如何计算不同时间间隔内有多少个孩子？

如何解决如何计算不同时间间隔内有多少个孩子？

我有一个数据集，其中包含幼儿园的孩子们什么时候送达和接送的信息。我想计算一下30分钟内有多少个孩子。因此，即有7.30-7.59 AM或16.00-16.29 PM。

在我的数据集中，有18个孩子，但这只是其中的一小部分。 问题是我在kommet_antal和afhentet_antal列表中的每个条目上都得到18的总和。

在底部，我写了一些有效的代码，但实际上并不是那么漂亮！

 [1] "08:09:00" "09:00:00" "07:37:00" "07:51:00"
 lst_afhentet
 [1] "15:38:00" "15:19:00" "15:56:00" "14:24:00"

W1M_antal <- for(i in 1:nrow(df_W1M)){      #W1m is the dataframe with week1 and mondays selected
  lst_kommet <- df_W1M$Kommet
  lst_afhentet <- df_W1M$Afhentet
  
  kommet_antal <- vector("list",21)      #21 timeintervals,open from 6.30 to 17.00
  afhentet_antal <- vector("list",21)
  
                            
  tid<- as.ITime("07:00")                  #initial start time,T1
  n <- 0
  for(k in 1:length(lst_kommet)){         #runs over when children is delivered
    if(k < tid){                          #If the time delivered is before time (tid),then count it
      n <- n + 1  
    } else n <- n
    for(t in 1:length(kommet_antal)){     #want to save number of kids delivered in the different intervals
      kommet_antal[t] <- n
    }
    
    tid = tid + as.ITime("00:30")     #add 30 min to the time so we have next time interval
  }
  
  tid <- as.ITime("07:00")            #Do the same for pick up
  m <- 0
  for(a in 1:length(lst_afhentet)){
    if(a < tid){
      m <- m + 1
    } else m <- m
    for(l in 1:length(afhentet_antal)){   #Save number of kids in intervals until they are picked up
      afhentet_antal[l] <- m
    }
    tid <- tid + as.ITime("00:30")
  }
}
tid
total_antal <- vector("list",21)
total_antal <- as.numeric(kommet_antal) - as.numeric(afhentet_antal) 
total_antal



This code works,and give me the correct number,but with datasets from a year with 5 days per week it is going to take a long time to count number of kids present.



T1 <- count(subset(Mandag,Kommet < "07:00")) - count(subset(Mandag,Afhentet <"07:00"))
T2 <- count(subset(Mandag,Kommet < "07:30")) - count(subset(Mandag,Afhentet <"07:30"))
T3 <- count(subset(Mandag,Kommet < "08:00")) - count(subset(Mandag,Afhentet <"08:00"))
T4 <- count(subset(Mandag,Kommet < "08:30")) - count(subset(Mandag,Afhentet <"08:30"))
T5 <- count(subset(Mandag,Kommet < "09:00")) - count(subset(Mandag,Afhentet <"09:00"))
T6 <- count(subset(Mandag,Kommet < "09:30")) - count(subset(Mandag,Afhentet <"09:30"))
T7 <- count(subset(Mandag,Kommet < "10:00")) - count(subset(Mandag,Afhentet <"10:00"))
T8 <- count(subset(Mandag,Kommet < "10:30")) - count(subset(Mandag,Afhentet <"10:30"))
T9 <- count(subset(Mandag,Kommet < "11:00")) - count(subset(Mandag,Afhentet <"11:00"))
T10 <- count(subset(Mandag,Kommet < "11:30")) - count(subset(Mandag,Afhentet <"11:30"))
T11 <- count(subset(Mandag,Kommet < "12:00")) - count(subset(Mandag,Afhentet <"12:00"))
T12 <- count(subset(Mandag,Kommet < "12:30")) - count(subset(Mandag,Afhentet <"12:30"))
T13 <- count(subset(Mandag,Kommet < "13:00")) - count(subset(Mandag,Afhentet <"13:00"))
T14 <- count(subset(Mandag,Kommet < "13:30")) - count(subset(Mandag,Afhentet <"13:30"))
T15 <- count(subset(Mandag,Kommet < "14:00")) - count(subset(Mandag,Afhentet <"14:00"))
T16 <- count(subset(Mandag,Kommet < "14:30")) - count(subset(Mandag,Afhentet <"14:30"))
T17 <- count(subset(Mandag,Kommet < "15:00")) - count(subset(Mandag,Afhentet <"15:00"))
T18 <- count(subset(Mandag,Kommet < "15:30")) - count(subset(Mandag,Afhentet <"15:30"))
T19 <- count(subset(Mandag,Kommet < "16:00")) - count(subset(Mandag,Afhentet <"16:00"))
T20 <- count(subset(Mandag,Kommet < "16:30")) - count(subset(Mandag,Afhentet <"16:30"))
T21 <- count(subset(Mandag,Kommet < "17:00")) - count(subset(Mandag,Afhentet <"17:00"))

#Laver output i dataframe

W <- c(rep("Week1",22*5),rep("Week2",rep("Week3",rep("Week4",22*5))
D <- c(rep("Monday",22*4),rep("Tuesday",rep("Wednesday",rep("Thursday",rep("Friday",22*4))
Time <- c(rep(1:22,20))
Value1 <- c(T1,T2,T3,T4,T5,T6,T7,T8,T9,T10,T11,T12,T13,T14,T15,T16,T17,T18,T19,T20,T21,rep(0,419))
Value <- do.call(rbind,Value1)

Output <- data.frame(W,D,Time,Value)
View(Output)  ```        

解决方法

lubridate软件包提供了一些有用的功能来解决此类问题。

您没有可复制的示例，因此我的帮助必须是通用的，您可以将原则应用于您的情况。

设置上下文：日期和时间在编程和R中也非常复杂。您正在检查时间是否落在两个端点之间，以进行接送。函数%within%负责这种类型的操作。一个示例是time_check %within% dropoff_pickup_intervals。当时间检查假设07:30介于学生的接送时间之间时，它将返回TRUE。

但是我们首先需要正确的日期和间隔格式。这是一些示例代码：（请注意，最初的大多数代码都是为了帮助创建示例，功能的实​​质是较低的）

library(lubridate)

# Make example data
# ----
# set seed for repeatable results for random processes
set.seed(20201023)

# set a time for start and ending for dropoffs and pickups using "ymd_hm" lubridate function
daystart <- ymd_hm("2020-01-01 07:00")
dayend <- ymd_hm("2020-01-01 16:00")

# Create a sequence of dates 
time_set <- seq(daystart,dayend,by="min")

# data frame sampling times from 
kinder <- data.frame(student_id=1:10,dropoff=sample(time_set,10),pickup=sample(time_set,10))

# remove dates that don't make sense
kinder <- kinder %>% filter(pickup > dropoff)
# --- Example data complete

# create time intervals for student arrival and leave times
dropoff_pickup <- interval(kinder$dropoff,kinder$pickup)

# Create sequence to check times every 30 minutes
time_checks <- seq(daystart,by="30 min")

# for every student check whether present at time checks
student_present <- sapply(k$intervals,function(x) time_checks %within% x)

# (Bonus: Make into a nice looking data frame)
df1 <- as.data.frame(t(student_present))
names(df1) <- substr(as.character(time_checks),12,16)
df1 <- cbind(k$id,df1)
df1

有关%within% function和lubridate的更多信息

,

谢谢 我可以运行您的示例并且它可以正常工作，然后我想可以运行一个count函数来计算在不同时间间隔内出现的孩子的实际数量。

但是我无法在我的数据上运行它，我试图更改它，但是当我想使用间隔函数时，我遇到了麻烦。即使我的变量是字符类型。该文档说应该可以正常工作-如果我理解正确的话。