根據R中的列匯總數據幀

Question

我正在嘗試針對每個城市的總和提供以下數據框：

> summary(dat1)
      Date                 City           Sales        
 Min.   :2010-06-18   Min.   : 1.00   Min.   :  667.4  
 1st Qu.:2011-02-18   1st Qu.:18.00   1st Qu.: 1138.6  
 Median :2011-10-28   Median :37.00   Median : 1507.5  
 Mean   :2011-10-29   Mean   :44.26   Mean   : 2065.4  
 3rd Qu.:2012-07-06   3rd Qu.:74.00   3rd Qu.: 2347.1  
 Max.   :2013-03-08   Max.   :99.00   Max.   :47206.6 

即我想找到具有相應的日期X城市觀察值的數據框，該數據框將顯示每個城市在每一天的銷售總額。

Answer 1

有幾種可能性。 僅舉幾例：

1. 函數aggregate（）：

   i） aggregate(Sales~Date+City, data=df, sum)

   ii） aggregate(df$Sales, list(df$Date,df$City), sum)

2. 函數tapply（）：

   i） tapply(df$Sales, list(df$Date, df$City), sum)

如果您的數據集很大，則函數tapply()尤其有用，因為聚合可能會阻塞非常大的數據集，但是tapply()通常會更tapply()處理這些數據。 另外， tapply()和aggregate()以不同的格式生成輸出，並且您可能希望選擇最適合進行進一步分析的輸出。

這些示例可以在下面顯示的模擬數據上進行測試：

df<-structure(list(Date = structure(c(4L, 2L, 4L, 2L, 3L, 4L, 3L, 
2L, 2L, 2L, 2L, 4L, 1L, 4L, 2L, 4L, 2L, 3L, 4L, 2L, 3L, 3L, 4L, 
3L, 4L, 2L, 2L, 2L, 3L, 1L, 1L, 4L, 2L, 4L, 1L, 2L, 1L, 2L, 3L, 
2L, 2L, 3L, 2L, 1L, 1L, 3L, 2L, 1L, 1L, 3L, 3L, 1L, 3L, 1L, 1L, 
1L, 3L, 2L, 3L, 1L, 3L, 3L, 2L, 2L, 4L, 2L, 1L, 3L, 3L, 1L, 4L, 
1L, 2L, 2L, 1L, 2L, 2L, 2L), .Label = c("2014-01-01", "2014-02-01", 
"2014-03-01", "2014-04-01"), class = "factor"), City = structure(c(1L, 
2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 
16L, 17L, 18L, 19L, 20L, 21L, 22L, 23L, 24L, 25L, 26L, 1L, 2L, 
3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 
17L, 18L, 19L, 20L, 21L, 22L, 23L, 24L, 25L, 26L, 1L, 2L, 3L, 
4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 
18L, 19L, 20L, 21L, 22L, 23L, 24L, 25L, 26L), .Label = c("a", 
"b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", 
"o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"), class = "factor"), 
    Sales = c(100, 100, 93, 92, 95, 115, 104, 106, 113, 94, 93, 
    98, 116, 85, 98, 97, 103, 110, 105, 104, 107, 86, 92, 94, 
    106, 115, 112, 92, 103, 100, 101, 97, 95, 110, 103, 92, 91, 
    98, 100, 93, 108, 87, 96, 101, 87, 111, 90, 94, 110, 95, 
    110, 101, 88, 99, 106, 117, 101, 120, 92, 86, 118, 104, 99, 
    89, 103, 102, 121, 99, 106, 99, 107, 105, 109, 110, 112, 
    94, 100, 112)), .Names = c("Date", "City", "Sales"), row.names = c(NA, 
    -78L), class = "data.frame")

Answer 2

請參閱aggregation功能

aggregate(Sales~Date+City, data=dat1, sum)