我有一个简单的数据集，其格式如下
df<- data.frame(c(10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20),
c(80, 80, 80, 80, 80, 80, 80, 80, 90, 90, 90, 90, 90, 90, 90, 90, 80, 80, 80, 80, 80, 80, 80, 80, 90, 90, 90, 90, 90, 90, 90, 90),
c(1, 1, 2, 2, 3, 3, 4, 4, 1, 1, 2, 2, 3, 3, 4, 4, 1, 1, 2, 2, 3, 3, 4, 4, 1, 1, 2, 2, 3, 3, 4, 4),
c(25, 75, 20, 40, 60, 50, 20, 10, 20, 30, 40, 60, 25, 75, 20, 40, 5, 5, 2, 4, 6, 5, 2, 1, 2, 3, 4, 6, 2, 7, 2, 4))
colnames(df)<-c("car_number", "year", "marker", "val")
我想要做的很简单，实际上：每car_number，我想找到与marker - 值和前面3个值相关联的值的平均值。因此，对于上面的示例数据，我想要的输出是
car=10, year=80 1: 50
car=10, year=80 2: 40
car=10, year=80 3: 45
car=10, year=80 4: 37.5
car=10, year=90 1: 31.25
car=10, year=90 2: 36.25
car=10, year=90 3: 35
car=10, year=90 4: 38.75
car=20, year=80 1: 5
car=20, year=80 2: 4
car=20, year=80 3: 4.5
car=20, year=80 4: 3.75
car=20, year=90 1: 3.125
car=20, year=90 2: 3.625
car=20, year=90 3: 3.375
car=20, year=90 4: 3.750
请注意，为了简化示例，上面的markers成对出现。实际数据并非如此，所以我认为一般解决方案将包含某种group_by（？）
欢迎任何有效的解决方案！
这是第二个示例数据集和输出：
df<- data.frame(c(10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20, 20),
c(80, 80, 80, 80, 80, 80, 80, 80, 90, 90, 90, 90, 90, 90, 90, 90, 80, 80, 80, 80, 80, 80, 80, 80, 90, 90, 90, 90, 90, 90, 90, 90),
c(1, 2, 2, 2, 3, 3, 4, 4, 1, 1, 2, 2, 3, 3, 3, 4, 1, 1, 1, 2, 3, 3, 4, 4, 4, 1, 2, 2, 3, 3, 3, 4),
c(25, 75, 20, 40, 60, 50, 20, 10, 20, 30, 40, 60, 25, 75, 20, 40, 5, 5, 2, 4, 6, 5, 2, 1, 2, 3, 4, 6, 2, 7, 2, 4))
colnames(df)<-c("car_number", "year", "marker", "val")
输出是（基于上述规则）
car=10, year=80 1: Mean{{25}] = 25
car=10, year=80 2: Mean[{40, 20, 75, 25}] = 40
car=10, year=80 3: Mean[{50, 60, 40, 20, 75, 25}] = 45
car=10, year=80 4: Mean[{10, 20, 50, 60, 40, 20, 75, 25}] = 37.5
car=10, year=90 1: Mean[{30, 20, 10, 20, 50, 60, 40, 20, 75}] = 36.11
car=10, year=90 2: Mean[{60, 40, 30, 20, 10, 20, 50, 60}] = 36.25
car=10, year=90 3: Mean[{20, 75, 25, 60, 40, 30, 20, 10, 20}] = 33.33
car=10, year=90 4: Mean[{40, 20, 75, 25, 60, 40, 30, 20}] = 38.75
car=20, year=80 1: Mean[{2, 5, 5}] = 4
car=20, year=80 2: Mean[{4, 2, 5, 5}] = 4
car=20, year=80 3: Mean[{5, 6, 4, 2, 5, 5}] = 4.5
car=20, year=80 4: Mean[{2, 1, 2, 5, 6, 4, 2, 5, 5}] = 3.55
car=20, year=90 1: Mean[{3, 2, 1, 2, 5, 6, 4}] = 3.29
car=20, year=90 2: Mean[{6, 4, 3, 2, 1, 2, 5, 6}] = 3.625
car=20, year=90 3: Mean[{2, 7, 2, 6, 4, 3, 2, 1, 2}] = 3.22
car=20, year=90 4: Mean[{4, 2, 7, 2, 6, 4, 3}] = 4
答案 0 :(得分：2)
第一个group_by按car_number，year，marker计算均值，并保留其重量（行数）。
group_by的第car_number个允许我们检索lag ging均值和权重以计算所需的均值：
library(purrr)
library(dplyr)
df %>%
arrange(car_number, year, marker) %>%
group_by(car_number, year, marker) %>%
summarise(mean_1 = mean(val, na.rm = TRUE), weight = n()) %>%
group_by(car_number) %>%
mutate(mean_2 = pmap_dbl(
list(mean_1, lag(mean_1), lag(mean_1, 2), lag(mean_1, 3),
weight, lag(weight), lag(weight, 2), lag(weight, 3)),
~ weighted.mean(c(..1, ..2, ..3, ..4),
c(..5, ..6, ..7, ..8),
na.rm = TRUE)
)) %>%
ungroup()
结果：
# # A tibble: 16 × 6
# car_number year marker mean_1 weight mean_2
# <dbl> <dbl> <dbl> <dbl> <int> <dbl>
# 1 10 80 1 50.0 2 50.000
# 2 10 80 2 30.0 2 40.000
# 3 10 80 3 55.0 2 45.000
# 4 10 80 4 15.0 2 37.500
# 5 10 90 1 25.0 2 31.250
# 6 10 90 2 50.0 2 36.250
# 7 10 90 3 50.0 2 35.000
# 8 10 90 4 30.0 2 38.750
# 9 20 80 1 5.0 2 5.000
# 10 20 80 2 3.0 2 4.000
# 11 20 80 3 5.5 2 4.500
# 12 20 80 4 1.5 2 3.750
# 13 20 90 1 2.5 2 3.125
# 14 20 90 2 5.0 2 3.625
# 15 20 90 3 4.5 2 3.375
# 16 20 90 4 3.0 2 3.750
编辑： purrr之前0.2.2.9000版本的替代语法：
df %>%
arrange(car_number, year, marker) %>%
group_by(car_number, year, marker) %>%
summarise(mean_1 = mean(val, na.rm = TRUE), weight = n()) %>%
group_by(car_number) %>%
mutate(mean_2 = pmap_dbl(
list(mean_1, lag(mean_1), lag(mean_1, 2), lag(mean_1, 3),
weight, lag(weight), lag(weight, 2), lag(weight, 3)),
function(a, b, c, d, e, f, g, h)
weighted.mean(c(a, b, c, d),
c(e, f, g, h),
na.rm = TRUE)
)) %>%
ungroup()
答案 1 :(得分：2)
这是一种方法，data.table修改了Frank在David Arenburg的回答here中的建议。
# aggregate data by car_number, year, and marker
dfNew <- setDT(df)[, .(val=mean(val)), by=.(car_number, year, marker)]
# calculate average of current a previous three values
dfNew[, val := rowMeans(dfNew[,shift(val, 0:3), by=car_number][, -1], na.rm=TRUE)]
第一行是标准聚合调用。第二行对链接答案中的rowMeans方法进行了一些更改。 rowMeans被提供了一个移位值的data.table，其中转移由car_number发生（感谢sotos捕获这个），它被链接到一个删除第一列的语句（使用-1），这是car_number列在链的第一部分返回。
返回
car_number year marker val
1: 10 80 1 50.000
2: 10 80 2 40.000
3: 10 80 3 45.000
4: 10 80 4 37.500
5: 10 90 1 31.250
6: 10 90 2 36.250
7: 10 90 3 35.000
8: 10 90 4 38.750
9: 20 80 1 5.000
10: 20 80 2 4.000
11: 20 80 3 4.500
12: 20 80 4 3.750
13: 20 90 1 3.125
14: 20 90 2 3.625
15: 20 90 3 3.375
16: 20 90 4 3.750
答案 2 :(得分：2)
只需在混合物中投入基础R溶液。我们可以使用Reduce和accumulate = TRUE以及tail(x, 4)制作自定义函数，以确保仅包含最后3个观察结果。所有这些在我们对car_type, year, marker设置的数据进行平均后，即
f1 <- function(x){
sapply(Reduce(c, x, accumulate = TRUE), function(i) mean(tail(i,4)))
}
dd <- aggregate(val ~ car_number+year+marker, df, mean)
dd <- dd[order(dd$car_number, dd$year, dd$marker),]
dd$new_avg <- with(dd, ave(val, car_number, FUN = f1))
dd
# car_number year marker val new_avg
#1 10 80 1 50.0 50.000
#5 10 80 2 30.0 40.000
#9 10 80 3 55.0 45.000
#13 10 80 4 15.0 37.500
#3 10 90 1 25.0 31.250
#7 10 90 2 50.0 36.250
#11 10 90 3 50.0 35.000
#15 10 90 4 30.0 38.750
#2 20 80 1 5.0 5.000
#6 20 80 2 3.0 4.000
#10 20 80 3 5.5 4.500
#14 20 80 4 1.5 3.750
#4 20 90 1 2.5 3.125
#8 20 90 2 5.0 3.625
#12 20 90 3 4.5 3.375
#16 20 90 4 3.0 3.750
答案 3 :(得分：0)
考虑df作为您的输入，您可以使用dplyr和zoo并尝试：
仅在car_number上进行分组，您可以尝试：
df %>%
group_by(car_number, year, marker) %>%
summarise(mm = mean(val)) %>%
group_by(car_number) %>%
mutate(rM=rollapply(mm, if_else(row_number() < 4, marker, 4), FUN=mean, align="right"))%>%
select(year, rM)
给出：
Source: local data frame [16 x 3]
Groups: car_number [2]
car_number year rM
<dbl> <dbl> <dbl>
1 10 80 50.000
2 10 80 40.000
3 10 80 45.000
4 10 80 37.500
5 10 90 31.250
6 10 90 36.250
7 10 90 35.000
8 10 90 38.750
9 20 80 5.000
10 20 80 4.000
11 20 80 4.500
12 20 80 3.750
13 20 90 3.125
14 20 90 3.625
15 20 90 3.375
16 20 90 3.750
答案 4 :(得分：0)
你可以这样做：
df %>%
group_by(car_number, year, marker) %>%
summarise(s = sum(val), w = n()) %>% # sum and number of values
group_by(car_number) %>%
mutate(S = cumsum(s) - cumsum(lag(s, 4, default=0))) %>% # sum of last four s
mutate(W = cumsum(w) - cumsum(lag(w, 4, default=0))) %>% # same for the weights
mutate(result = S/W)
第二个例子的输出：
# Source: local data frame [16 x 8]
# Groups: car_number [2]
#
# car_number year marker s w S W result
# <dbl> <dbl> <dbl> <dbl> <int> <dbl> <int> <dbl>
# 1 10 80 1 25 1 25 1 25.000000
# 2 10 80 2 135 3 160 4 40.000000
# 3 10 80 3 110 2 270 6 45.000000
# 4 10 80 4 30 2 300 8 37.500000
# 5 10 90 1 50 2 325 9 36.111111
# 6 10 90 2 100 2 290 8 36.250000
# 7 10 90 3 120 3 300 9 33.333333
# 8 10 90 4 40 1 310 8 38.750000
# 9 20 80 1 12 3 12 3 4.000000
# 10 20 80 2 4 1 16 4 4.000000
# 11 20 80 3 11 2 27 6 4.500000
# 12 20 80 4 5 3 32 9 3.555556
# 13 20 90 1 3 1 23 7 3.285714
# 14 20 90 2 10 2 29 8 3.625000
# 15 20 90 3 11 3 29 9 3.222222
# 16 20 90 4 4 1 28 7 4.000000
修改：
使用包RcppRoll可能效率更高，您可以尝试：S = roll_sum(c(0, 0, 0, s), 4)（W也一样）。