Question

我制作了这个虚拟数据集：

df = data.frame(Order = "Order1", 
       Condition = c("P", "A", "B", "C", "D", "E", "F"),
       Value = c(500, -10, -5,0,0, -10,0))

假设条件的som属于不同的组。

list = list( Group1 = c("A", "B"), 
      Group2 = c("C", "D"), 
      Group3 = c("E","F"))

我需要按条件汇总它们，我得到每个组的总和和计数。

预期产出：

Order   P   Group1 Group2 Group3 Group1n Group2n Group3n
Order1 500   -15     0      -10    2        0       1

我在想的是：

df %>% group_by(Order) %>% summarise(Group1 = sum(Value[Condition == "A" | Condition == "B" ]),
                                 Group2 = sum(Value[Condition == "C" | Condition == "D" ] ),
                                 Group3 = sum(Value[Condition == "E" | Condition == "F"]),
                                 Group1n = length(Value[Condition == "A" | Condition == "B" ]),
                                 Group2n = length(Value[Condition == "C" | Condition == "D" ]),
                                 Group3n = length(Value[Condition == "E" | Condition == "F" ]))

我的输出：

  # A tibble: 1 x 7
Order  Group1 Group2 Group3 Group1n Group2n Group3n
 <fct>   <dbl>  <dbl>  <dbl>   <dbl>   <int>   <int>
 Order1  -15.0      0  -10.0    2       2       2

但我无法将计数正确。还有一种有效的方法，我可以放弃列表而不是明确写出条件== A或B ......等。

谢谢

Answer 1

这可以给你你想要的东西：

df %>% 
  group_by(Order) %>% 
  summarise(Group1 = sum(Value[Condition == "A" | Condition == "B" ]),
            Group2 = sum(Value[Condition == "C" | Condition == "D" ] ),
            Group3 = sum(Value[Condition == "E" | Condition == "F"]),
            Group1n = sum(Condition == "A" | Condition == "B"),
            Group2n = sum(Condition == "C" | Condition == "D"),
            Group3n = sum(Condition == "E" | Condition == "F"))

你可以多清理一下。此版本也只计算非零值（与每个组中的所有行相对。）

# don't rename "list"
list_of_groups = list( Group1 = c("A", "B"), 
  Group2 = c("C", "D"), 
  Group3 = c("E","F"))
df %>% 
  group_by(Order) %>% 
  summarise(Group1 = sum(Value[Condition %in% list_of_groups$Group1]),
            Group2 = sum(Value[Condition %in% list_of_groups$Group2] ),
            Group3 = sum(Value[Condition %in% list_of_groups$Group3]),
            Group1n = sum(Condition %in% list_of_groups$Group1 & Value != 0),
            Group2n = sum(Condition %in% list_of_groups$Group2 & Value != 0),
            Group3n = sum(Condition %in% list_of_groups$Group3 & Value != 0))

利用您的群组列表 - 如果您更改群组，那么您就不必修复所有内容（如果相关）。

Answer 2

我建议这是一个整洁的解决方案：

group_map = data.frame(Condition = unlist(list), Group = rep(names(list), lengths(list)), stringsAsFactors = FALSE)

result = df %>% mutate(Condition = as.character(Condition)) %>%
    inner_join(group_map) %>%
    group_by(Group, Order) %>%
    summarize(sums = sum(Value), n_nonzero = sum(Value != 0))

result
# # A tibble: 3 x 4
# # Groups:   Group [?]
#    Group  Order  sums n_nonzero
#    <chr> <fctr> <dbl>     <int>
# 1 Group1 Order1   -15         2
# 2 Group2 Order1     0         0
# 3 Group3 Order1   -10         1

如果您需要宽格式，可以使用data.table重新整形多列：

library(data.table)
setDT(result)
data.table::dcast(Order ~ Group, data = result, value.var = c("sums", "n_nonzero"))
#     Order sums_Group1 sums_Group2 sums_Group3 n_nonzero_Group1 n_nonzero_Group2 n_nonzero_Group3
# 1: Order1         -15           0         -10                2                0                1

根据R中的条件汇总和创建新变量

2 个答案: