根據其他列向我的資料框添加一列

我希望在我的名為周末的資料框中添加一個新列，如果當天是否在周末，則包含 TRUE 或 FASLE。這可以基于日期或作業日列。我該怎么做？

資料片段

      Substation  Date       cluster  Time     Value  weekday
1     511016      2013-01-17       1 00:00 0.6215941 Thursday
2     511029      2013-01-17       1 00:00 0.5677445 Thursday
3     511030      2013-01-17       1 00:00 0.6065458 Thursday
4     511033      2013-01-08       2 00:00 0.3090885  Tuesday
5     511034      2013-01-17       1 00:00 0.5263230 Thursday
6     511035      2013-01-19       1 00:00 0.5267718 Saturday

uj5u.com熱心網友回復：

像這樣的東西？你data用你的資料框替換的地方

library(tidyverse)
data %>% mutate(weekend = ifelse(weekday %in% c("Monday", "Tuesday", "Wednesday", "Thursday", "Friday"), F, T))

或者

library(tidyverse)
data %>% mutate(weekend = ifelse(weekday %in% c("Saturday", "Sunday"), T, F))

uj5u.com熱心網友回復：

這可以通過?strptime（其中包括%-codes format(..)）通知：

     '%u' Weekday as a decimal number (1-7, Monday is 1).

準備，以防列中沒有Date物件dat$Date：

dat$Date <- as.Date(dat$Date)

代碼：

dat$is_weekend <- format(dat$Date, "%u") %in% 6:7
dat
#   Substation       Date cluster  Time     Value  weekday is_weekend
# 1     511016 2013-01-17       1 00:00 0.6215941 Thursday      FALSE
# 2     511029 2013-01-17       1 00:00 0.5677445 Thursday      FALSE
# 3     511030 2013-01-17       1 00:00 0.6065458 Thursday      FALSE
# 4     511033 2013-01-08       2 00:00 0.3090885  Tuesday      FALSE
# 5     511034 2013-01-17       1 00:00 0.5263230 Thursday      FALSE
# 6     511035 2013-01-19       1 00:00 0.5267718 Saturday       TRUE

資料

dat <- structure(list(Substation = c(511016L, 511029L, 511030L, 511033L, 511034L, 511035L), Date = structure(c(15722, 15722, 15722, 15713, 15722, 15724), class = "Date"), cluster = c(1L, 1L, 1L, 2L, 1L, 1L), Time = c("00:00", "00:00", "00:00", "00:00", "00:00", "00:00"), Value = c(0.6215941, 0.5677445, 0.6065458, 0.3090885, 0.526323, 0.5267718), weekday = c("Thursday", "Thursday", "Thursday", "Tuesday", "Thursday", "Saturday"), is_weekend = c(FALSE, FALSE, FALSE, FALSE, FALSE, TRUE)), row.names = c("1",  "2", "3", "4", "5", "6"), class = "data.frame")

uj5u.com熱心網友回復：

另一種可能的解決方案，基于lubridate::wday：

library(tidyverse)
library(lubridate)

df %>% 
  mutate(weekend = wday(ymd(Date)) %in% c(1,7))

#>   Substation       Date cluster  Time     Value  weekday weekend
#> 1     511016 2013-01-17       1 00:00 0.6215941 Thursday   FALSE
#> 2     511029 2013-01-17       1 00:00 0.5677445 Thursday   FALSE
#> 3     511030 2013-01-17       1 00:00 0.6065458 Thursday   FALSE
#> 4     511033 2013-01-08       2 00:00 0.3090885  Tuesday   FALSE
#> 5     511034 2013-01-17       1 00:00 0.5263230 Thursday   FALSE
#> 6     511035 2013-01-19       1 00:00 0.5267718 Saturday    TRUE

uj5u.com熱心網友回復：

isWeekend這是使用from 的另一種可能的解決方案timeDate：

library(timeDate)

df$is_weekend <- isWeekend(as.Date(df$Date), wday = 1:5)

輸出

  Substation       Date cluster  Time     Value  weekday is_weekend
1     511016 2013-01-17       1 00:00 0.6215941 Thursday      FALSE
2     511029 2013-01-17       1 00:00 0.5677445 Thursday      FALSE
3     511030 2013-01-17       1 00:00 0.6065458 Thursday      FALSE
4     511033 2013-01-08       2 00:00 0.3090885  Tuesday      FALSE
5     511034 2013-01-17       1 00:00 0.5263230 Thursday      FALSE
6     511035 2013-01-19       1 00:00 0.5267718 Saturday       TRUE

資料

df <- structure(list(Substation = c(511016L, 511029L, 511030L, 511033L, 
511034L, 511035L), Date = c("2013-01-17", "2013-01-17", "2013-01-17", 
"2013-01-08", "2013-01-17", "2013-01-19"), cluster = c(1L, 1L, 
1L, 2L, 1L, 1L), Time = c("00:00", "00:00", "00:00", "00:00", 
"00:00", "00:00"), Value = c(0.6215941, 0.5677445, 0.6065458, 
0.3090885, 0.526323, 0.5267718), weekday = c("Thursday", "Thursday", 
"Thursday", "Tuesday", "Thursday", "Saturday"), is_weekend = c(FALSE, 
FALSE, FALSE, FALSE, FALSE, TRUE)), row.names = c(NA, -6L), class = "data.frame")

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