在滚动中替换NA意味着没有先前观察值等于窗口大小的观测值

Question

structure(list(Score = c(2, 5, 1, 36, 69, 8, 54, 25, 2, 2, 2, 
5, 5, 4, 1), ID = c(1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 
2)), .Names = c("Score", "ID"), row.names = c(NA, -15L), class = 
"data.frame")

假设我采用group_by（window = 3），上述数据的滚动平均值。

new %>% group_by(ID) %>% mutate(roll_mean = rollapply(ID,3,mean,align ='right', 
                                                             fill=NA))

上述陈述的输出是，

structure(list(Score = c(2, 5, 1, 36, 69, 8, 54, 25, 2, 2, 2, 
5, 5, 4, 1), ID = c(1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 
2), roll_mean = c(NA, NA, 1, 1, 1, 1, 1, NA, NA, 2, 2, 2, 2, 
2, 2)), .Names = c("Score", "ID", "roll_mean"), class = c("grouped_df", 
"tbl_df", "tbl", "data.frame"), row.names = c(NA, -15L), vars = "ID", labels 
= structure(list(
ID = c(1, 2)), row.names = c(NA, -2L), class = "data.frame", vars = "ID", 
drop = TRUE, .Names = "ID"), indices = list(
0:6, 7:14), drop = TRUE, group_sizes = 7:8, biggest_group_size = 8L)

我想通过滚动平均值来填写NA，无论窗口大小是多少。我希望输出为：

structure(list(Score = c(2L, 5L, 1L, 36L, 69L, 8L, 54L, 25L, 
2L, 2L, 2L, 5L, 5L, 4L, 1L), ID = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), Roll3 = c(2, 3.5, 2.666666667, 
14, 35.33333333, 37.66666667, 43.66666667, 25, 13.5, 9.666666667, 
2, 3, 4, 4.666666667, 3.333333333)), .Names = c("Score", "ID", 
"Roll3"), class = "data.frame", row.names = c(NA, -15L))

[注意：在我的原始数据中，我有2000万观察结果。如果可以使用data.table执行上述操作，那真的很棒]