如何按日期分组data.frame？ [英] How to group a data.frame by date?

问题描述

我有一个.csv文件，格式如下：

I have a .csv file in the following format:

Date       ,     Time  , Value
1899-01-01 ,  4:00:00  ,    1
1899-01-01 ,  4:01:00  ,    2
1899-01-01 ,  4:02:00  ,    3
1899-01-01 ,  4:03:00  ,    4
1899-01-01 ,  4:04:00  ,    5
1900-08-22 , 22:00:00  ,  101
1900-08-22 , 22:01:00  ,  102
2013-08-29 ,  4:00:00  , 1000
2013-02-29 ,  4:02:00  , 1001
2013-02-29 ,  4:03:00  , 1002

是否有可能按日期分组以产生以下格式的 data.table ：

Is it possible to group by date to produce a data.table in the the following format:

Date      , Vector(variable length)
1899-02-28, c(1,2,3,4,5)
1900-08-22, c(101,102)
1900-08-22, c(1000,1001,1002)

这是我到目前为止（经过一天的尝试之后）最好的：

This is the best that I have so far (after a day of attempts):

raw <- read.csv(pathName, header = TRUE, stringsAsFactors = FALSE)
groupedByDate <- split(raw, raw$Date)

但是，这似乎产生了一个非常宽的表格，每个日期都有一列，这与我的内容不太接近

However, this seems to produce a very wide table with one column for each date, which is not very close to what I want.

推荐答案

在<$ c $上使用聚合怎么办？ c> data.frame 名为 mydf，如下所示：

What about using aggregate on a data.frame named "mydf" as follows:

> temp <- aggregate(Value ~ Date, mydf, as.vector) 
> temp
         Date         Value
1 1899-01-01  1, 2, 3, 4, 5
2 1900-08-22       101, 102
3 2013-02-29     1001, 1002
4 2013-08-29           1000

值列现在为列表，其中包含您的向量。

The "Value" column is now a list which contains your vectors.

> temp$Value
$`0`
[1] 1 2 3 4 5

$`1`
[1] 101 102

$`2`
[1] 1001 1002

$`3`
[1] 1000

使用分割可能要查找的内容是：

What you were probably looking for with split is:

> split(mydf$Value, mydf$Date)
$`1899-01-01 `
[1] 1 2 3 4 5

$`1900-08-22 `
[1] 101 102

$`2013-02-29 `
[1] 1001 1002

$`2013-08-29 `
[1] 1000

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