如何使用 R 处理数据

Question

我正在尝试进行预测分析，其中 x 年的指标将预测 x+1 年。

我想使用 R 在 SQL 中执行相当于 UPDATE 查询的操作。如果我有这样的数据：

x <- c("Randy Watson", "Cleo McDowell", "Darryl Jenks", "Jaffe Joffer",
       "Randy Watson", "Cleo McDowell", "Darryl Jenks", "Jaffe Joffer",
       "Randy Watson", "Cleo McDowell", "Darryl Jenks", "Jaffe Joffer")
y <- c("2012", "2012", "2012", "2012", 
       "2013", "2013", "2013", "2013", 
       "2014", "2014", "2014", "2014")
z <- c(100, 50, 75, 0, 
       110, 75, 0, 25, 
       125, 25, 10, 50)

df <- data.frame(x, y, z)

colnames(df) <- c("Name", "Year", "Sales")

print(df)

            Name Year Sales
1   Randy Watson 2012   100
2  Cleo McDowell 2012    50
3   Darryl Jenks 2012    75
4   Jaffe Joffer 2012     0
5   Randy Watson 2013   110
6  Cleo McDowell 2013    75
7   Darryl Jenks 2013     0
8   Jaffe Joffer 2013    25
9   Randy Watson 2014   125
10 Cleo McDowell 2014    25
11  Darryl Jenks 2014    10
12  Jaffe Joffer 2014    50

我希望最终输出看起来像：

print(df)
           Name YearX YearX1
1  Randy Watson   100   110
2 Cleo McDowell    50    75
3  Darryl Jenks    75     0
4  Jaffe Joffer     0    25

...

如何在 R 中做到这一点？我知道如何在 SQL 中做到这一点（尽管我宁愿不使用 sqldf 除非它是最好的方法）。

谢谢。

编辑： 下面的解决方案并不是我想要的。虽然如果只有两年它可以工作，但我的数据有 10 年。我不需要 Name、Year1、Year2、Year3 等...我只需要 Name、YearX、YearX+1。抱歉，如果我对此不清楚。

Answer 1

像这样

# I took the liberty of rearranging your working example a bit   
df <- data.frame(
           Name = c("Randy Watson", "Cleo McDowell", "Darryl Jenks", "Jaffe Joffer",
                    "Randy Watson", "Cleo McDowell", "Darryl Jenks", "Jaffe Joffer"),
           Year = c("2013", "2013", "2013", "2013", "2014", "2014", "2014", "2014"),
           Sales = c(100, 50, 75, 0, 110, 75, 0, 25))

reshape(df, idvar = "Name", timevar = "Year", direction = "wide")

           Name Sales.2013 Sales.2014
1  Randy Watson        100        110
2 Cleo McDowell         50         75
3  Darryl Jenks         75          0
4  Jaffe Joffer          0         25

或密切关注您的问题

df_wide <- reshape(df, idvar = "Name", timevar = "Year", direction = "wide")

colnames(df_wide) <- c("Name", "Year0", "Year1") 

print(df_wide)
           Name Year0 Year1
1  Randy Watson   100   110
2 Cleo McDowell    50    75
3  Darryl Jenks    75     0
4  Jaffe Joffer     0    25

一些应该给你相同结果的替代方法

library(reshape)
cast(df, Name ~ Year)

xtabs(Sales  ~  Name + Year, data = df)