R 如何在数据帧上使用强制转换？

我有一个如下所示的数据帧：

  year income    group
1  2008  27907  Under25
2  2009  25522  Under25
3  2010  26777  Under25
4  2008  58809 Age25_34
5  2009  57239 Age25_34
6  2010  58558 Age25_34
7  2008  75677 Age35_44
8  2009  74900 Age35_44
9  2010  74136 Age35_44
10 2008  78537 Age45_54
11 2009  77460 Age45_54
12 2010  76266 Age45_54
13 2008  69009 Age55_64
14 2009  67586 Age55_64
15 2008  44402 Age65_74
16 2009  46147 Age65_74
17 2010  48595 Age65_74
18 2008  32747   Over75
19 2009  31272   Over75
20 2010  31638   Over75

> str(df)
'data.frame':   20 obs. of  3 variables:
 $ year  : int  2008 2009 2010 2008 2009 2010 2008 2009 2010 2008 ...
 $ income: int  27907 25522 26777 58809 57239 58558 75677 74900 74136 78537 ...
 $ group : Factor w/ 7 levels "Age25_34","Age35_44",..: 7 7 7 1 1 1 2 2 2 3 ...

我想使用cast来查找组的平均值。此外，我想从这个df创建一个广泛的data.frame，其中第一列是年份，下面几列是不同群体的收入。比如说

year     under25     Age25_34     Age35_44     Age45_54    ...
2008     27907        58809        75677        78537      ...
2009     25522        57239        74900        77460      ...
...

当我尝试强制转换时，出现以下错误：

铸造（df，收入组，平均值） 使用组作为值列。使用值参数强制转换以覆盖此选择

[.data.frame

中出错（数据、变量、drop=FALSE）： 选择未定义的列

我对cast命令有什么错

如何将其转换为示例中所示的宽格式

下面列出了我的R版本信息

> unlist(R.Version())
   platform                            arch                              os 
"x86_64-pc-mingw32"                  "x86_64"                       "mingw32" 

 system                          status                           major 
"x86_64, mingw32"                   ""                             "2" 

minor                            year                           month 
"13.1"                          "2011"                            "07" 

day                         svn rev                        language 
"08"                         "56322"                             "R" 
version.string 
"R version 2.13.1 (2011-07-08)" 

为什么不使用tapply

with(df, tapply(income, list(year, group), mean))

---

（感谢Ramnath的好评）

请用

cast

cast(df, year ~ group, mean, value = 'income')

  year Age25_34 Age35_44 Age45_54 Age55_64 Age65_74 Over75 Under25
1 2008    58809    75677    78537    69009    44402  32747   27907
2 2009    57239    74900    77460    67586    46147  31272   25522
3 2010    58558    74136    76266      NaN    48595  31638   26777

创建数据框：

year<-c(2008,2009, 2010,2008,2009, 2010, 2008,2009, 2010,2008, 2009, 2010, 2008, 2009, 2008, 2009, 2010, 2008,2009,2010)
income<-c(27907,25522, 26777,58809, 57239, 58558, 75677,74900, 74136, 78537,77460,76266, 69009,67586, 44402, 46147,48595,32747, 31272,31638)
group<-c("Under25","Under25","Under25","Age25_34","Age25_34","Age25_34","Age35_44","Age35_44","Age35_44","Age45_54","Age45_54","Age45_54","Age55_64","Age55_64","Age65_74","Age65_74","Age65_74","Over75","Over75","Over75")

demographic_data<-data.frame(year, income,group)
demographic_data

str(demographic_data)

---

year没有给出正确的输出，因为它跨年份计算平均值。请查看提供的示例输出，以及描述
…第一列是year…
完美！我会使用
with
语句来避免重复多次
df
。是的，这对于获得我想要的平均值非常有用。原因是重点是我经常使用melt，我认为这是一个很好的补充，但效果很好。谢谢。谢谢！我确实试过了，但没有在收入上加引号。这是我的问题！
year<-c(2008,2009, 2010,2008,2009, 2010, 2008,2009, 2010,2008, 2009, 2010, 2008, 2009, 2008, 2009, 2010, 2008,2009,2010)
income<-c(27907,25522, 26777,58809, 57239, 58558, 75677,74900, 74136, 78537,77460,76266, 69009,67586, 44402, 46147,48595,32747, 31272,31638)
group<-c("Under25","Under25","Under25","Age25_34","Age25_34","Age25_34","Age35_44","Age35_44","Age35_44","Age45_54","Age45_54","Age45_54","Age55_64","Age55_64","Age65_74","Age65_74","Age65_74","Over75","Over75","Over75")

demographic_data<-data.frame(year, income,group)
demographic_data

str(demographic_data)

library(reshape)
melted_demographic_data<-melt(demographic_data,id=c("group","year")) 
melted_demographic_data

groupmeans<-cast(melted_demographic_data,group~variable, mean)
groupmeans
yearmeans<-cast(melted_demographic_data,year~variable, mean)
yearmeans