blocks|key|1817155|text|这是一个对向量起作用的切分：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1817156|#'+Seconds+difference+without+weekends
#'
#'+@param+a,+b+POSIXt
#'+@param+weekends+'character',+day+of+the+week+(see
#'+++[base::strptime()]+for+the+"%25w"+argument),+"0"+is+Sunday,+"6"+is
#'+++Saturday;+defaults+to+`c("0","6")`:+Saturday+and+Sunday
#'+@param+units+'character',+legal+values+for+[base::units()],+such+as
#'+++"secs",+"mins",+"hours"
#'+@return+'difftime'+object
#'+@md
secs_no_weekend+<-+function(a,+b,+weekends+=+c("0",+"6"),+units+=+"secs")+{
++out+<-+mapply(function(a0,+b0)+{
++++astart+<-+as.POSIXct(format(a0,+"%25Y-%25m-%25d+00:00:00"))
++++aend+<-+as.POSIXct(format(a0,+"%25Y-%25m-%25d+24:00:00"))
++++bstart+<-+as.POSIXct(format(b0,+"%25Y-%25m-%25d+00:00:00"))
++++days+<-+seq.POSIXt(astart,+bstart,+by+=+"day")
++++ndays+<-+length(days)
++++if+(ndays+==+1)+{
++++++d+<-+b0+-+a0
++++++units(d)+<-+"secs"
++++}+else+{
++++++d+<-+rep(60+*+60+*+24,+ndays)+#+secs
++++++d[1]+<-+`units<-`(aend+-+a0,+"secs")
++++++d[ndays]+<-+`units<-`(b0+-+bstart,+"secs")
++++++wkend+<-+format(days,+"%25w")
++++++d[+wkend+%25in%25+weekends+]+<-+0
++++}
++++sum(pmax(0,+d))
++},+a,+b)
++out+<-+structure(out,+class+=+"difftime",+units+=+units)
++out
}|code-block|syntax|javascript|1817157|1817158|测试/验证：|1817159|1817160|以下是本月(2019年6月)的日历，格式为ISO-8601+(右)和US/not-ISO+(左)：|1817161|week+<-+c("Mon","Tue","Wed","Thu","Fri","Sat","Sun")
#+sunfirst+<-+...+calculated
monfirst+<-+tibble(dt+=+seq(as.Date("2019-06-01"),+as.Date("2019-06-30"),+by="days"))+%25>%25
++mutate(
++++dow+=+factor(format(dt,+format+=+"%25a"),+levels+=+week),
++++dom+=+as.integer(format(dt,+format+=+"%25e")),
++++wom+=+format(dt,+format+=+"%25V")+#+%25U+for+sunfirst,+%25V+for+monfirst
++)+%25>%25
++select(-dt)+%25>%25
++spread(dow,+dom)+%25>%25
++select(-wom)
monfirst+<-+rbind(monfirst,+NA)
cbind(sunfirst,+++`+`="+++++",++++++++monfirst+++++++++++++++++++)
#+++Sun+Mon+Tue+Wed+Thu+Fri+Sat+++++++Mon+Tue+Wed+Thu+Fri+Sat+Sun
#+1++NA++NA++NA++NA++NA++NA+++1++++++++NA++NA++NA++NA++NA+++1+++2
#+2+++2+++3+++4+++5+++6+++7+++8+++++++++3+++4+++5+++6+++7+++8+++9
#+3+++9++10++11++12++13++14++15++++++++10++11++12++13++14++15++16
#+4++16++17++18++19++20++21++22++++++++17++18++19++20++21++22++23
#+5++23++24++25++26++27++28++29++++++++24++25++26++27++28++29++30
#+6++30++NA++NA++NA++NA++NA++NA++++++++NA++NA++NA++NA++NA++NA++NA|1817162|一些数据和预期。(为了简单起见，我在这里使用dplyr，上面的函数不需要它。)|offset|length|style|CODE|1817163|dh+<-++43200+#+day-half,+60*60*12
d1+<-++86400+#+day=1,+60*60*24
d4+<-+345600+#+days=4,+4*d1
d5+<-+432000+#+days=5
d7+<-+432000+#+7+days+minus+weekend
d+<-+tribble(
++~x+++++++++++++++++++,+~y+++++++++++++++++++,+~expect,+~description
,+"2019-06-03+12:00:00",+"2019-06-03+12:00:05",++++++5+,+"same+day"
,+"2019-06-03+12:00:00",+"2019-06-04+12:00:05",+++d1%2B5+,+"next+day"
,+"2019-06-03+12:00:00",+"2019-06-07+12:00:05",+++d4%2B5+,+"4d+%2B+5"
,+"2019-06-03+12:00:00",+"2019-06-08+12:00:05",++d4%2Bdh+,+"start+weekday,+end+weekend,+no+5"
,+"2019-06-03+12:00:00",+"2019-06-09+12:00:05",++d4%2Bdh+,+"start+weekday,+end+weekend%2B,+no+5,+same"
,+"2019-06-03+12:00:00",+"2019-06-10+12:00:05",+++d7%2B5+,+"start/end+weekday,+1+full+week"
,+"2019-06-02+12:00:00",+"2019-06-03+12:00:05",+++dh%2B5+,+"start+weekend,+end+weekday,+1/2+day"
,+"2019-06-02+12:00:00",+"2019-06-08+12:00:05",+++++d7+,+"start/end+weekend,+no+5"
)+%25>%25+mutate_at(vars(x,+y),+as.POSIXct)
(out+<-+secs_no_weekend(d$x,+d$y))
#+Time+differences+in+secs
#+[1]++++++5++86405+345605+388800+388800+432005++43205+432000
all(out+==+d$expect)
#+[1]+TRUE|1817164|entityMap^0|0|0|0|0|0|0|0|M|5|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|11|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|12|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|13|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|14|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|15|8|@$Q|16|R|17|S|T]]|9|@]|A|$]]|$1|U|3|V|5|D|7|18|8|@]|9|@]|A|$E|F]]|$1|W|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|X|$]]

Here's a cut that works on vectors:

<pre class="lang-r prettyprint-override"><code>#' Seconds difference without weekends
#'
#' @param a, b POSIXt
#' @param weekends 'character', day of the week (see
#' [base::strptime()] for the "%w" argument), "0" is Sunday, "6" is
#' Saturday; defaults to `c("0","6")`: Saturday and Sunday
#' @param units 'character', legal values for [base::units()], such as
#' "secs", "mins", "hours"
#' @return 'difftime' object
#' @md
secs_no_weekend &lt;- function(a, b, weekends = c("0", "6"), units = "secs") {
 out &lt;- mapply(function(a0, b0) {
 astart &lt;- as.POSIXct(format(a0, "%Y-%m-%d 00:00:00"))
 aend &lt;- as.POSIXct(format(a0, "%Y-%m-%d 24:00:00"))
 bstart &lt;- as.POSIXct(format(b0, "%Y-%m-%d 00:00:00"))
 days &lt;- seq.POSIXt(astart, bstart, by = "day")
 ndays &lt;- length(days)
 if (ndays == 1) {
 d &lt;- b0 - a0
 units(d) &lt;- "secs"
 } else {
 d &lt;- rep(60 * 60 * 24, ndays) # secs
 d[1] &lt;- `units&lt;-`(aend - a0, "secs")
 d[ndays] &lt;- `units&lt;-`(b0 - bstart, "secs")
 wkend &lt;- format(days, "%w")
 d[ wkend %in% weekends ] &lt;- 0
 }
 sum(pmax(0, d))
 }, a, b)
 out &lt;- structure(out, class = "difftime", units = units)
 out
}
</code></pre>

<hr>

<h2>Testing/validation:</h2>

Perhaps this will be updated as examples come in that do not match my assumptions.

For perspective, here is this month's (June 2019) calendar, in ISO-8601 (right) and US/not-ISO (left):

<pre class="lang-r prettyprint-override"><code>week &lt;- c("Mon","Tue","Wed","Thu","Fri","Sat","Sun")
# sunfirst &lt;- ... calculated
monfirst &lt;- tibble(dt = seq(as.Date("2019-06-01"), as.Date("2019-06-30"), by="days")) %&gt;%
 mutate(
 dow = factor(format(dt, format = "%a"), levels = week),
 dom = as.integer(format(dt, format = "%e")),
 wom = format(dt, format = "%V") # %U for sunfirst, %V for monfirst
 ) %&gt;%
 select(-dt) %&gt;%
 spread(dow, dom) %&gt;%
 select(-wom)
monfirst &lt;- rbind(monfirst, NA)
cbind(sunfirst, ` `=" ", monfirst )
# Sun Mon Tue Wed Thu Fri Sat Mon Tue Wed Thu Fri Sat Sun
# 1 NA NA NA NA NA NA 1 NA NA NA NA NA 1 2
# 2 2 3 4 5 6 7 8 3 4 5 6 7 8 9
# 3 9 10 11 12 13 14 15 10 11 12 13 14 15 16
# 4 16 17 18 19 20 21 22 17 18 19 20 21 22 23
# 5 23 24 25 26 27 28 29 24 25 26 27 28 29 30
# 6 30 NA NA NA NA NA NA NA NA NA NA NA NA NA
</code></pre>

Some data and expectations. (I use <code>dplyr</code> here for simplicity/readability, the function above does not require it.)

<pre class="lang-r prettyprint-override"><code>dh &lt;- 43200 # day-half, 60*60*12
d1 &lt;- 86400 # day=1, 60*60*24
d4 &lt;- 345600 # days=4, 4*d1
d5 &lt;- 432000 # days=5
d7 &lt;- 432000 # 7 days minus weekend
d &lt;- tribble(
 ~x , ~y , ~expect, ~description
, "2019-06-03 12:00:00", "2019-06-03 12:00:05", 5 , "same day"
, "2019-06-03 12:00:00", "2019-06-04 12:00:05", d1+5 , "next day"
, "2019-06-03 12:00:00", "2019-06-07 12:00:05", d4+5 , "4d + 5"
, "2019-06-03 12:00:00", "2019-06-08 12:00:05", d4+dh , "start weekday, end weekend, no 5"
, "2019-06-03 12:00:00", "2019-06-09 12:00:05", d4+dh , "start weekday, end weekend+, no 5, same"
, "2019-06-03 12:00:00", "2019-06-10 12:00:05", d7+5 , "start/end weekday, 1 full week"
, "2019-06-02 12:00:00", "2019-06-03 12:00:05", dh+5 , "start weekend, end weekday, 1/2 day"
, "2019-06-02 12:00:00", "2019-06-08 12:00:05", d7 , "start/end weekend, no 5"
) %&gt;% mutate_at(vars(x, y), as.POSIXct)
(out &lt;- secs_no_weekend(d$x, d$y))
# Time differences in secs
# [1] 5 86405 345605 388800 388800 432005 43205 432000
all(out == d$expect)
# [1] TRUE
</code></pre>

blocks|key|1602827|text|这里有一个使用lubridate和其他tidyverse包的解决方案。lubridate的好处在于，它将非常无缝地处理一系列与时间相关的奇怪问题，从时区到闰年，再到夏令时的转换。(如果您关心这些，只需确保您的数据具有时区。)|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1602828|我在这里使用的概念是lubridate中的intervals+(使用%25--%25操作符创建)。从字面上看，interval就像它听起来那样:一个非常有用的类，基本上有一个开始日期时间和一个结束日期时间。|1602829|我生成了两个数据集:一个用于开始和结束时间，另一个用于周末开始和结束时间，每个数据集都有自己的interval列。在周末数据集中，请注意开始时间和结束时间被任意设置为星期六和星期日的年份。您应该使用对您有意义的值来设置它们，或者从数据中找出一种方法来设置它们。:)|1602830|从那里，我们将使用lubridate的intersect函数找到您的间隔和周末间隔之间的重叠，因此稍后我们可以将相关的周末秒数加起来并减去它们。|1602831|但首先，我们使用来自tidyr的crossing，以确保我们在weekends数据集中每个周末检查您的每个间隔。它只运行两个数据集的笛卡尔乘积(参见this+SO+answer)。|1602832|最后，我们使用int_length来计算周末秒数，对每个间隔的周末秒数求和，对每个间隔的总秒数进行计数，然后从总秒数中减去周末秒数。瞧！我们总共有几秒钟，不包括周末。|1602833|这个解决方案的另一个好处是它非常灵活。我把周末定义为周六0:00到周一0:00+...但你可以删除周五晚上，周一凌晨，任何你喜欢的和满足你的分析要求的时间。|1602834|library(dplyr)
library(tidyr)
library(tibble)
library(lubridate)+#+makes+dates+and+times+easier!

test+<-+tribble(
++++++++++++~start_time,+++++++++++++~end_time,
++"2019-05-22+12:35:42",+"2019-05-23+12:35:42",+#+same+week+no+weekends
++"2019-05-22+12:35:42",+"2019-05-26+12:35:42",+#+ends+during+weekend
++"2019-05-22+12:35:42",+"2019-05-28+12:35:42",+#+next+week+full+weekend
++"2019-05-26+12:35:42",+"2019-05-29+12:35:42",+#+starts+during+weekend
++"2019-05-22+12:35:42",+"2019-06-05+12:35:42"++#+two+weeks+two+weekends
)+%25>%25+
++mutate(
++++id+=+row_number(),
++++timespan+=+start_time+%25--%25+end_time
++)

weekend_beginnings+<-+ymd_hms("2019-05-18+00:00:00")+%2B+weeks(0:51)
weekend_endings+<-+ymd_hms("2019-05-20+00:00:00")+%2B+weeks(0:51)
weekends+<-+weekend_beginnings+%25--%25+weekend_endings

final_answer+<-+crossing(test,+weekends)+%25>%25+
++mutate(
++++weekend_intersection+=+intersect(timespan,+weekends),
++++weekend_seconds+=+int_length(weekend_intersection)
++)+%25>%25+
++group_by(id,+start_time,+end_time,+timespan)+%25>%25+
++summarise(
++++weekend_seconds+=+sum(weekend_seconds,+na.rm+=+TRUE)
++)+%25>%25+
++mutate(
++++total_seconds+=+int_length(timespan),
++++weekday_seconds+=+total_seconds+-+weekend_seconds
++)

glimpse(final_answer)|code-block|syntax|javascript|1602835|entityMap|0|LINK|mutability|MUTABLE|url|https://rawgit.com/rstudio/cheatsheets/master/lubridate.pdf|1|https://stackoverflow.com/questions/43228379/cartesian-product-with-dplyr-r^0|7|9|J|9|Z|9|0|L|9|Y|4|A|9|0|0|0|J|9|0|A|5|G|8|V|8|22|E|1|0|7|A|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1D|8|@$9|1E|A|1F|B|C]|$9|1G|A|1H|B|C]]|D|@$9|1I|A|1J|1|1K]]|E|$]]|$1|H|3|I|5|6|7|1L|8|@]|D|@]|E|$]]|$1|J|3|K|5|6|7|1M|8|@$9|1N|A|1O|B|C]]|D|@]|E|$]]|$1|L|3|M|5|6|7|1P|8|@$9|1Q|A|1R|B|C]|$9|1S|A|1T|B|C]|$9|1U|A|1V|B|C]]|D|@$9|1W|A|1X|1|1Y]]|E|$]]|$1|N|3|O|5|6|7|1Z|8|@$9|20|A|21|B|C]]|D|@]|E|$]]|$1|P|3|Q|5|6|7|22|8|@]|D|@]|E|$]]|$1|R|3|S|5|T|7|23|8|@]|D|@]|E|$U|V]]|$1|W|3|-4|5|6|7|24|8|@]|D|@]|E|$]]]|X|$Y|$5|Z|10|11|E|$12|13]]|14|$5|Z|10|11|E|$12|15]]]]

Here's a solution using <code>lubridate</code> and other <code>tidyverse</code> packages. The nice thing about <code>lubridate</code> is that it will handle pretty seamlessly a host of oddball issues with time, from time zones to leap years to the switch to and from daylight saving time. (If you care about those, just make sure your data has time zones.)

The concept I'm using here is that of <code>intervals</code> in <a href="https://rawgit.com/rstudio/cheatsheets/master/lubridate.pdf" rel="nofollow noreferrer">lubridate</a> (created using the <code>%--%</code> operator). An interval is literally what it sounds like: a very useful class that basically has a start datetime and an end datetime.

I generate two datasets: one for your start and end times, and another for weekend start and end times, each with its own interval column. In the weekend dataset, note that the start and end times are arbitrarily set to a year of Saturdays and Sundays. You should set those with values that make sense for you, or work out a way to set it from the data. :)

From there, we'll find the overlap between your intervals and the weekend intervals using lubridate's <code>intersect</code> function, so later we can count up the relevant weekend seconds and subtract them out.

But first we use <code>crossing</code> from <code>tidyr</code> to make sure we're checking every one of your intervals against every weekend in the <code>weekends</code> dataset. It just runs a Cartesian product of the two datasets (see <a href="https://stackoverflow.com/questions/43228379/cartesian-product-with-dplyr-r">this SO answer</a>).

Finally we use <code>int_length</code> to count up the weekend seconds, sum up the weekend seconds for each of your intervals, count up the total seconds for each, and subtract the weekend seconds from the total seconds. And voila! We have total seconds, excluding weekends.

The other nice thing about this solution is that it's extremely flexible. I've defined weekends as 0:00 Saturday to 0:00 Monday... but you could remove Friday evenings, Monday wee hours, whatever strikes your fancy and meets your analytical requirements.

<pre class="lang-r prettyprint-override"><code>library(dplyr)
library(tidyr)
library(tibble)
library(lubridate) # makes dates and times easier!

test &lt;- tribble(
 ~start_time, ~end_time,
 "2019-05-22 12:35:42", "2019-05-23 12:35:42", # same week no weekends
 "2019-05-22 12:35:42", "2019-05-26 12:35:42", # ends during weekend
 "2019-05-22 12:35:42", "2019-05-28 12:35:42", # next week full weekend
 "2019-05-26 12:35:42", "2019-05-29 12:35:42", # starts during weekend
 "2019-05-22 12:35:42", "2019-06-05 12:35:42" # two weeks two weekends
) %&gt;% 
 mutate(
 id = row_number(),
 timespan = start_time %--% end_time
 )

weekend_beginnings &lt;- ymd_hms("2019-05-18 00:00:00") + weeks(0:51)
weekend_endings &lt;- ymd_hms("2019-05-20 00:00:00") + weeks(0:51)
weekends &lt;- weekend_beginnings %--% weekend_endings

final_answer &lt;- crossing(test, weekends) %&gt;% 
 mutate(
 weekend_intersection = intersect(timespan, weekends),
 weekend_seconds = int_length(weekend_intersection)
 ) %&gt;% 
 group_by(id, start_time, end_time, timespan) %&gt;% 
 summarise(
 weekend_seconds = sum(weekend_seconds, na.rm = TRUE)
 ) %&gt;% 
 mutate(
 total_seconds = int_length(timespan),
 weekday_seconds = total_seconds - weekend_seconds
 )

glimpse(final_answer)
</code></pre>

If I have a dataframe with 2 columns which are YMD HMS, how do I calculate the difference in seconds between the two excluding weekends?

col 2 - col 1 = time in seconds; need to exclude the weekend seconds

<pre><code>Dates1 &lt;- as.POSIXct("2011-01-30 12:00:00") + rep(0, 10)
Dates2 &lt;- as.POSIXct("2011-02-04") + seq(0, 9, 1)
df &lt;- data.frame(Dates1 = Dates1, Dates2 = Dates2)
</code></pre>

I need it to give me (388800 - 43200) = 345600; The reason why I am subtracting 43200 is because that is a Sunday weekend time from Noon until Midnight which the clock stops.

Calculate seconds between 2 timestamps in R excluding weekends

如果我有一个包含两列的dataframe，这两列都是YMD HMS，我如何计算这两列之间的差值(不包括周末)？COL2-COL1=以秒为单位的时间；需要排除周末秒Dates1 <- as.POSIXct("2011-01-30 12:00:00") + rep(0, 10)Dates2 <- as.POSIXct("2...

问计算R中2个时间戳之间的秒数，不包括周末
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问计算R中2个时间戳之间的秒数，不包括周末
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