在data.frame上重构魔法

时间:2013-01-24 18:27:53

标签: r dataframe reshape

我目前正在学习使用data.frame,并对如何重新排序它们感到困惑。

目前,我有一个data.frame显示:

  • 第1栏:商店名称
  • 第2栏:产品
  • 第3栏:本店购买此产品的次数

或在视觉上像这样:

+---+-----------+-------+----------+--+
|   | Shop.Name | Items | Product  |  |
+---+-----------+-------+----------+--+
| 1 | Shop1     |     2 | Product1 |  |
| 2 | Shop1     |     4 | Product2 |  |
| 3 | Shop2     |     3 | Product1 |  |
| 4 | Shop3     |     2 | Product1 |  |
| 5 | Shop3     |     1 | Product4 |  |
+---+-----------+-------+----------+--+

我想要实现的是以下“以商店为中心”的结构:

  • 第1栏:商店名称
  • 第2栏:为product1
  • 销售的商品
  • 第3栏:为product2销售的商品
  • 第4栏:为product3销售的商品 ...

如果某个商店/产品没有专线(因为没有销售),我想创建一个0。

+---+-------+-------+-------+-------+-------+-----+--+--+
|   | Shop  | Prod1 | Prod2 | Prod3 | Prod4 | ... |  |  |
+---+-------+-------+-------+-------+-------+-----+--+--+
| 1 | Shop1 |     2 |     4 |     0 |     0 | ... |  |  |
| 2 | Shop2 |     3 |     0 |     0 |     0 | ... |  |  |
| 3 | Shop3 |     2 |     0 |     0 |     1 | ... |  |  |
+---+-------+-------+-------+-------+-------+-----+--+--+

3 个答案:

答案 0 :(得分:12)

到目前为止,答案在某种程度上起作用,但没有完全回答你的问题。特别是,它们没有解决没有商店销售特定产品的情况的问题。根据您的示例输入和所需的输出,没有商店出售“Product3”。实际上,“Product3”甚至没有出现在您的来源data.frame中。此外,它们没有解决每个Shop + Product组合具有多行的可能情况。

到目前为止,这是您的数据的修改版本和两个解决方案。我为“Shop1”和“Product1”的组合添加了另一行。请注意,我已将您的产品转换为factor变量,其中包含变量可以采用的级别,即使这些级别实际上都没有该级别。

mydf <- data.frame(
  Shop.Name = c("Shop1", "Shop1", "Shop2", "Shop3", "Shop3", "Shop1"),
  Items = c(2, 4, 3, 2, 1, 2),
  Product = factor(
    c("Product1", "Product2", "Product1", "Product1", "Product4", "Product1"),
    levels = c("Product1", "Product2", "Product3", "Product4")))
    来自“reshape2”的
  1. dcast

    library(reshape2)
    dcast(mydf, formula = Shop.Name ~ Product, value="Items", fill=0)
    # Using Product as value column: use value.var to override.
    # Aggregation function missing: defaulting to length
    # Error in .fun(.value[i], ...) : 
    #   2 arguments passed to 'length' which requires 1
    

    什?突然不起作用。这样做:

    dcast(mydf, formula = Shop.Name ~ Product, 
          fill = 0, value.var = "Items", 
          fun.aggregate = sum, drop = FALSE)
    #   Shop.Name Product1 Product2 Product3 Product4
    # 1     Shop1        4        4        0        0
    # 2     Shop2        3        0        0        0
    # 3     Shop3        2        0        0        1
    
  2. 让我们去上学。来自“重塑”的cast

    library(reshape)
    cast(mydf, formula = Shop.Name ~ Product, value="Items", fill=0)
    # Aggregation requires fun.aggregate: length used as default
    #   Shop.Name Product1 Product2 Product4
    # 1     Shop1        2        1        0
    # 2     Shop2        1        0        0
    # 3     Shop3        1        0        1
    

    的Eh。不是你想要的......试试这个:

    cast(mydf, formula = Shop.Name ~ Product, 
         value = "Items", fill = 0, 
         add.missing = TRUE, fun.aggregate = sum)
    #   Shop.Name Product1 Product2 Product3 Product4
    # 1     Shop1        4        4        0        0
    # 2     Shop2        3        0        0        0
    # 3     Shop3        2        0        0        1
    
  3. 让我们回到基础。来自基地R的xtabs

    xtabs(Items ~ Shop.Name + Product, mydf)
    #          Product
    # Shop.Name Product1 Product2 Product3 Product4
    #     Shop1        4        4        0        0
    #     Shop2        3        0        0        0
    #     Shop3        2        0        0        1
    

    或者,如果您更喜欢data.frame(请注意您的“Shop.Name”变量已转换为row.names的{​​{1}}):

    data.frame

答案 1 :(得分:1)

使用dcast库中的reshape2

library(reshape2)

> df <- data.frame(Shop.Name=rep(c("Shop1","Shop2","Shop3"),each=3),
+                  Items=rpois(9,5),
+                  Product=c(rep(c("Prod1","Prod2","Prod3","Prod4"),2),"Prod5")
+ )
> df
  Shop.Name Items Product
1     Shop1     6   Prod1
2     Shop1     5   Prod2
3     Shop1     6   Prod3
4     Shop2     5   Prod4
5     Shop2     6   Prod1
6     Shop2     6   Prod2
7     Shop3     4   Prod3
8     Shop3     7   Prod4
9     Shop3     5   Prod5
> dcast(df,Shop.Name ~ Product,value.var="Items",fill=0)
  Shop.Name Prod1 Prod2 Prod3 Prod4 Prod5
1     Shop1     6     5     6     0     0
2     Shop2     6     6     0     5     0
3     Shop3     0     0     4     7     5

答案 2 :(得分:0)

如果您出于任何原因想要使用原始的重塑包:

Shop.Name <- c("Shop1", "Shop1", "Shop2", "Shop3", "Shop3")
Items <- c(2,4,3,2,1)
Product <- c("Product1", "Product2", "Product1", "Product1", "Product4")
(df <- data.frame(Shop.Name, Items, Product))

cast(df, formula = Shop.Name ~ Product, value="Items", fill=0)