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我有一个按星期几划分的 df/zoo/xts/whatever。我想为每个条目按周进一步拆分。

一个例子是星期五，有一个 id 列表，每个 id 都有相关的时间。这些时间可能是一年中的任何一个星期五。我想在那个星期五制作一个新的 df，其中包含每个 id 以及每周的计数（按顺序）。

它看起来像下面这样，其中每个 w 列是不同的星期五计数：

    id w1 w2 w3 w4
1 id_1  1  2  2  8
2 id_2  3  1  5  2
3 id_3  7  4 10  7

输入：

structure(list(id = c("id_1", "id_2", "id_3"), w1 = c(1, 3, 7
), w2 = c(2, 1, 4), w3 = c(2L, 5L, 10L), w4 = c(8L, 2L, 7L)), .Names = c("id", 
"w1", "w2", "w3", "w4"), row.names = c(NA, 3L), class = "data.frame")

这似乎是聚合的成熟，但我不能完全正确的语法。我尝试过的其他事情如下：

# Applies sum to everything, which doesnt make sense in this context
apply.weekly(friday, sum)

# I considered doing something like getting the unique weeks with:
as.numeric(unique(format(friday[,2], "%U")))
# and then generating each week, getting the counts for each user, and then making a new df from this process. But this seems very inefficient.

编辑：来自 str(data[1:20,]) 的输出：

'data.frame':   20 obs. of  2 variables:
 $ id  : num  1 2 3 4 5 1 2 3 3 2 ...
 $ time: POSIXct, format: "2011-04-25 14:00:00" "2011-04-28 20:00:00" "2011-05-03 06:00:00" "2011-05-06 14:00:00" ...

来自 dput(data[1:20,]) 的输出：

structure(list(id = c(1, 2, 3, 4, 5, 1, 2, 3, 3, 2, 1, 4, 3, 
2, 1, 4, 3, 2, 1, 7), time = structure(c(1303754400, 1304035200, 
1304416800, 1304704800, 1304920800, 1305252000, 1305428400, 1305522000, 
1305774000, 1306404000, 1306422000, 1308261600, 1308290400, 1308340800, 
1308542400, 1308715200, 1308722400, 1308844800, 1309575600, 1309730400
), class = c("POSIXct", "POSIXt"))), .Names = c("id", "time"), row.names = c(1L, 
2L, 3L, 4L, 5L, 6L, 7L, 8L, 
9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L), class = "data.frame")
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1 回答 1

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如果我理解您想要什么，您需要为一周中的某一天（以便您识别）和一年中的一周（以便您最终为每个列添加单独的列）添加额外的列。使用data您给出的dput()：

data$day.of.week <- format(data$time, "%A")
data$week.of.year <- format(data$time, "%U")

现在你想有效地重塑数据，所以使用reshape2包（不是唯一的方法，而是我最熟悉的方法）

library("reshape2")

dcast(data[data$day.of.week=="Friday",], id~week.of.year, 
    value_var="time", fun.aggregate=length)

在该示例中，我对数据进行了子集化以获取星期五。如果您想每天都做，但每天分开，plyr包可以帮助进行迭代。

library("plyr")

dlply(data, .(day.of.week), dcast, id~week.of.year, 
    value_var="time", fun.aggregate=length)

这两个的结果是：

> dcast(data[data$day.of.week=="Friday",], id~week.of.year, value_var="time", fun.aggregate=length)
  id 18 24 26
1  1  0  0  1
2  2  0  1  0
3  4  1  0  0

> dlply(data, .(day.of.week), dcast, id~week.of.year, value_var="time", fun.aggregate=length)
$Friday
  id 18 24 26
1  1  0  0  1
2  2  0  1  0
3  4  1  0  0

$Monday
  id 17
1  1  1

$Saturday
  id 19
1  2  1

$Sunday
  id 19 20 25 27
1  1  0  0  1  0
2  3  0  1  0  0
3  5  1  0  0  0
4  7  0  0  0  1

$Thursday
  id 17 19 21 24 25
1  1  0  1  1  0  0
2  2  1  0  1  0  1
3  3  0  0  0  1  0
4  4  0  0  0  1  0

$Tuesday
  id 18 25
1  3  1  1
2  4  0  1

$Wednesday
  id 20
1  3  1

attr(,"split_type")
[1] "data.frame"
attr(,"split_labels")
  day.of.week
1      Friday
2      Monday
3    Saturday
4      Sunday
5    Thursday
6     Tuesday
7   Wednesday
于 2011-11-13T05:40:57.760 回答