2 
Name of member Allowance Type             Expenditure Type  Date          Amount, £

Adam Afriyie Office running costs (IEP/AOE) Incidentals     07/03/2009 111.09
Adam Afriyie Office running costs (IEP/AOE) Incidentals     11/05/2009 111.09
Adam Afriyie Office running costs (IEP/AOE) Incidentals     11/05/2009 51.75
Adam Holloway   Office running costs (IEP/AOE)  Incidentals  10/01/2009  35
Adam Holloway   Office running costs (IEP/AOE)  Incidentals  10/01/2009  413.23
Adam Holloway   Office running costs (IEP/AOE)  Incidentals  10/01/2009  9.55
Adam Holloway   Office running costs (IEP/AOE   IT equipment 07/03/2009 890.01
Adam Holloway   Communications Expenditure   Publications   12/04/2009  1774
Adam Holloway   Office running costs (IEP/AOE)  Incidentals  12/08/2009  1.1
Adam Holloway   Office running costs (IEP/AOE   Incidentals  12/08/2009  64.31
Adam Holloway   Office running costs (IEP/AOE)  Incidentals  12/08/2009  64.31

嗨，我是 R 新手，也是编程新手。这是议员在特定时间段内支出的子集。我想小计每位议员的费用，我使用了另一篇文章中的代码

> aggregate(cbind(bsent, breturn, tsent, treturn, csales) ~ yname, data = foo, 
 +           FUN = sum)

并将其编辑为我自己的情况。

我的代码：

expenses2 <- aggregate(cbind(Amount..Â.) ~ Name.of.member, data = expenses, FUN = sum)

现在虽然这段代码确实做了某种聚合，但数字不匹配。例如，可以计算出 Adam Afriyie 的费用为 273.93 英镑，但此代码给出的结果为 12697。我不知道这个数字代表什么。有人可以帮助我并告诉我我做错了什么吗？

先感谢您

4

3 回答 3

2

我将该文本拉入编辑器。然后制作有效的标题名称并放回显然已被空格替换的选项卡并读入 R 获取此对象：

    MPexp <- structure(list(Name_of_member = c("Adam Afriyie", "Adam Afriyie", 
    "Adam Afriyie", "Adam Holloway", "Adam Holloway", "Adam Holloway", 
    "Adam Holloway", "Adam Holloway", "Adam Holloway", "Adam Holloway", 
    "Adam Holloway"), Allowance_Type = c("Office running costs (IEP/AOE)", 
    "Office running costs (IEP/AOE)", "Office running costs (IEP/AOE)", 
    " Office running costs (IEP/AOE)", " Office running costs (IEP/AOE)", 
    " Office running costs (IEP/AOE)", " Office running costs (IEP/AOE", 
    " Communications Expenditure", " Office running costs (IEP/AOE)", 
    " Office running costs (IEP/AOE", " Office running costs (IEP/AOE)"
    ), Expenditure_Tyoe = c("Incidentals", "Incidentals", "Incidentals", 
    "Incidentals", "Incidentals", "Incidentals", "IT equipment", 
    "Publications", "Incidentals", "Incidentals", "Incidentals"), 
        Date = c("07/03/09", "11/05/09", "11/05/09", "10/01/09", 
        "10/01/09", "10/01/09", "07/03/09", "12/04/09", "12/08/09", 
        "12/08/09", "12/08/09"), Amount = c(111.09, 111.09, 51.75, 
        35, 413.23, 9.55, 890.01, 1774, 1.1, 64.31, 64.31)), .Names = c("Name_of_member", 
    "Allowance_Type", "Expenditure_Tyoe", "Date", "Amount"), 
class = "data.frame", row.names = c(NA, 
    -11L))

现在这应该会产生带有聚合的预期结果：

> aggregate(MPexp$Amount, MPexp["Name_of_member"], sum)
  Name_of_member       x
1   Adam Afriyie  273.93
2  Adam Holloway 3251.51

再次阅读您的问题让我意识到您正在使用 aggregate.formula 所以这也适用于该数据：

> aggregate(Amount ~ Name_of_member, data=MPexp, FUN=sum)
  Name_of_member  Amount
1   Adam Afriyie  273.93
2  Adam Holloway 3251.51
于 2010-12-02T17:06:26.103 回答
1

另一种方法使用plyr

library(plyr)

#Using data from mropa's answer
> ddply(df, .(name), summarise, sum = sum(amount))
           name     sum
1  Adam Afriyie  273.93
2 Adam Holloway 3251.51
于 2010-12-03T02:50:34.147 回答
1

仅使用您的姓名列和最后一个金额列：

df <- data.frame(name = c(rep("Adam Afriyie", 3), rep("Adam Holloway", 8)),
                 amount = c(111.09, 111.09, 51.75, 35,
                   413.23, 9.55, 890.01, 1774, 1.1, 64.31, 64.31)
                 )

版本 1

aggregate(df$amount, by = list(name = df$name), FUN = "sum")

第 2 版

aggregate(amount ~ name, data = df, FUN = "sum")

输出：

1  Adam Afriyie  273.93
2  Adam Holloway 3251.51
于 2010-12-02T17:02:37.073 回答