按唯一标识符聚合并将相关值连接成字符串 [英] Aggregating by unique identifier and concatenating related values into a string

问题描述

我有一个我认为可以通过 aggregate 或 reshape 来满足的需求，但我不太清楚.

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(brand)和随附的 ID 号(id).此数据为长格式，因此名称可以有多个 ID.我想按名称 (brand) 去除重复并将多个可能的 id 连接成一个由注释分隔的字符串.

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 

应该变成:

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

我将如何做到这一点?

推荐答案

让我们调用你的 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

另一种使用 aggregate 的替代方法是:

Another alternative using aggregate is:

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

这会产生相同的结果，现在 id 不再是 list.感谢@Frank 评论.要查看每列的 class，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，另一种选择是使用 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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