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r - 如何在R中通过ID计算分类法术的数量和持续时间

问题描述

我有一个纵向数据集，每月记录一个人的就业状况，持续 45 个月。我希望能够创建两个变量以添加到此数据集中：1）每个人“失业”的总持续时间 2）失业法术的数量

理想情况下，它也会跳过 NA 而不会中断咒语

我创建了一个示例数据集以使事情变得简单：


    ID <- c(1:10, 1:10, 1:10)
    date <- c("2006-09-01", "2006-09-01", "2006-09-01", "2006-09-01", "2006-09-01", "2006-09-01", "2006-09-01", 
              "2006-09-01", "2006-09-01", "2006-09-01", "2006-10-01", "2006-10-01", "2006-10-01", "2006-10-01", 
              "2006-10-01", "2006-10-01", "2006-10-01", "2006-10-01", "2006-10-01", "2006-10-01", "2006-11-01", 
              "2006-11-01", "2006-11-01", "2006-11-01", "2006-11-01", "2006-11-01", "2006-11-01", "2006-11-01", 
              "2006-11-01", "2006-11-01")
    act <- c("Unemployed", "Employment", "Education", "Education", "Education", "Education", "Education", 
             "Education", "Education", "Unemployed", "Education", "Unemployed", "Unemployed", "Unemployed", 
             "Education", "Education", "Employment", "Education", "Education", "NA", "Unemployed", 
             "Unemployed", "NA", "Unemployed", "Education", "Employment", "Employment", "NA", "Education", 
             "Unemployed")
    df <- data.frame(ID, date, act)
    df[order(ID),]

       ID       date        act
    1   1 2006-09-01 Unemployed
    11  1 2006-10-01  Education
    21  1 2006-11-01 Unemployed
    2   2 2006-09-01 Employment
    12  2 2006-10-01 Unemployed
    22  2 2006-11-01 Unemployed
    3   3 2006-09-01  Education
    13  3 2006-10-01 Unemployed
    23  3 2006-11-01         NA
    4   4 2006-09-01  Education
    14  4 2006-10-01 Unemployed
    24  4 2006-11-01 Unemployed
    5   5 2006-09-01  Education
    15  5 2006-10-01  Education
    25  5 2006-11-01  Education
    6   6 2006-09-01  Education
    16  6 2006-10-01  Education
    26  6 2006-11-01 Employment
    7   7 2006-09-01  Education
    17  7 2006-10-01 Employment
    27  7 2006-11-01 Employment
    8   8 2006-09-01  Education
    18  8 2006-10-01  Education
    28  8 2006-11-01         NA
    9   9 2006-09-01  Education
    19  9 2006-10-01  Education
    29  9 2006-11-01  Education
    10 10 2006-09-01 Unemployed
    20 10 2006-10-01         NA
    30 10 2006-11-01 Unemployed

我尝试了 Roland 在计算 R 中的持续时间时提出的解决方案，但我不确定如何调整它以通过 ID 为我提供结果并处理 NA。


    library(data.table)
    setDT(df)
    df[, date := as.POSIXct(date, format = "%Y-%m-%d", tz = "GMT")]

    glimpse(df)
    df$act <- ifelse(df$act == "Unemployed",1,-1)
    df[, run := cumsum(c(1, diff(act) != 0))]

    df1 <- df[, list(act = unique(act), 
                               duration = difftime(max(date), min(date), unit = "weeks")), 
                        by = run]
    df1
        run act duration
     1:   1   1  0 weeks
     2:   2  -1  0 weeks
     3:   3   1  0 weeks
     4:   4  -1  0 weeks
     5:   5   1  0 weeks
     6:   6  -1  0 weeks
     7:   7   1  0 weeks
     8:   8  -1  0 weeks
     9:   9   1  0 weeks
    10:  10  -1  0 weeks
    11:  11   1  0 weeks

我所追求的是实现这一目标（这里的持续时间是几个月，但可以是几周或几天）：

    ID spell_count duration
1    1           2        2
2    2           1        2
3    3           1        1
...
10  10           1        2

任何帮助，任何链接/文献/示例将不胜感激。

谢谢你。

标签： rdurationlongitudinal

解决方案

我只使用你的第一个代码块，然后在整个持续时间内，我这样做：


    library(data.table)
    setDT(df)
    df_duration = df[act=="Unemployed",.(duration = .N),by = ID]

失业法术的数量有点棘手：


    df_spell_count = df[order(ID,date)]
    df_spell_count <- df_spell_count[!(is.na(act)|act=="NA")]
    df_spell_count[,previous_act := shift(act,1),by = ID]
    df_spell_count<-df_spell_count[act =="Unemployed" & (previous_act!="Unemployed" | is.na(previous_act))]
    df_spell_count<-df_spell_count[,.(spell_count =.N),by = ID]

如果你想合并这两件事，只需：

df_stats <- merge(df_duration,df_spell_count, by = "ID", all.x = TRUE,all.y = TRUE)

请注意，此 df 不包含那些没有失业期的用户的行。

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