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))
假设某些条件属于不同的组。
list = list( Group1 = c("A", "B"),
Group2 = c("C", "D"),
Group3 = c("E","F"))
我需要按条件聚合它们,在那里我得到每个组的总和和计数。
预期输出:
Order P Group1 Group2 Group3 Group1n Group2n Group3n
Order1 500 -15 0 -10 2 0 1
我在想这样的事情:
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
但我无法正确计数..还有一种有效的方法可以让我传递列表而不是明确写出条件 ==A 或 B ...等。
谢谢
这应该给你你想要的:
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。这个版本也只计算非零值(而不是每个组中的所有行。)
# 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))
利用您的组列表 - 这样,如果您更改组(如果相关),则不必修复所有内容。
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