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我有一个列表，其中包含大约 100,000 次一起订购的项目，我已将它们粘贴到一列中，这样我就可以计算每个组合出现的次数。

4845   Curly Fries California Burger   1
4846   French Fries California Burger  1
4847   Hamburger California Burger     1
4848   $1 Fountain Drinks Curly Fries  1
4849   $1 Fountain Drinks Curly Fries  1
4850   California Burger Curly Fries   1
4851   Curly Fries Curly Fries         1

我探索了聚合函数，它给了我以下错误：

aggregate(t1$count,list(t1$pc), sum) <br>
Error in sort.list(y) : 'x' must be atomic for 'sort.list'
Have you called 'sort' on a list? <br>

我也尝试过 ddply 的变体：

ddply(t1,t1$pc,transform,occurances=sum(t1$count))

但我得到这个错误

Error in UseMethod("as.quoted") : 
no applicable method for 'as.quoted' applied to an object of class "c('matrix', 'list')"

我假设我得到了这个，因为我试图基本上按字符值“分组”。我也探索tapply并recast基于类似问题的答案，但无济于事。

我怎样才能得到这个组合计数？

供考虑，单独列出的项目示例（再次为格式问题道歉）：

                   Var1                     Var2 Var3
>2               Onion Rings              Onion Rings    1
>3  Pineapple Cheddar Burger              Onion Rings    1
>4               Onion Rings Pineapple Cheddar Burger    1
>5  Pineapple Cheddar Burger Pineapple Cheddar Burger    1
>5              Onion Rings              Onion Rings     1
>6  Pineapple Cheddar Burger              Onion Rings    1
>7               Onion Rings Pineapple Cheddar Burger    1
>8  Pineapple Cheddar Burger Pineapple Cheddar Burger    1
>9             Fountain Soda            Fountain Soda    1
>10             French Fries            Fountain Soda    1
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2 回答 2

4

该table()功能在这里很有帮助：

with(t1, table(pc)) ## or equivalently table(t1$pc)

这假设pc是您要计算其出现次数的因子变量。（如果它不是一个因素，它将被强制为一个因素。）

于 2013-02-26T19:54:28.697 回答
1

您最初的方法非常接近我认为您想要的方法。将它们组合成一个因素肯定会奏效，前提是您以相同的顺序组合它们，这样您就不会得到“薯条，汉堡”和“汉堡，薯条”。

可能有一种更简单的方法可以做你想做的事，但我没有想到那是什么。不过，我认为这可以满足您的需求：

# Let's assume your data looks like this:
> df
                       Var1                      Var2 Var3
1               Onion Rings               Onion Rings    1
2  Pineapple Cheddar Burger               Onion Rings    1
3               Onion Rings  Pineapple Cheddar Burger    1
4  Pineapple Cheddar Burger  Pineapple Cheddar Burger    1
5               Onion Rings               Onion Rings    1
6  Pineapple Cheddar Burger               Onion Rings    1
7               Onion Rings  Pineapple Cheddar Burger    1
8  Pineapple Cheddar Burger  Pineapple Cheddar Burger    1
9             Fountain Soda             Fountain Soda    1
10             French Fries             Fountain Soda    1

# Now, for each row
#     1. sort the Var1 and Var2,
#     2. combine the sorted vars, and
#     3. convert them back into a factor

df$sortcomb <- as.factor(apply(df[,1:2], 1, function(x) paste(sort(x), collapse=", ")))

table(df$sortcomb) # then use table as per normal

ddply(df, .(sortcomb), summarize, count=length(sortcomb)) # or ddply
于 2013-02-27T01:26:50.973 回答