如何计算工作期限

Question

我拥有为期 1 年的每月员工就业状况数据。数据框包括 4 个变量：人员 ID、国家/地区、月份和该特定月份的主要活动（就业、失业、非活动、其他）。我在这里举了一个例子：

ID <- c(1:10, 1:10)
country <- c("AT", "BE", "CH", "CZ", "HR", "SO", "SV", "RU", "GR", "GE", "AT", "BE", "CH", "CZ", "HR", 
             "SO", "SV", "RU", "GR", "GE")
month <- c("Jan", "Feb", "Mar", "Apr", "May", "Aug", "Dec", "Nov", "Sep", "Jan", "Jun", "Jul", "Oct",
           "Jan", "Feb", "Mar", "Apr", "May", "Aug", "Dec")
act <- c("Unemployed", "Employed", "Other", "Other", "Inactive", "Unemployed", "Employed", 
         "Employed", "Employed", "Unemployed", "Other", "Unemployed", "Unemployed", "Unemployed", 
         "Other", "Other", "Employed", "Other", "Other", "NA")
df <- data.frame(ID, country, month, act)
df[order(ID),]

   ID country month        act
1   1      AT   Jan Unemployed
11  1      AT   Jun      Other
21  1      AT   Nov Unemployed
2   2      BE   Feb   Employed
12  2      BE   Jul Unemployed
22  2      BE   Sep Unemployed
3   3      CH   Mar      Other
13  3      CH   Oct Unemployed
23  3      CH   Jan         NA
4   4      CZ   Apr      Other
14  4      CZ   Jan Unemployed
24  4      CZ   Jun Unemployed
5   5      HR   May   Inactive
15  5      HR   Feb      Other
25  5      HR   Jul      Other
6   6      SO   Aug Unemployed
16  6      SO   Mar      Other
26  6      SO   Oct   Employed
7   7      SV   Dec   Employed
17  7      SV   Apr   Employed
27  7      SV   Nov   Employed
8   8      RU   Nov   Employed
18  8      RU   May      Other
28  8      RU   Jan         NA
9   9      GR   Sep   Employed
19  9      GR   Aug      Other
29  9      GR   Jun   Inactive
10 10      GE   Jan Unemployed
20 10      GE   Dec         NA
30 10      GE   Aug Unemployed

我的目标是创建一个新的数据框，其中每一行代表一个就业期但条件是就业期之前和之后必须是失业期。这样我就能够只包括人们从失业到就业再回到失业的就业咒语，并计算这些咒语的持续时间。理想情况下，最后会有 4 个变量：PersID、国家、拼写持续时间、开始月份、结束月份。它应该是这样的：

   ID country spell_duration starting ending
1   1      AT              5      Jan    May
11  1      AT              5      Jun    Oct
2   2      BE              7      Feb    Aug
12  2      BE              6      Jul    Dec
3   3      CH             10      Mar    Dec
13  3      CH              1      Oct    Oct
4   4      CZ              8      Apr    Nov
14  4      CZ              5      Jan    May
5   5      HR              5      May    Sep
15  5      HR              4      Feb    May
6   6      SO              2      Aug    Sep
16  6      SO              6      Mar    Aug
7   7      SV              1      Dec    Dec
17  7      SV              9      Apr    Dec
8   8      RU              8      Nov    Dec
18  8      RU              7      May    Nov
9   9      GR              3      Sep    Nov
19  9      GR              2      Aug    Sep
10 10      GE              8      Jan    Aug
20 10      GE              1      Dec    Dec

我已经找到了 Maria (How to calculate number and duration of categorical spells by ID in R) 的解决方案，但她的问题不同。我不想要工作的总时长，我也不需要咒语的数量

Answer 1

我使用data.table 包进行操作，我猜循环正在工作。

编辑： 剩下一个额外的“}”，我对其进行了编辑。我试过了，效果很好。

EDIT2：我也添加了“setDT(df)”。

library(data.table)

 df <- fread(paste("ID country month        act
    1      AT   Jan Unemployed
    1      AT   Jun      Other
    1      AT   Nov Unemployed
    2      BE   Feb   Employed
    2      BE   Jul Unemployed
    2      BE   Sep Unemployed
    3      CH   Mar      Other
    3      CH   Oct Unemployed
    3      CH   Jan         NA
    4      CZ   Apr      Other
    4      CZ   Jan Unemployed
    4      CZ   Jun Unemployed
    5      HR   May   Inactive
    5      HR   Feb      Other
    5      HR   Jul      Other
    6      SO   Aug Unemployed
    6      SO   Mar      Other
    6      SO   Oct   Employed
    7      SV   Dec   Employed
    7      SV   Apr   Employed
    7      SV   Nov   Employed
    8      RU   Nov   Employed
    8      RU   May      Other
    8      RU   Jan         NA
    9      GR   Sep   Employed
    9      GR   Aug      Other
    9      GR   Jun   Inactive
    10      GE   Jan Unemployed
    10      GE   Dec         NA
    10      GE   Aug Unemployed", collapse = '\n'))
setDT(df)
df[, monthInt := match(month, month.abb)]

df <- df[order(ID,monthInt)]

finalDt <- data.table()
for (i in unique(df[, ID])) {
  
  tempT <- df[ID == i]
  
  for (tim in 1:(nrow(tempT)-1)) {
    timT <- data.table(ID = tempT[tim,ID],
                       country = tempT[tim, country],
                       spell_duration = tempT[tim+1, monthInt] - tempT[tim, monthInt],
                       starting = month.abb[tempT[tim, monthInt]],
                       ending = month.abb[tempT[tim+1, monthInt]-1])
    
    finalDt <- rbind(finalDt,timT)
    
  }
}

Answer 2

没想太多，第一时间就想到了。不过非常麻烦。我确信有更优雅的解决方案，但这不需要任何额外的包。

data <- df
Empl_spells <- data.frame(ID = c(), Start = c(), End = c())

for(user in unique(data$ID)){
  # subset per user
  user_dat <- data[data$ID == user,]
  # initiate a list to store where changes occur and a counter for
  # entries to this list
  if(nrow(user_dat) > 2){
    Changes_data <- list()
    entry <- 1
    # for every row, check if it switches from employed to unemployed
    # or the opposite. Mark with "break" if some other entry interrupts
    for(i in 2:nrow(user_dat)){
      if(user_dat$act[i] == "Employed" &
         user_dat$act[i-1] == "Unemployed"){
        Changes_data[[entry]] <- c("Start", i)
        entry <- entry + 1
      }else if(user_dat$act[i] == "Unemployed" &
               user_dat$act[i-1] == "Employed"){
        Changes_data[[entry]] <- c("End", i)
        entry <- entry + 1
      }else if(user_dat$act[i] != "Employed" &
               user_dat$act[i] != "Unemployed"){
        Changes_data[[entry]] <- c("Break", i)
        entry <- entry + 1
      }
    }
    # see where to an "End" follows a "Start" immediately in the new list
    Changes_df <- do.call(rbind.data.frame, Changes_data)
    EmplToUnempl <- which(Changes_df[-nrow(Changes_df), 1] == "Start" & Changes_df[-1, 1] == "End")
    if(length(EmplToUnempl) >= 1){
      append <- data.frame(ID = user,
                           Start = user_dat$month[as.numeric(Changes_df[EmplToUnempl, 2])],
                           End = user_dat$month[as.numeric(Changes_df[EmplToUnempl + 1, 2])-1])
      # append the data to the data.frame for all of the people
      Empl_spells <- rbind(Empl_spells, append)
    }
  }
}

由于我没有您的数据，因此我没有对此进行测试。这是你想要的吗？

编辑（矢量化；可能会更快）：

data <- df

users <- unique(data$ID)
calculate <- function(user){
  # subset per user
  user_dat <- data[data$ID == user,]
  # initiate a list to store where changes occur and a counter for
  # entries to this list
  if(nrow(user_dat) > 2){
    Changes_data <- list()
    entry <- 1
    # for every row, check if it switches from employed to unemployed
    # or the opposite. Mark with "break" if some other entry interrupts
    for(i in 2:nrow(user_dat)){
      if(user_dat$act[i] == "Employed" &
         user_dat$act[i-1] == "Unemployed"){
        Changes_data[[entry]] <- c("Start", i)
        entry <- entry + 1
      }else if(user_dat$act[i] == "Unemployed" &
               user_dat$act[i-1] == "Employed"){
        Changes_data[[entry]] <- c("End", i)
        entry <- entry + 1
      }else if(user_dat$act[i] != "Employed" &
               user_dat$act[i] != "Unemployed"){
        Changes_data[[entry]] <- c("Break", i)
        entry <- entry + 1
      }
    }
    # see where to an "End" follows a "Start" immediately in the new list
    Changes_df <- do.call(rbind.data.frame, Changes_data)
    EmplToUnempl <- which(Changes_df[-nrow(Changes_df), 1] == "Start" & Changes_df[-1, 1] == "End")
    if(length(EmplToUnempl) >= 1){
      append <- data.frame(ID = user,
                           Start = user_dat$month[as.numeric(Changes_df[EmplToUnempl, 2])],
                           End = user_dat$month[as.numeric(Changes_df[EmplToUnempl + 1, 2])-1])
      # append the data to the data.frame for all of the people
      return(append)
    }
  }
}

empl_spells <- lapply(users, FUN = calculate)
Empl_spells <- do.call(rbind.data.frame, empl_spells)

编辑#2（计算持续时间）：

MonthToNumeric <- function(x){
  which(c("Jan", "Feb", "Mar", "Apr", "May", "Jun",
          "Jul", "Aug", "Sep", "Oct", "Nov", "Dec") == x)
}

calcDuration <- function(Start, End){
  return(MonthToNumeric(End) - MonthToNumeric(Start) + 1)
}

Empl_spells$Duration <- mapply(FUN = calcDuration, Start = Empl_spells[, 2], End = Empl_spells[, 3])