在R中仅折叠一些长到宽格式的变量

Question

我对 R 比较陌生，每次需要“重塑”数据时，我都感到非常困惑。我的数据如下所示：

有：

  ID ever_smoked alcoholic        medication dosage
1  1          no        no humira/adalimumab   40mg
2  1          no        no        prednisone   15mg
3  1          no        no      azathioprine   30mg
4  1          no        no            rowasa    9mg
5  2         yes        no            lialda   20mg
6  2         yes        no    mercaptopurine     1g
7  2         yes        no            asacol 1600mg

想要：

 ID  ever_smoked  alcoholic  medication
1  1          no        no   humira/adalimumab, prednisone, azathioprine, rowasa
2  2         yes        no   lialda, mercaptopurine, asacol

  dosage                  most_recent_med     most_recent_dose
1 40mg, 15mg, 30mg, 9mg   rowasa              9mg
2 20mg, 1g, 1600mg        asacol              1600mg

（请注意，它应该是 2 个观察值和 7 个变量）。

本质上，我想 (1) 只折叠几个变量，(2) 保留其他变量的第一行，并且 (3) 根据一些变量的最后条目创建 2 个新变量变量。

重现代码：

have <- data.frame(ID = c(1, 1, 1, 1, 2, 2, 2),
    ever_smoked = c("no", "no", "no", "no", "yes", "yes", "yes"), 
    alcoholic = c("no", "no", "no", "no", "no", "no", "no"),
    medication = c("humira/adalimumab", "prednisone", "azathioprine", "rowasa", "lialda", "mercaptopurine", "asacol"),
    dosage = c("40mg", "15mg", "30mg", "9mg", "20mg", "1g", "1600mg"), stringsAsFactors = FALSE)

want <- data.frame(ID = c(1, 2),
    ever_smoked = c("no", "yes"), 
    alcoholic = c("no", "no"),
    medication = c("humira/adalimumab, prednisone, azathioprine, rowasa", "lialda, mercaptopurine, asacol"),
    dosage = c("40mg, 15mg, 30mg, 9mg", "20mg, 1g, 1600mg"),
    most_recent_med = c("rowasa", "asacol"),
    most_recent_dose = c("9mg", "1600mg"), stringsAsFactors = FALSE)

谢谢。

Answer 1

这里有一些不同的方法：

1) sqldf

library(sqldf)
sqldf("select ID, 
              ever_smoked, 
              alcoholic, 
              group_concat(medication) as medication,
              group_concat(dosage) as dosage, 
              medication as last_medication, 
              dosage as last_doage
        from have
        group by ID")

给予：

  ID ever_smoked alcoholic                                       medication             dosage last_medication last_doage
1  1          no        no humira/adalimumab,prednisone,azathioprine,rowasa 40mg,15mg,30mg,9mg          rowasa        9mg
2  2         yes        no                     lialda,mercaptopurine,asacol     20mg,1g,1600mg          asacol     1600mg

2) 数据表

library(data.table)
have_dt <- data.table(have)
have_dt[, list(medication = toString(medication),
               dosage = toString(dosage),
               last_medication = medication[.N],
               last_dosage = dosage[.N]),
           by = "ID,ever_smoked,alcoholic"]

给予：

   ID ever_smoked alcoholic                                          medication                dosage last_medication last_dosage
1:  1          no        no humira/adalimumab, prednisone, azathioprine, rowasa 40mg, 15mg, 30mg, 9mg          rowasa         9mg
2:  2         yes        no                      lialda, mercaptopurine, asacol      20mg, 1g, 1600mg          asacol      1600mg

3) 基地 - 由

do.call("rbind", by(have, have$ID, with, data.frame(
     ID = ID[1], 
     ever_smoked = ever_smoked[1], 
     alcoholic = alcoholic[1],
     medication = toString(medication),
     dosage = toString(dosage),
     last_medication = tail(medication, 1),
     last_dosage = tail(dosage, 1))))

给予：

  ID ever_smoked alcoholic                                          medication                dosage last_medication last_dosage
1  1          no        no humira/adalimumab, prednisone, azathioprine, rowasa 40mg, 15mg, 30mg, 9mg          rowasa         9mg
2  2         yes        no                      lialda, mercaptopurine, asacol      20mg, 1g, 1600mg          asacol      1600mg

请注意，这也可以写成：

do.call("rbind", by(have, have$ID, function(x) with(x, data.frame(
     ID = ID[1], 
     ever_smoked = ever_smoked[1], 
     alcoholic = alcoholic[1],
     medication = toString(medication),
     dosage = toString(dosage),
     last_medication = tail(medication, 1),
     last_dosage = tail(dosage, 1)))))

4) 基数 - 聚合

aggregate(. ~ ID + ever_smoked + alcoholic, have,
  function(x) c(values = toString(x), last = as.character(tail(x, 1))))

给予：

  ID ever_smoked alcoholic                                   medication.values medication.last         dosage.values dosage.last
1  1          no        no humira/adalimumab, prednisone, azathioprine, rowasa          rowasa 40mg, 15mg, 30mg, 9mg         9mg
2  2         yes        no                      lialda, mercaptopurine, asacol          asacol      20mg, 1g, 1600mg      1600mg

请注意，这会返回一个 2 x 5 的数据框，其中最后两列各是 2 列矩阵，这比展平形式更便于索引，但如果首选展平则：do.call("data.frame", DF)

Answer 2

这是一个汇总过程，你可以使用summarise_all并传递两个函数来汇总每一列：一个用toString折叠列，一个用last取最后一行：

have %>% 
    group_by(ID, ever_smoked, alcoholic) %>% 
    summarise_all(funs(toString(.), most_recent = last(.)))

# A tibble: 2 x 7
# Groups:   ID, ever_smoked [?]
#     ID ever_smoked alcoholic                                 medication_toString       dosage_toString medication_most_recent dosage_most_recent
#  <dbl>       <chr>     <chr>                                               <chr>                 <chr>                  <chr>              <chr>
#1     1          no        no humira/adalimumab, prednisone, azathioprine, rowasa 40mg, 15mg, 30mg, 9mg                 rowasa                9mg
#2     2         yes        no                      lialda, mercaptopurine, asacol      20mg, 1g, 1600mg                 asacol             1600mg

_{假设 ever_smoked 和 酒精 对于此处的每个 ID 都是唯一的。}