值（Fn =当地时间午夜之后的F值），每天我要计算Z =（Fn-F）/ Fn

但是，它需要在Cadadian / Mountain中为0:29:05选择Fn，并且每天每一小时的z都需要使用当天的Fn进行计算。

虚拟数据：

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现在，当使用分钟数据集时，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我的hour-1数据示例。以下是我的尝试...并且我认为在这种情况下，也许变异是正确的选择？

 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)

谢谢。

Answer 1

  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，但您不应命名R对象F，因为它通常是为FALSE值保留的。

使用dplyr从特定时间中选择数据，然后使用该数据来计算

1 个答案: