datetime - separar - obtener el dia de una fecha en r

Calcule el número de días de la semana entre 2 fechas en R

Question

Intento escribir una función R para calcular el número de días de la semana entre dos fechas. Por ejemplo, Nweekdays (''01 / 30/2011 '','' 02/04/2011 '') sería igual a 5.

Similar a esta pregunta . ¡Gracias!

/ editar: @J. La respuesta de Winchester es genial, pero me preguntaba si alguien podría pensar en una forma de vectorizar esto, para que funcione en 2 columnas de fechas. ¡Gracias! / edit 2: ¡Gracias de nuevo!

Answer 1

Escribí este, pero la otra respuesta es mejor :)

Nweekdays <- function(a,b)
{
dates <- as.Date(as.Date(a,"%m/%d/%y",origin="1900-01-01"):as.Date(b,"%m/%d/%y",origin="1900-01-01"),origin="1900-01-01")
days <- format(dates,"%w")[c(-1,-length(dates))]
return(sum(!days%in%c(0,6)))
}
Nweekdays(''01/30/2011'',''02/04/2011'')
[1] 3

EDITAR: Calcula cuántos días de la semana se encuentran entre los dos días especificados.

Editar:

Tomando el consejo de J. Winchesters, la función podría simplificarse como:

 Nweekdays <- function(a,b)
{
dates <- as.numeric((as.Date(a,"%m/%d/%y")):(as.Date(b,"%m/%d/%y")))
dates <- dates[- c(1,length(dates))]
return(sum(!dates%%7%in%c(0,6)))
}

Algunos resultados:

> Nweekdays(''01/30/2011'',''02/04/2011'')
[1] 4
>
> Nweekdays(''01/30/2011'',''01/30/2011'')
[1] 0
>
> Nweekdays(''01/30/2011'',''01/25/2011'')
[1] 3

Tenga en cuenta que esto es independiente de la configuración regional. (Sobre ese tema, ¿cómo cambio la configuración regional de todos modos?)

Answer 2

Date1 <- as.Date("2011-01-30")
Date2 <- as.Date("2011-02-04")
sum(!weekdays(seq(Date1, Date2, "days")) %in% c("Saturday", "Sunday"))

EDITAR: Y Zach dijo, vamos a Vectorize :)

Dates1 <- as.Date("2011-01-30") + rep(0, 10)
Dates2 <- as.Date("2011-02-04") + seq(0, 9)
Nweekdays <- Vectorize(function(a, b)
sum(!weekdays(seq(a, b, "days")) %in% c("Saturday", "Sunday")))
Nweekdays(Dates1, Dates2)

Answer 3

Utilizo el siguiente enfoque: primero un ayudante:

weekDays <- function(UPPER = TRUE) {
days <- c(''MONDAY'', ''TUESDAY'', ''WEDNESDAY'',
''THURSDAY'', ''FRIDAY'', ''SATURDAY'',
''SUNDAY'')
if(!UPPER) return(.Internal(tolower(days)))
days
}

... y ahora la función principal:

NumWeekDays <- function(dd, Xdays = c(''saturday'', ''sunday'')) {
# a function to count the number of non-Xdays in a month
# >
# first check if Xdays is of correct format
stopifnot( all(.Internal(tolower(Xdays)) %in% weekDays(UP = FALSE)))
# >
# a helper function to find the number of non-X days between two dates
NonXDays <- function(startDate, endDate, Xdays) {
sum(!(.Internal(tolower(weekdays(seq(startDate, endDate, ''day'')))) %in%
.Internal(tolower(Xdays))))
}
startDate <- as.Date(as.yearmon(index(dd)), frac = 0)
endDate <- as.Date(as.yearmon(index(dd)), frac = 1)
vapply(1:nrow(dd),
FUN = function(i) NonXDays(startDate[i],
endDate[i],
Xdays = c(''saturday'', ''sunday'')),
FUN.VALUE = numeric(1))
}

Ejemplo:

set.seed(1)
dx <- apply.monthly(xts(rnorm(600), order.by = Sys.Date() + 1:600), mean)
R> NumWeekDays(dx)
[1] 23 21 22 23 20 23 22 20 22 22 21 22 23 21 22 22 21 23 21 21

Answer 4

Estas funciones modificadas tienen en cuenta las diferencias de fechas, ya sean positivas o negativas, mientras que la solución aceptada da cuenta de la diferencia de fechas positiva.

library("dplyr")
e2 <- structure(list(date.pr = structure(c(16524, 16524, 16524, 16524, 16524, 16524, 16524, 16524, 16524, 16524, 16545, 5974), class = "Date"),
date.po = structure(c(16524, 16525, 16526, 16527, 16528, 16529, 16530, 16531, 16538, 16545, 16524, 15974), class = "Date")),
.Names = c("date.1", "date.2"), class = c("tbl_df", "data.frame"), row.names = c(NA, -12L))

1. Solución dependiente de la Nweekdays() regional: la función Nweekdays() se adapta de @J. La solución de Won. Funciona para locale = "English_United States.1252"

Nweekdays <- Vectorize(
function(a, b)
{
ifelse(a < b,
return(sum(!weekdays(seq(a, b, "days")) %in% c("Saturday", "Sunday")) - 1),
return(sum(!weekdays(seq(b, a, "days")) %in% c("Saturday", "Sunday")) - 1))
})

a. Configuración regional en inglés

> Sys.setlocale(category="LC_ALL", locale = "English_United States.1252")
[1] "LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252"
> Sys.getlocale()
[1] "LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252"
> e2 %>%
mutate(wkd1 = format(date.1, "%A"),
wkd2 = format(date.2, "%A"),
ndays_with_wkends = ifelse((date.2 > date.1), (date.2 - date.1), (date.1 - date.2)),
ndays_no_wkends = Nweekdays(date.1, date.2))
Source: local data frame [12 x 6]
date.1 date.2 wkd1 wkd2 ndays_with_wkends ndays_no_wkends
(date) (date) (chr) (chr) (dbl) (dbl)
1 2015-03-30 2015-03-30 Monday Monday 0 0
2 2015-03-30 2015-03-31 Monday Tuesday 1 1
3 2015-03-30 2015-04-01 Monday Wednesday 2 2
4 2015-03-30 2015-04-02 Monday Thursday 3 3
5 2015-03-30 2015-04-03 Monday Friday 4 4
6 2015-03-30 2015-04-04 Monday Saturday 5 4
7 2015-03-30 2015-04-05 Monday Sunday 6 4
8 2015-03-30 2015-04-06 Monday Monday 7 5
9 2015-03-30 2015-04-13 Monday Monday 14 10
10 2015-03-30 2015-04-20 Monday Monday 21 15
11 2015-04-20 2015-03-30 Monday Monday 21 15
12 1986-05-11 2013-09-26 Sunday Thursday 10000 7143

segundo. Configuración regional china

> Sys.setlocale(category="LC_ALL", locale = "chinese")
[1] "LC_COLLATE=Chinese (Simplified)_People''s Republic of China.936;LC_CTYPE=Chinese (Simplified)_People''s Republic of China.936;LC_MONETARY=Chinese (Simplified)_People''s Republic of China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People''s Republic of China.936"
> Sys.getlocale()
[1] "LC_COLLATE=Chinese (Simplified)_People''s Republic of China.936;LC_CTYPE=Chinese (Simplified)_People''s Republic of China.936;LC_MONETARY=Chinese (Simplified)_People''s Republic of China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People''s Republic of China.936"
> e2 %>%
mutate(wkd1 = format(date.1, "%A"),
wkd2 = format(date.2, "%A"),
ndays_with_wkends = ifelse((date.2 > date.1), (date.2 - date.1), (date.1 - date.2)),
ndays_no_wkends = Nweekdays(date.1, date.2))
Source: local data frame [12 x 6]
date.1 date.2 wkd1 wkd2 ndays_with_wkends ndays_no_wkends
(date) (date) (chr) (chr) (dbl) (dbl)
1 2015-03-30 2015-03-30 ÐÇÆÚÒ» ÐÇÆÚÒ» 0 0
2 2015-03-30 2015-03-31 ÐÇÆÚÒ» ÐÇÆÚ¶þ 1 1
3 2015-03-30 2015-04-01 ÐÇÆÚÒ» ÐÇÆÚÈý 2 2
4 2015-03-30 2015-04-02 ÐÇÆÚÒ» ÐÇÆÚËÄ 3 3
5 2015-03-30 2015-04-03 ÐÇÆÚÒ» ÐÇÆÚÎå 4 4
6 2015-03-30 2015-04-04 ÐÇÆÚÒ» ÐÇÆÚÁù 5 5
7 2015-03-30 2015-04-05 ÐÇÆÚÒ» ÐÇÆÚÈÕ 6 6
8 2015-03-30 2015-04-06 ÐÇÆÚÒ» ÐÇÆÚÒ» 7 7
9 2015-03-30 2015-04-13 ÐÇÆÚÒ» ÐÇÆÚÒ» 14 14
10 2015-03-30 2015-04-20 ÐÇÆÚÒ» ÐÇÆÚÒ» 21 21
11 2015-04-20 2015-03-30 ÐÇÆÚÒ» ÐÇÆÚÒ» 21 21
12 1986-05-11 2013-09-26 ÐÇÆÚÈÕ ÐÇÆÚËÄ 10000 10000

2. Solución independiente de Nweekdays() regional: la función Nweekdays() se adapta de la solución de @Sacha Epskamp. Funciona para todas las configuraciones regionales, sin embargo @Sacha Epskamp usó c(0,6) para eliminar los fines de semana, que es diferente de esta solución que usa c(2,3) para extraer los fines de semana.

Nweekdays <- Vectorize(
function(a, b) {
return(sum(!(((as.numeric(b:a)) %% 7) %in% c(2,3))) - 1) # 2: Saturday and 3: Sunday
})

a. Configuración regional en inglés

> Sys.setlocale(category="LC_ALL", locale = "English_United States.1252")
[1] "LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252"
> Sys.getlocale()
[1] "LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252"
> e2 %>%
mutate(wkd1 = format(date.1, "%A"),
wkd2 = format(date.2, "%A"),
ndays_with_wkends = ifelse((date.2 > date.1), (date.2 - date.1), (date.1 - date.2)),
ndays_no_wkends = Nweekdays(date.1, date.2))
Source: local data frame [12 x 6]
date.1 date.2 wkd1 wkd2 ndays_with_wkends ndays_no_wkends
(date) (date) (chr) (chr) (dbl) (dbl)
1 2015-03-30 2015-03-30 Monday Monday 0 0
2 2015-03-30 2015-03-31 Monday Tuesday 1 1
3 2015-03-30 2015-04-01 Monday Wednesday 2 2
4 2015-03-30 2015-04-02 Monday Thursday 3 3
5 2015-03-30 2015-04-03 Monday Friday 4 4
6 2015-03-30 2015-04-04 Monday Saturday 5 4
7 2015-03-30 2015-04-05 Monday Sunday 6 4
8 2015-03-30 2015-04-06 Monday Monday 7 5
9 2015-03-30 2015-04-13 Monday Monday 14 10
10 2015-03-30 2015-04-20 Monday Monday 21 15
11 2015-04-20 2015-03-30 Monday Monday 21 15
12 1986-05-11 2013-09-26 Sunday Thursday 10000 7143

segundo. Configuración regional china

> Sys.setlocale(category="LC_ALL", locale = "chinese")
[1] "LC_COLLATE=Chinese (Simplified)_People''s Republic of China.936;LC_CTYPE=Chinese (Simplified)_People''s Republic of China.936;LC_MONETARY=Chinese (Simplified)_People''s Republic of China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People''s Republic of China.936"
> Sys.getlocale()
[1] "LC_COLLATE=Chinese (Simplified)_People''s Republic of China.936;LC_CTYPE=Chinese (Simplified)_People''s Republic of China.936;LC_MONETARY=Chinese (Simplified)_People''s Republic of China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People''s Republic of China.936"
> e2 %>%
mutate(wkd1 = format(date.1, "%A"),
wkd2 = format(date.2, "%A"),
ndays_with_wkends = ifelse((date.2 > date.1), (date.2 - date.1), (date.1 - date.2)),
ndays_no_wkends = Nweekdays(date.1, date.2))
Source: local data frame [12 x 6]
date.1 date.2 wkd1 wkd2 ndays_with_wkends ndays_no_wkends
(date) (date) (chr) (chr) (dbl) (dbl)
1 2015-03-30 2015-03-30 ÐÇÆÚÒ» ÐÇÆÚÒ» 0 0
2 2015-03-30 2015-03-31 ÐÇÆÚÒ» ÐÇÆÚ¶þ 1 1
3 2015-03-30 2015-04-01 ÐÇÆÚÒ» ÐÇÆÚÈý 2 2
4 2015-03-30 2015-04-02 ÐÇÆÚÒ» ÐÇÆÚËÄ 3 3
5 2015-03-30 2015-04-03 ÐÇÆÚÒ» ÐÇÆÚÎå 4 4
6 2015-03-30 2015-04-04 ÐÇÆÚÒ» ÐÇÆÚÁù 5 4
7 2015-03-30 2015-04-05 ÐÇÆÚÒ» ÐÇÆÚÈÕ 6 4
8 2015-03-30 2015-04-06 ÐÇÆÚÒ» ÐÇÆÚÒ» 7 5
9 2015-03-30 2015-04-13 ÐÇÆÚÒ» ÐÇÆÚÒ» 14 10
10 2015-03-30 2015-04-20 ÐÇÆÚÒ» ÐÇÆÚÒ» 21 15
11 2015-04-20 2015-03-30 ÐÇÆÚÒ» ÐÇÆÚÒ» 21 15
12 1986-05-11 2013-09-26 ÐÇÆÚÈÕ ÐÇÆÚËÄ 10000 7143