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我有一个数据框，其中包含按国家/地区列出的 GDP 值以及随附的日期列。以下代码复制了两个国家（法国和德国）和六年（2005-2010）的示例数据集：

df <- structure(list(geo = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 
                2L, 2L, 2L, 2L, 2L), .Label = c("DE", "FR"), class = "factor"), 
                 date = structure(c(12784, 13149, 13514, 13879, 14245, 14610, 
                 12784, 13149, 13514, 13879, 14245, 14610), class = "Date"), 
                 GDP = c(2147975, 2249584.4, 2373993.1, 2382892.6, 2224501.8, 
                2371033.2, 1557584.8, 1621633.2, 1715655.4, 1713157.1, 1636336.3, 
               1707966.5)), .Names = c("geo", "date", "GDP"), row.names = c(NA, 
              -12L), class = "data.frame")

现在我想计算一个额外的列，显示每年的百分比差异。我尝试以下方法：

library(quantmod) 
# provides the Delt() function to calculate percent differences

df$dtGDP <- as.numeric(Delt(df$GDP))

这是错误的，因为它使用 2010 年的 DE 值计算 2005 年的 FR 值。有没有办法应用“每个因子水平”函数？

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2 回答 2

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> df$dtGDP <-with(df, ave(GDP, geo, FUN=Delt))
> df
   geo       date     GDP        dtGDP
1   DE 2005-01-01 2147975           NA
2   DE 2006-01-01 2249584  0.047304741
3   DE 2007-01-01 2373993  0.055302971
4   DE 2008-01-01 2382893  0.003748747
5   DE 2009-01-01 2224502 -0.066469970
6   DE 2010-01-01 2371033  0.065871558
7   FR 2005-01-01 1557585           NA
8   FR 2006-01-01 1621633  0.041120329
9   FR 2007-01-01 1715655  0.057979943
10  FR 2008-01-01 1713157 -0.001456178
11  FR 2009-01-01 1636336 -0.044841655
12  FR 2010-01-01 1707966  0.043774742
于 2012-11-21T00:55:12.493 回答
2

试试这个：

foo <- aggregate(GDP~geo, df, function(x) list(Delt(x)))
df <- cbind(df, dtGDP = as.numeric(unlist(foo[,-1])))
df

假设你已经运行了这个：

library(quantmod) 
df <- structure(list(geo = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 2L, 
                2L, 2L, 2L, 2L, 2L), .Label = c("DE", "FR"), class = "factor"), 
                date = structure(c(12784, 13149, 13514, 13879, 14245, 14610, 
                12784, 13149, 13514, 13879, 14245, 14610), class = "Date"), 
                GDP = c(2147975, 2249584.4, 2373993.1, 2382892.6, 2224501.8, 
                2371033.2, 1557584.8, 1621633.2, 1715655.4, 1713157.1, 1636336.3, 
                1707966.5)), .Names = c("geo", "date", "GDP"), row.names = c(NA, 
                -12L), class = "data.frame")
于 2012-11-20T23:12:17.023 回答