通过唯一标识符聚合并将相关值连接到字符串中 [英] Aggregating by unique identifier and concatenating related values into a string

问题描述

我想我可以通过聚合或 reshape 来满足，但是我不能完全满足找出答案。

I have a need that I imagine could be satisfied by aggregate or reshape, but I can't quite figure out.

我有一个名称列表（品牌）和随附的ID号（ id ）。此数据采用长格式，因此名称可以有多个ID。我想按名称（ brand ）进行去复制，然后将多个可能的 id 连接到一个

I have a list of names (brand), and accompanying ID number (id). This data is in long form, so names can have multiple ID's. I'd like to de-dupicate by the name (brand) and concatenate the multiple possible id's into a string separated by a comment.

例如：

brand            id 
RadioShack       2308
Rag & Bone       4466
Ragu             1830
Ragu             4518
Ralph Lauren     1638
Ralph Lauren     2719
Ralph Lauren     2720
Ralph Lauren     2721
Ralph Lauren     2722 

应变为：

RadioShack       2308
Rag & Bone       4466
Ragu             1830,4518
Ralph Lauren     1638,2719,2720,2721,2722

我将如何实现？

推荐答案

让我们调用您的data.frame DF

Let's call your data.frame DF

> aggregate(id ~ brand, data = DF, c)
         brand                           id
1   RadioShack                         2308
2   Rag & Bone                         4466
3         Ragu                   1830, 4518
4 Ralph Lauren 1638, 2719, 2720, 2721, 2722

使用 aggregate 的另一种替代方法是：

Another alternative using aggregate is:

result <- aggregate(id ~ brand, data = DF, paste, collapse = ",")

这将产生相同的结果，现在 id 不再是列表。感谢@Frank评论。要查看每列的类，请尝试：

This produces the same result and now id is not a list anymore. Thanks to @Frank comment. To see the class of each column try:

> sapply(result, class)
      brand          id 
   "factor" "character"

正如@DavidArenburg在评论中提到的，另一种替代方法是使用 toString 函数：

As mentioned by @DavidArenburg in the comments, another alternative is using the toString function:

aggregate(id ~ brand, data = DF, toString)

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