我有一个数据集如下作为样本。我的实际数据集有5000列：
# Define Adstock Rate
adstock_rate = 0.50
lag_number = 3
# Create Data
advertising = c(117.913, 120.112, 125.828, 115.354, 177.090, 141.647, 137.892, 0.000, 0.000, 0.000, 0.000,
0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 158.511, 109.385, 91.084, 79.253, 102.706,
78.494, 135.114, 114.549, 87.337, 107.829, 125.020, 82.956, 60.813, 83.149, 0.000, 0.000,
0.000, 0.000, 0.000, 0.000, 129.515, 105.486, 111.494, 107.099, 0.000, 0.000, 0.000,
0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000,
134.913, 123.112, 178.828, 112.354, 100.090, 167.647, 177.892, 0.000, 0.000, 0.000, 0.000,
0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 112.511, 155.385, 123.084, 89.253, 67.706,
23.494, 122.114, 112.549, 65.337, 134.829, 123.020, 81.956, 23.813, 65.149, 0.000, 0.000,
0.000, 0.000, 0.000, 0.000, 145.515, 154.486, 121.494, 117.099, 0.000, 0.000, 0.000,
0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000
)
advertising2 = c(43.913, 231.112, 76.828, 22.354, 98.090, 123.647, 90.892, 0.000, 0.000, 0.000, 0.000,
0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 234.511, 143.385, 78.084, 89.253, 12.706,
34.494, 56.114, 78.549, 12.337, 67.829, 42.020, 90.956, 23.813, 83.149, 0.000, 0.000,
0.000, 0.000, 0.000, 0.000, 52.515, 76.486, 89.494, 12.099, 0.000, 0.000, 0.000,
0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000,
67.913, 12.112, 45.828, 78.354, 89.090, 90.647, 23.892, 0.000, 0.000, 0.000, 0.000,
0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 78.511, 23.385, 43.084, 67.253, 33.706,
56.494, 78.114, 98.549, 45.337, 31.829, 67.020, 87.956, 94.813, 65.149, 0.000, 0.000,
0.000, 0.000, 0.000, 0.000, 55.515, 32.486, 78.494, 33.099, 0.000, 0.000, 0.000,
0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000, 0.000
)
Region = c(500, 500, 500, 500, 500, 500, 500, 500,500, 500, 500, 500,500, 500, 500, 500,500, 500, 500, 500,500, 500, 500, 500,
500, 500, 500, 500,500, 500, 500, 500,500, 500, 500, 500,500, 500, 500, 500,500, 500, 500, 500,500, 500, 500, 500, 500, 500,
500, 500,
501, 501, 501, 501, 501, 501, 501, 501,501, 501, 501, 501,501, 501, 501, 501,501, 501, 501, 501,501, 501, 501, 501,
501, 501, 501, 501,501, 501, 501, 501,501, 501, 501, 501,501, 501, 501, 501,501, 501, 501, 501,501, 501, 501, 501, 501, 501,
501, 501)
advertising_dataset<-data.frame(cbind(Region, advertising, advertising2))
我的数据集如下所示：
head(advertising_dataset, 15)
Region advertising advertising2
1 500 117.913 43.913
2 500 120.112 231.112
3 500 125.828 76.828
4 500 115.354 22.354
5 500 177.090 98.090
6 500 141.647 123.647
7 500 137.892 90.892
8 500 0.000 0.000
9 500 0.000 0.000
10 500 0.000 0.000
11 500 0.000 0.000
12 500 0.000 0.000
13 500 0.000 0.000
14 500 0.000 0.000
15 500 0.000 0.000
然后，只为 1 列创建for循环，然后在Region之后创建group_by函数。
foo <- function(df_, lag_val = 1) {
df_$adstocked_advertising = df_$advertising
for (i in (1 + lag_val):nrow(df_)) {
df_$adstocked_advertising[i] = df_$advertising[i] + adstock_rate *
df_$adstocked_advertising[i - lag_val]
}
return(df_)
}
adv_2 <- data.frame(advertising_dataset %>%
group_by(Region) %>%
do(foo(data.frame(.), lag_val = 3)))
如何将上述功能（包括adv_2）应用于2：ncol（advertising_dataset）的所有列，而不仅仅是advertising列？
我的最终列数最终会加倍，因为将为每个现有列创建一个新修订的列。
我有一种感觉，就像上面的函数一样：
data.frame(advertising_dataset[1],
apply(advertising_dataset[2:ncol(advertising_dataset)],2, foo) )
任何帮助都会很棒，谢谢！
答案 0 :(得分：2)
我们可以将accumulate与mutate_all
library(tidyverse)
out <- advertising_dataset %>%
group_by(Region) %>%
mutate_all(funs(adstocked = accumulate(., ~ .y + adstock_rate * .x)))
out
# A tibble: 104 x 5
# Groups: Region [2]
# Region advertising advertising2 advertising_adstocked advertising2_adstocked
# <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 500 118. 43.9 118. 43.9
# 2 500 120. 231. 179. 253.
# 3 500 126. 76.8 215. 203.
# 4 500 115. 22.4 223. 124.
# 5 500 177. 98.1 289. 160.
# 6 500 142. 124. 286. 204.
# 7 500 138. 90.9 281. 193.
# 8 500 0 0 140. 96.4
# 9 500 0 0 70.2 48.2
#10 500 0 0 35.1 24.1
# ... with 94 more rows
检查OP解决方案的输出
head(out[[4]])
#[1] 117.9130 179.0685 215.3623 223.0351 288.6076 285.9508
head(adv_2[[4]])
#[1] 117.9130 179.0685 215.3623 223.0351 288.6076 285.9508
我们可以针对不同的foo
lag_val
foo1 <- function(dot, lag_val = 1) {
tmp <- dot
for(i in (1 + lag_val): length(tmp)) {
tmp[i] <- tmp[i] + adstock_rate * tmp[i - lag_val]
}
return(tmp)
}
advertising_dataset %>%
group_by(Region) %>%
mutate_all(funs(adstocked = foo1(., lag_val = 1)))
# A tibble: 104 x 5
# Groups: Region [2]
# Region advertising advertising2 advertising_adstocked advertising2_adstocked
# <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 500 118. 43.9 118. 43.9
# 2 500 120. 231. 179. 253.
# 3 500 126. 76.8 215. 203.
# 4 500 115. 22.4 223. 124.
# 5 500 177. 98.1 289. 160.
# 6 500 142. 124. 286. 204.
# 7 500 138. 90.9 281. 193.
# 8 500 0 0 140. 96.4
# 9 500 0 0 70.2 48.2
#10 500 0 0 35.1 24.1
# ... with 94 more rows
- 更改lag_val
advertising_dataset %>%
group_by(Region) %>%
mutate_all(funs(adstocked = foo1(., lag_val = 2)))
# A tibble: 104 x 5
# Groups: Region [2]
# Region advertising advertising2 advertising_adstocked advertising2_adstocked
# <dbl> <dbl> <dbl> <dbl> <dbl>
# 1 500 118. 43.9 118. 43.9
# 2 500 120. 231. 120. 231.
# 3 500 126. 76.8 185. 98.8
# 4 500 115. 22.4 175. 138.
# 5 500 177. 98.1 269. 147.
# 6 500 142. 124. 229. 193.
# 7 500 138. 90.9 273. 165.
# 8 500 0 0 115. 96.3
# 9 500 0 0 136. 82.3
#10 500 0 0 57.3 48.2
# ... with 94 more rows
答案 1 :(得分：0)
使用data.table，您可以使用j语句中的lapply将任何函数应用于列的子集。但是，这些函数将作为向量作用于列，因此您可能希望将F1编辑为任何向量的通用。但是，基本上，你所要做的就是：
foo()