你如何计算R中不同年份的相同月份之间的差异

Question

我有这个数据帧叫做df。行是几个月，rowname是年。我想订购从1月到12月的月份，并且想要计算不同年份的相同月份之间的差异（％）。例如，我想知道1月，2月等在2009年和2008年之间的差异百分比。在所有的月份中都这样做。

这是我的DF：

df <- structure(list(YEAR = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 1L, 
2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 
6L, 1L, 2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L, 
4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 6L, 1L, 
2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 6L, 1L, 2L, 3L, 4L, 5L, 
6L), .Label = c("2008", "2009", "2010", "2011", "2012", "2013" 
), class = "factor"), M = structure(c(1L, 1L, 1L, 1L, 1L, 1L, 
2L, 2L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 
4L, 4L, 5L, 5L, 5L, 5L, 5L, 5L, 6L, 6L, 6L, 6L, 6L, 6L, 7L, 7L, 
7L, 7L, 7L, 7L, 8L, 8L, 8L, 8L, 8L, 8L, 9L, 9L, 9L, 9L, 9L, 9L, 
10L, 10L, 10L, 10L, 10L, 10L, 11L, 11L, 11L, 11L, 11L, 11L, 12L, 
12L, 12L, 12L, 12L, 12L), .Label = c("Apr", "Aug", "Dec", "Feb", 
"Jan", "Jul", "Jun", "Mar", "May", "Nov", "Oct", "Sep"), class = "factor"), 
    Freq = c(93221016, 124800455, 224127360, 287150001, 318228530, 
    387573710, 98811936, 171940117, 239581603, 294965702, 336269471, 
    406584525, 112958413, 215853263, 282293439, 314483537, 355561387, 
    386086538, 89354868, 109900379, 206640377, 268944957, 322896485, 
    356774443, 91007916, 113469678, 220743958, 284697404, 324823553, 
    373885187, 96887316, 158230269, 242175673, 284271058, 335464023, 
    397269760, 90091044, 143862802, 232512479, 262275285, 324988644, 
    388064866, 93936288, 139665422, 213302607, 297847827, 329044914, 
    386372600, 99646750, 139195786, 229651074, 277779620, 324395065, 
    397346365, 106477407, 197698621, 256559666, 242683830, 347193478, 
    430880720, 100909236, 185392147, 258317251, 238847338, 349017727, 
    422523576, 96888876, 170467493, 240815506, 285132804, 324063033, 
    389471906)), .Names = c("YEAR", "M", "Freq"), row.names = c(NA, 
-72L), class = "data.frame") 

是否有一个简单的方法来做到这一点，可能是包R'

Answer 1

此命令将增加年之间与不同的新列（百分比）：

transform(df, diff = ave(Freq, M, FUN = function(x) 
    c(0, (diff(x)/head(x, -1)) * 100))) 

Answer 2

试试这个：

library(zoo) 

z <- read.zoo(df, index = 1:2, FUN = function(y, m) as.yearmon(paste(y, m), "%Y %b")) 
diff(z, 12, arithmetic = FALSE) 

或略微更加紧凑（只有##行已经改变）：

library(zoo) 
library(gsubfn) 
z <- fn$read.zoo(df, index = 1:2, FUN = ~ as.yearmon(paste(y, m), "%Y %b")) ## 
diff(z, 12, arithmetic = FALSE) 

增加了紧凑的形式。

Answer 3

在dplyr：

library(dplyr) 
df %.% 
    arrange(M, YEAR) %.% 
    group_by(M) %.% 
    mutate(lag_Freq = lag(Freq), z = (Freq - lag_Freq)/lag_Freq) 

Source: local data frame [72 x 5] 
Groups: M 

    YEAR M  Freq lag_Freq   z 
1 2008 Apr 93221016  NA   NA 
2 2009 Apr 124800455 93221016 0.33875879 
3 2010 Apr 224127360 124800455 0.79588576 
4 2011 Apr 287150001 224127360 0.28119120 
5 2012 Apr 318228530 287150001 0.10823099 
6 2013 Apr 387573710 318228530 0.21791000 
7 2008 Aug 98811936  NA   NA 
8 2009 Aug 171940117 98811936 0.74007437 
9 2010 Aug 239581603 171940117 0.39340142 
10 2011 Aug 294965702 239581603 0.23117008 
11 2012 Aug 336269471 294965702 0.14002906 
12 2013 Aug 406584525 336269471 0.20910329