Question

这是我的第一个Stack Overflow帖子。我researched extensively，但没有找到类似的帖子。

我试图根据两个条件来估算NA值的中位数。

这是我的代码：

#Create sample of original data for reproducibility
Date<-c("2009-05-01","2009-05-02","2009-05-03","2009-06-01","2009-06-02",
        "2009-06-03", "2010-05-01","2010-05-02","2010-05-03","2010-06-01",
        "2010-06-02","2010-06-03","2011-05-01","2011-05-02","2011-05-03",
        "2011-06-01","2011-06-02","2011-06-03")
Month<- c("May","May","May","June","June","June",
          "May","May","May","June","June","June",
          "May","May","May","June","June","June")
DayType<- c("Monday","Tuesday","Wednesday","Monday","Tuesday","Wednesday",
            "Monday","Tuesday","Wednesday","Monday","Tuesday","Wednesday",
            "Monday","Tuesday","Wednesday","Monday","Tuesday","Wednesday")
Qty<- c(NA,NA,NA,NA,NA,NA,
        1,2,1,10,15,13,
        3,2,5,20,14,16)

#Combine into dataframe
Example<-data.frame(Date,Month,DayType,Qty)

#Test output
Example

# Make a separate dataframe to calculate the median value based on day of the month
test1 <- ddply(Example,. (DayType,Month),summarize,median=median(Qty,na.rm=TRUE))

这按预期工作。 Test1输出看起来像这样：

DayType   Month  Median
Monday    June   15.0
Monday    May    2.0
Tuesday   June   14.5
Tuesday   May    2.0
Wednesday June   14.5
Wednesday May    3.0

第二步，将原始数据集中的“ NA”值替换为在test1中计算出的中位数。这就是我的问题所在。

Example$Qty[is.na(Example$Qty)] <- test1$median[match(Example$DayType,test1$DayType,Example$Month,test1$Month)][is.na(Example$Qty)]

示例

Match []仅匹配每天的中间值，而不是逐月每天的中间值。对于整个集合，输出是相同的七个重复值。我还没有弄清楚如何同时在两列上进行匹配。

Output:
Date         DayType   Month   GSEvtQty
2009-05-01   Monday    May     15.0    *should be 2.0, matching to June
2009-05-02   Tuesday   May     14.5    *should be 2.0, matching to June
2009-05-03   Wednesday May     14.5    *should be 3.0, matching to June
2009-06-01   Monday    June    15.0    *imputes correctly
2009-06-02   Tuesday   June    14.5    *imputes correctly
2009-06-03   Wednesday June    14.5    *imputes correctly
2010-05-01   Monday    May     1.0     
2010-05-02   Tuesday   May     2.0  
2010-05-03   Wednesday May     1.0 
2010-06-01   Monday    June    10.0
2010-06-02   Tuesday   June    15.0  
2010-06-03   Wednesday June    13.0

我也尝试使用％in％：

Example$Qty[is.na(Example$Qty)] <- test1$median[Example$DayType %in% test1$DayType & Example$Month %in% test1$Month][is.na(Example$Qty)]

但这不能正确匹配，只能输出有限数量的值，而不能输出整个NA系列。

通过@Jaap的巧妙建议，通过Zoo软件包使用na.aggregate：

setDT(Example)[, Value := na.aggregate("Qty", FUN = median), by = c("DayType","Month")]

由于某些原因不能改变NA：

Output:
 Date         Month   DayType   Qty
 2009-05-01   May     Monday    NA
 2009-05-02   May     Tuesday   NA
 2009-05-03   May     Wednesday NA
 2009-06-01   June    Monday    NA

任何建议将不胜感激！感谢您长期担任此职位，并希望将来能为您提供帮助。

Answer 1

这就是创建merge的目的。

info$GSEvtQty[is.na(info$GSEvtQty)]<- merge(info[is.na(info$GSEvtQty,)], test1, by=c("DayType", "Month"))[,"GSEvtQty"]

根据多个条件将NA替换为中位数

1 个答案: