Question

我有时间格式为m / d / Y H / m / s的数据。由于PosixIt错误，我无法执行mutate功能。请帮忙。

我的数据：

structure(list(time = "9/26/2017 6:19:00", panel = 230, mpp = 2.68627), row.names = c(NA, 
-1L), class = c("tbl_df", "tbl", "data.frame"))

我的代码：

library(dplyr)
library(ggplot2)

data2%>% #filtering 230 and 231
  select(time,panel,mpp)%>%
  filter(data2,panel=="230")
filter(data2,panel=="231")
data_230<-filter(data2,panel=="230")
data_231<-filter(data2,panel=="231")
data_230
data_231

install.packages("lubridate")
library(lubridate)
#data_230$ts <- strptime(data_230$time, "%m/%d/%Y %H:%M:%S")

#data_230$tsf <- ymd_hms(data_230$ts)

library(dplyr)


  #data_230%>%
  data_230$ts <- as.POSIXct(strptime(data_230$time, "%m/%d/%Y %H:%M:%S"))
    data_230$tsf <- ymd_hms(data_230$ts)

    data_230 %>% mutate(ts=strptime(time, "%m/%d/%Y %H:%M:%S"),tsf=ymd_hms(ts)) %>% mutate(datehour= format(time,"%m-%d-%Y %H"), date1= format(time,"%m-%d-%Y"), month=format(time,"%m-%Y")) %>% 
    group_by(datehour) %>% mutate(hourlyP=mean(mpp)) %>% distinct(datehour, .keep_all = TRUE) %>% 
    group_by(date1) %>% mutate(dailyP=sum(hourlyP)) %>% distinct(date1, .keep_all = TRUE) %>% 
    group_by(month) %>% summarise(monthlyP=sum(dailyP))

Answer 1

dplyr函数可以最好地将日期时间数据作为POSIXct对象处理。因此，我建议您将代码修改如下：

data_230 <- filter(data2, panel=="230")

data_230$time <-  as.POSIXct(strptime(
                                      data_230$time,
                                      "%m/%d/%Y %H:%M:%S")
                             ) # convert your character column to POSIXct object
# then the rest of your code:
data_230 %>% 
    mutate(ts = as.POSIXct(strptime(time, "%m/%d/%Y %H:%M:%S"),
           tsf = ymd_hms(ts))) %>%
    mutate(datehour = format(time,"%m-%d-%Y %H"),
           date1 = format(time,"%m-%d-%Y"),
           month = format(time,"%m-%Y")) %>% 
    group_by(datehour) %>% 
    mutate(hourlyP=mean(mpp)) %>%
    distinct(datehour, .keep_all = TRUE) %>% 
    group_by(date1) %>% 
    mutate(dailyP=sum(hourlyP)) %>% 
    distinct(date1, .keep_all = TRUE) %>% 
    group_by(month) %>% 
    summarise(monthlyP=sum(dailyP))

尽管我真的不明白为什么需要ts和tsf，但是代码会运行并产生以下输出：

# A tibble: 1 x 2
  month   monthlyP
  <chr>      <dbl>
1 09-2017     2.69

由于POSIXIt错误，无法执行变异

1 个答案: