Question

我有一个示例数据框“数据”如下：

X          Y  Month   Year    income
2281205 228120  3   2011    1000
2281212 228121  9   2010    1100
2281213 228121  12  2010    900
2281214 228121  3   2011    9000
2281222 228122  6   2010    1111
2281223 228122  9   2010    3000
2281224 228122  12  2010    1889
2281225 228122  3   2011    778
2281243 228124  12  2010    1111
2281244 228124  3   2011    200
2281282 228128  9   2010    7889
2281283 228128  12  2010    2900
2281284 228128  3   2011    3400
2281302 228130  9   2010    1200
2281303 228130  12  2010    2000
2281304 228130  3   2011    1900
2281352 228135  9   2010    2300
2281353 228135  12  2010    1333
2281354 228135  3   2011    2340

我使用ddply计算每个Y

的收入

x <- ddply(data, .(Y), summarize, freq=length(Y), tot=sum(income))
#Now, I also need to find out the X for each Y depending upon the following conditions: 
a. If Y consists of observations of months 9 (2010), 12 (2010), and 3 (2011), then the x corresponds to months 9(2010) i.e. for Y =228121 x=2281212
b. If Y consists of observations of month 6 (2010), 9 (2010), 12(2010) , and 3 (2011)  then the x corresponds to months 6 (2010) i.e. for Y =228122 x=2281222. 
c. If Y consists of observations of month 12 (2010), 3 (2011) then the x corresponds to months 12 (2010) i.e. for Y =228124 x=2281243. 
d. If Y consists of observations of month 12 (2010), 3 (2011) then the x corresponds to months 12 (2010) i.e. for Y =228124 x=2281243. 
e. If Y consists of only one observation then the x corresponds to month of that observation i.e. for Y =228120 x=2281205.

这里的要点是，如果我对每个Y有多个观察，我选择x对应于第6个月（2010）（如果可用），但如果不可用，我选择接近6（2010）的月份（例如.9（2010））。请注意，如果我只有一个观察，我将为该观察选择x。

请建议如何在ddply中加入这些条件。

Answer 1

此解决方案假定@DWin选择了最早的X值作为sugessted。 Month和Year变量转换为日期格式，然后选择最早作为选择X的条件。

library(zoo) #necessary for date manipulation
x <- ddply(data, .(Y), summarize, freq=length(Y), tot=sum(income),
 X=X[as.yearmon(paste(Month,Year,sep="/"),format="%m/%Y")==min(as.yearmon(paste(Month,Year,sep="/"),format="%m/%Y"))])

       Y freq   tot       X
1 228120    1  1000 2281205
2 228121    3 11000 2281212
3 228122    4  6778 2281222
4 228124    2  1311 2281243
5 228128    3 14189 2281282
6 228130    3  5100 2281302
7 228135    3  5973 2281352

编辑 - 假设在2010-06之前可能存在值

的解决方案

library(zoo)
#new column containing dates made from Month and Year
data$Time<-as.Date(as.yearmon(paste(data$Month,data$Year,sep="/"),format="%m/%Y"))

#calculated difference between new date column and 2010-06
data$Time.dif<-abs(as.numeric(data$Time-as.Date("2010-06-01")))

#now selects X when Time.dif is smallest (0 in case of 2010-06)
x <- ddply(data, .(Y), summarize, freq=length(Y), tot=sum(income), 
           X=X[Time.dif==min(Time.dif)])

使用ddply生成满足多个条件的值

1 个答案:

编辑 - 假设在2010-06之前可能存在值