我的数据如下:

我想让它看起来像这样:

我想使用%>% - chaining在tidyverse中执行此操作.

df <- 
structure(list(id = c(2L, 2L, 4L, 5L, 5L, 5L, 5L), start_end = structure(c(2L, 
1L, 2L, 2L, 1L, 2L, 1L), .Label = c("end", "start"), class = "factor"), 
    date = structure(c(6L, 7L, 3L, 8L, 9L, 10L, 11L), .Label = c("1979-01-03", 
    "1979-06-21", "1979-07-18", "1989-09-12", "1991-01-04", "1994-05-01", 
    "1996-11-04", "2005-02-01", "2009-09-17", "2010-10-01", "2012-10-06"
    ), class = "factor")), .Names = c("id", "start_end", "date"
), row.names = c(3L, 4L, 7L, 8L, 9L, 10L, 11L), class = "data.frame")

我尝试过的:

data.table::dcast( df, formula …

我想转换这样的表 (*)：

set.seed(1)
mydata <- data.frame(ID=rep(1:4, each=3), R=rep(1:3, times=4), FIXED=rep(runif(4), each=3), AAA=rnorm(12), BBB=rbinom(12,12,0.5), CCC=runif(12))

ID R    FIXED    AAA  BBB   CCC
 1 1    0.26   -0.83   8   0.82
 1 2    0.26    1.59   5   0.64
 1 3    0.26    0.32   6   0.78
 2 1    0.37   -0.82   6   0.55
 2 2    0.37    0.48   6   0.52
 2 3    0.37    0.73   4   0.78
 3 1    0.57    0.57   8   0.02
 3 2    0.57   -0.30   7   0.47
 3 3    0.57    1.51   7   0.73
 4 1    0.90    0.38   4   0.69 …

我想将 data.tabledcast函数放入一个函数中，该函数可以处理聚合函数的自定义数量/顺序。这就是为什么我需要将聚合函数作为参数传递给dcast函数。参数需要在外部定义dcast。我怎么能这样做呢？

这很好用，但我想在 dcast 之外定义聚合函数。

dt = data.table(x = sample(5, 20, TRUE), y = sample(2, 20, TRUE), 
                z = sample(letters[1:2], 20, TRUE), d1 = runif(20), d2 = 1L

dcast(dt, x + y ~ z, fun = list(sum, min), value.var = "d1")

我尝试了这个方法：

func <- list(sum, min)

dcast(dt, x + y ~ z, fun = func, value.var = "d1")

然后我收到此错误消息：

eval(expr, envir, enclos) 中的错误：找不到函数“func”

我有一个数据框df，看起来像这样...

"ID","ReleaseYear","CriticPlayerPrefer","n","CountCriticScores","CountUserScores"
"1",1994,"Both",1,5,283
"2",1994,"Critics",0,0,0
"3",1994,"Players",0,0,0
"4",1995,"Both",3,17,506
"5",1995,"Critics",0,0,0
"6",1995,"Players",0,0,0
"7",1996,"Both",18,163,3536
"8",1996,"Critics",2,18,97
"9",1996,"Players",3,20,79

我想翻转数据框，使列如下所示：

"ReleaseYear","Both","Critics","Players"

每个栏位Both',Critics andPlayers would be then` 的值。

当我尝试运行此...

require(dcast)
chartData.CriticPlayerPreferByYear <- dcast(
    data = df,
    formula = ReleaseYear ~ CriticPlayerPrefer,
    fill = 0,
    value.var = n
)

...我得到这个错误：

Error in match(x, table, nomatch = 0L) : 
  'match' requires vector arguments

这里有什么问题？我如何解决它？

我有一个df,其数据如下:

sub = c("X001","X002", "X001","X003","X002","X001","X001","X003","X002","X003","X003","X002") 
month = c("201506", "201507", "201506","201507","201507","201508", "201508","201507","201508","201508", "201508", "201508") 
tech = c("mobile", "tablet", "PC","mobile","mobile","tablet", "PC","tablet","PC","PC", "mobile", "tablet") 
brand = c("apple", "samsung", "dell","apple","samsung","apple", "samsung","dell","samsung","dell", "dell", "dell")

revenue = c(20, 15, 10,25,20,20, 17,9,14,12, 9, 11)

df = data.frame(sub, month, brand, tech, revenue)

我想使用sub和month作为密钥,每个订户每月获得一行,显示该月份该订户的技术和品牌的唯一值的收入总和.这个例子很简单,列数较少,因为我有一个巨大的数据集,我决定尝试这样做data.table.

我已经设法为一个catagorical列做了这个,使用这个:技术或品牌:

df1 <- dcast(df, sub + month ~ tech,  fun=sum, value.var = "revenue")

但我想为两个或更多的caqtogorical列做这个,到目前为止我已经尝试过这个:

df2 <- dcast(df, sub + month ~ tech+brand,  fun=sum, value.var = "revenue")

它只是连接了catogorical列的唯一值和总和,但我不希望这样.我想为所有catogorical列的每个独特值分隔列.

我是R的新手,非常感谢任何帮助.

试图解决这个问题.假设你有一个data.table:

dt <- data.table (person=c('bob', 'bob', 'bob'), 
                  door=c('front door', 'front door', 'front door'),
                  type=c('timeIn', 'timeIn', 'timeOut'),
                  time=c(
as.POSIXct('2016 12 02 06 05 01', format = '%Y %m %d %H %M %S'),
as.POSIXct('2016 12 02 06 05 02', format = '%Y %m %d %H %M %S'),
as.POSIXct('2016 12 02 06 05 03', format = '%Y %m %d %H %M %S')                     )
)

我想将它转动为这样

person        door        timeIn             timeOut

bob           front door  min(<date/time>) max(<date/time>)

我似乎无法为dcast.data.table获得正确的语法.我试过了

dcast.data.table(
  dt, person + door ~ type, 
  value.var …

我正在尝试为具有多个组的数据框的多列找到不包括 NA 的方法

airquality <- data.frame(City = c("CityA", "CityA","CityA",
                                  "CityB","CityB","CityB",
                                  "CityC", "CityC"),
                         year = c("1990", "2000", "2010", "1990", 
                                  "2000", "2010", "2000", "2010"),
                         month = c("June", "July", "August",
                                   "June", "July", "August",
                                   "June", "August"),
                         PM10 = c(runif(3), rnorm(5)),
                         PM25 = c(runif(3), rnorm(5)),
                         Ozone = c(runif(3), rnorm(5)),
                         CO2 = c(runif(3), rnorm(5)))
airquality

所以我得到一个带有数字的名称列表，所以我知道要选择哪些列：

nam<-names(airquality)
namelist <- data.frame(matrix(t(nam)));namelist

我想按城市和年份计算 PM25、臭氧和二氧化碳的平均值。这意味着我需要第 1,2,4,6:7 列）

acast(datadf, year ~ city, mean, na.rm=TRUE)

但这并不是我真正想要的，因为它包含了我不需要的东西的平均值，而且它不是数据帧格式。我可以转换它然后删除，但这似乎是一种非常低效的方法。

有没有更好的办法？

我正在使用包中的dcast函数 library(reshape2)来投射一个简单的三列表格

df = data.table(id  = 1:1e6, 
             var = c('continent','subcontinent',...), 
             val = c('America','Caribbean',...)````

bydcast(df, id ~ var, value.var ='val')并自动将值转换为计数，即

id     continent   subcontinent
 1     1           1
 2     1           1

但是，如果我将大小减少到 10000 行，它会正确输出

id     continent   subcontinent
 1     America     Caribbean
 2     Europe      West Europe

这是一个错误还是我需要以某种方式更改代码？请帮忙。谢谢！

我正在使用 dcast 转置下表

date               event          user_id
25-07-2020         Create          3455
25-07-2020         Visit           3567
25-07-2020         Visit           3567
25-07-2020         Add             3567
25-07-2020         Add             3678
25-07-2020         Add             3678
25-07-2020         Create          3567
24-07-2020         Edit            3871

我正在使用 dcast 转置以将我的事件作为列并计算 user_id

dae_summ <- dcast(ahoy_events, date ~ event, value.var="user_id")

但我没有得到唯一的用户 ID。它多次计算相同的 user_id。我该怎么做才能让一个 user_id 在同一日期和事件中只计算一次。

使用 dcast 时，如何根据列“Col”指定列顺序？

df <- dcast(x, ID ~ ColumnName, value.var = "Answer")

我需要解决方案不特定于数据，因为 x 可以是任何问题的结果（因此 Col 可以是 1-3 或 1-2 等）。下面是 x 的两个虚拟示例。

ID        Answer        ColumnName        Col
1         Anduin        First Name        1
1         Wrynn         Surname           2
1         Alliance      Faction           3
2         Sylvanas      First Name        1
2         Windrunner    Surname           2
2         Horde         Faction           3



ID        Answer        ColumnName        Col
1         The Kirin Tor Quest             1
1         90            Level             2
2         Emissary      Quest             1
2         38            Level             2