r - 按年份简化数据框并计算变化百分比

Question

我有两个问题: 您建议阅读哪些资源来提高数据操作能力？我一直在处理更大的数据集，并且一直在努力适应——我感觉自己碰壁了，不知道该去哪里寻找(许多在线资源在没有打好基础的情况下变得过于复杂)。

例如，我正在尝试解决这个问题。我有一个包含数百万行的 df，我正在尝试简化它并分析趋势。我有一个 dput 示例。我试图隔离每个 ID 并获取给定年份的最小值。 (有些 ID 的年份对其他人不可用)。简化该数据后，我尝试添加百分比变化列。鉴于这是一个 20 多年的时间序列，我可以在此时忽略月份，因为一年的最小值与另一年的最小值相比应该会产生合理的百分比变化。

谢谢!

输入:

structure(list(ID = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L), .Label = c("a", "b"), class = "factor"), Date = structure(c(1L, 
2L, 3L, 4L, 5L, 6L, 10L, 12L, 14L, 7L, 8L, 9L, 11L, 13L, 5L, 
6L, 10L, 12L, 14L, 7L, 8L, 9L, 11L, 13L, 15L, 16L), .Label = c("2/21/2009", 
"2/22/2009", "2/23/2009", "2/24/2009", "2/25/2009", "2/26/2009", 
"3/2/2011", "3/3/2011", "3/4/2011", "3/5/2010", "3/5/2011", "3/6/2010", 
"3/6/2011", "3/7/2010", "3/7/2011", "3/8/2011"), class = "factor"), 
    Year = c(2009L, 2009L, 2009L, 2009L, 2009L, 2009L, 2010L, 
    2010L, 2010L, 2011L, 2011L, 2011L, 2011L, 2011L, 2009L, 2009L, 
    2010L, 2010L, 2010L, 2011L, 2011L, 2011L, 2011L, 2011L, 2011L, 
    2011L), Value = c(10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 
    20, 21, 22, 5, 6, 7, 8, 8, 9, 10, 11, 12, 15, 23, 25, 27)), .Names = c("ID", 
"Date", "Year", "Value"), class = "data.frame", row.names = c(NA, 
-26L))

预期输出:

structure(list(ID = structure(c(1L, 1L, 1L, 2L, 2L, 2L), .Label = c("a", 
"b"), class = "factor"), Date = structure(c(1L, 4L, 5L, 2L, 4L, 
3L), .Label = c("2/21/2009", "2/25/2009", "3/2/2011", "3/5/2010", 
"3/6/2011"), class = "factor"), Year = c(2009L, 2010L, 2011L, 
2009L, 2010L, 2011L), Value = c(10, 16, 5, 6, 8, 10), Percent.Increase = c(NA, 
0.6, -0.6875, NA, 0.333333333, 0.25)), .Names = c("ID", "Date", 
"Year", "Value", "Percent.Increase"), class = "data.frame", row.names = c(NA, 
-6L))

Answer 1

按“ID”、“年份”分组后，我们切片每个组中的min“值”行，然后按“ID”分组，我们创建“Percent.Increase”，方法是从“Value”的lag 中减去“Value”，然后除以“Value”的lag。

res <-  df1 %>%
         group_by(ID, Year) %>%
         slice(which.min(Value)) %>% 
         group_by(ID) %>%
         mutate(Percent.Increase = (Value-lag(Value))/lag(Value))