Question

我有一个如下所示的数据框：

_________________id ________________text______
    1   | 7821             | "some text here"
    2   | 7821             |  "here as well"
    3   | 7821             |  "and here"
    4   | 567              |   "etcetera"
    5   | 567              |    "more text"
    6   | 231              |   "other text"

我想按ID对文本进行分组，因此我可以运行聚类算法：

________________id___________________text______
    1   | 7821             | "some text here here as well and here"
    2   | 567              |   "etcetera more text"
    3   | 231              |   "other text"

有没有办法做到这一点？我从数据库表导入，我有很多数据，所以我不能手动完成。

Answer 1

您实际上是在寻找aggregate，而不是merge，并且应该在SO上显示许多示例，以展示不同的聚合选项。这是最基本和最直接的方法，使用公式方法指定aggregate的哪些列。

以下是您可以复制并粘贴的数据

mydata <- structure(list(id = c(7821L, 7821L, 7821L, 567L, 567L, 231L), 
    text = structure(c(6L, 3L, 1L, 2L, 4L, 5L), .Label = c("and here", 
    "etcetera", "here as well", "more text", "other text", "some text here"
    ), class = "factor")), .Names = c("id", "text"), class = "data.frame", 
    row.names = c(NA, -6L))

这是汇总输出。

aggregate(text ~ id, mydata, paste, collapse = " ")
#     id                                 text
# 1  231                           other text
# 2  567                   etcetera more text
# 3 7821 some text here here as well and here

当然，还有data.table，它具有非常紧凑的语法（并且速度极快）：

> library(data.table)
> DT <- data.table(mydata)
> DT[, paste(text, collapse = " "), by = "id"]
     id                                   V1
1: 7821 some text here here as well and here
2:  567                   etcetera more text
3:  231                           other text

R：按索引合并文本文档

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