Question

我想使用Beer和Wine以及Spirits和行总数SUM的组合，例如：

df <- data.frame(
        Beer = c(1L, 1L, 1L, 3L, 4L, 0L, 0L, 3L),
        Wine = c(1L, 1L, 0L, 0L, 0L, 1L, 1L, 2L),
     Spirits = c(0L, 1L, 0L, 0L, 0L, 1L, 2L, 1L),
         SUM = c(2L, 3L, 1L, 3L, 4L, 2L, 3L, 6L)
)

如何创建新列COMBINE，将SUM和列名称粘贴在一起，其值大于0.这样的事情，但任何大于5的SUM都被视为5 +。

Beer    Wine    Spirits    SUM     COMBINE
1        1        0        2       2Beer&Wine
1        1        1        3       3Beer&Wine&Spirits
1        0        0        1       1Beer
1        0        0        1       1Beer
3        0        0        3       3Beer
4        0        0        4       4Beer
0        1        1        2       2Wine&Spirits
0        1        1        2       2Wine&Spirits
3        2        1        6       5+Beer&Wine&Spirits

对于某些添加的上下文，所有这一切的最终结果是我希望计算COMBINE中的因子，尽管这不是我正在努力的部分。

COMBINE    Count
1Beer        2
2Beer        0
3Beer        1
4Beer        1
5+Beer       0
1Wine        0
2Wine        0
.
.
.
2Wine&Spirits 2
.....

Answer 1

使用ifelse直接解决方案：

d$COMBINE <- with(d, gsub("&$", "", 
                          paste0(ifelse(SUM > 5, "5+", SUM),
                                 ifelse(Beer > 0, "Beer&", ""),
                                 ifelse(Wine > 0, "Wine&", ""),
                                 ifelse(Spirits > 0, "Spirits", ""))))

  Beer Wine Spirits SUM             COMBINE
1    1    1       0   2          2Beer&Wine
2    1    1       1   3  3Beer&Wine&Spirits
3    1    0       0   1               1Beer
4    3    0       0   3               3Beer
5    4    0       0   4               4Beer
6    0    1       1   2       2Wine&Spirits
7    0    1       2   3       3Wine&Spirits
8    3    2       1   6 5+Beer&Wine&Spirits

计算您可以使用的因素：table(d$COMBINE)

Answer 2

您也可以使用聚合：

执行此操作

d$sum5 = pmin(5, d$Beer + d$Wine + d$Spirits)
d$count = 1
aggregate(count ~ (Beer>0) + (Wine>0) + (Spirits>0) + sum5, data=d, FUN=sum)

  Beer > 0  Wine > 0  Spirits > 0  sum5  count
1     TRUE     FALSE        FALSE     1      1
2     TRUE      TRUE        FALSE     2      1
3    FALSE      TRUE         TRUE     2      1
4     TRUE     FALSE        FALSE     3      1
5    FALSE      TRUE         TRUE     3      1
6     TRUE      TRUE         TRUE     3      1
7     TRUE     FALSE        FALSE     4      1
8     TRUE      TRUE         TRUE     5      1

忽略总和（举例说明并非一切都是1）：

aggregate(count ~ (Beer>0) + (Wine>0) + (Spirits>0), data=d, FUN=sum)

  Beer > 0  Wine > 0  Spirits > 0  count
1     TRUE     FALSE        FALSE      3
2     TRUE      TRUE        FALSE      1
3    FALSE      TRUE         TRUE      2
4     TRUE      TRUE         TRUE      2

创建合并大于0的列名的新列

2 个答案: