我正在使用類似于以下內容的廣泛資料集：
我正在尋找一個函式，我可以迭代具有相似名稱但名稱不同的列集。就函式本身而言，為了簡單起見，我將創建一個取兩列平均值的函式。
avg <- function(data, scorecol, distcol) {
ScoreDistanceAvg = (scorecol distcol)/2
data$ScoreDistanceAvg <- ScoreDistanceAvg
return(data)
}
avg(data = dat, scorecol = dat$ScoreGame0, distcol = dat$DistanceGame0)
如何將新函式應用于名稱重復但數字不同的列集？也就是說，如何創建一個取 ScoreGame0 和 DistanceGame0 均值的列，然后創建一個取 ScoreGame5 和 DistanceGame5 均值的列，等等？這將是最終輸出：
當然，我可以多次運行該函式，但由于我的完整資料集要大得多，我該如何自動化這個程序呢？我想它涉及應用，但我不確定如何將應用與這樣的重復模式一起使用。此外，我想它可能涉及重寫函式以更好地自動化列的命名。
資料：
structure(list(Player = c("Lebron James", "Lebron James", "Lebron James",
"Lebron James", "Lebron James", "Lebron James", "Lebron James",
"Lebron James", "Lebron James", "Lebron James", "Lebron James",
"Lebron James", "Steph Curry", "Steph Curry", "Steph Curry",
"Steph Curry", "Steph Curry", "Steph Curry", "Steph Curry", "Steph Curry",
"Steph Curry", "Steph Curry", "Steph Curry", "Steph Curry"),
Game = c(0L, 1L, 2L, 3L, 4L, 5L, 0L, 1L, 2L, 3L, 4L, 5L,
0L, 1L, 2L, 3L, 4L, 5L, 0L, 1L, 2L, 3L, 4L, 5L), ScoreGame0 = c(32L,
32L, 32L, 32L, 32L, 32L, 44L, 44L, 44L, 44L, 44L, 44L, 45L,
45L, 45L, 45L, 45L, 45L, 76L, 76L, 76L, 76L, 76L, 76L), ScoreGame5 = c(27L,
27L, 27L, 27L, 27L, 27L, 12L, 12L, 12L, 12L, 12L, 12L, 76L,
76L, 76L, 76L, 76L, 76L, 32L, 32L, 32L, 32L, 32L, 32L), DistanceGame0 = c(12L,
12L, 12L, 12L, 12L, 12L, 79L, 79L, 79L, 79L, 79L, 79L, 18L,
18L, 18L, 18L, 18L, 18L, 88L, 88L, 88L, 88L, 88L, 88L), DistanceGame5 = c(13L,
13L, 13L, 13L, 13L, 13L, 34L, 34L, 34L, 34L, 34L, 34L, 42L,
42L, 42L, 42L, 42L, 42L, 54L, 54L, 54L, 54L, 54L, 54L)), class = "data.frame", row.names = c(NA,
-24L))
uj5u.com熱心網友回復：
稍微重寫您的函式并通過在列上使用mapply它grep。sort使這更加安全。
avg <- function(scorecol, distcol) {
(scorecol distcol)/2
}
mapply(avg, dat[sort(grep('ScoreGame', names(dat)))], dat[sort(grep('DistanceGame', names(dat)))])
# ScoreGame0 ScoreGame5
# [1,] 22.0 20
# [2,] 22.0 20
# [3,] 22.0 20
# [4,] 22.0 20
# [5,] 22.0 20
# [6,] 22.0 20
# [7,] 61.5 23
# [8,] 61.5 23
# [9,] 61.5 23
# [10,] 61.5 23
# [11,] 61.5 23
# [12,] 61.5 23
# [13,] 31.5 59
# [14,] 31.5 59
# [15,] 31.5 59
# [16,] 31.5 59
# [17,] 31.5 59
# [18,] 31.5 59
# [19,] 82.0 43
# [20,] 82.0 43
# [21,] 82.0 43
# [22,] 82.0 43
# [23,] 82.0 43
# [24,] 82.0 43
看看有什么grep嘗試
grep('DistanceGame', names(dat), value=TRUE)
# [1] "DistanceGame0" "DistanceGame5"
uj5u.com熱心網友回復：
這是一個帶有 forloop 和的解決方案dplyr：
library(dplyr)
library(readr)
game_num <- names(dat) %>%
readr::parse_number() %>%
na.omit()
for(i in unique(game_num)) {
avg <- paste0("ScoreDistanceAvg", i)
score <- paste0("ScoreGame", i)
distance <- paste0("DistanceGame", i)
dat[[avg]] <- (dat[[score]] dat[[distance]])/2
}
這使：
Player Game ScoreGame0 ScoreGame5 DistanceGame0 DistanceGame5 ScoreDistanceAvg0 ScoreDistanceAvg5
1 Lebron James 0 32 27 12 13 22.0 20
2 Lebron James 1 32 27 12 13 22.0 20
3 Lebron James 2 32 27 12 13 22.0 20
4 Lebron James 3 32 27 12 13 22.0 20
5 Lebron James 4 32 27 12 13 22.0 20
6 Lebron James 5 32 27 12 13 22.0 20
7 Lebron James 0 44 12 79 34 61.5 23
8 Lebron James 1 44 12 79 34 61.5 23
9 Lebron James 2 44 12 79 34 61.5 23
10 Lebron James 3 44 12 79 34 61.5 23
11 Lebron James 4 44 12 79 34 61.5 23
12 Lebron James 5 44 12 79 34 61.5 23
13 Steph Curry 0 45 76 18 42 31.5 59
uj5u.com熱心網友回復：
在基礎 R 中：
cols_used <- names(df[, -(1:2)])
f <- sub("[^0-9] ", 'ScoreDistance', cols_used)
data.frame(lapply(split.default(df[cols_used], f), rowMeans))
ScoreDistance0 ScoreDistance5
1 22.0 20
2 22.0 20
3 22.0 20
4 22.0 20
5 22.0 20
6 22.0 20
7 61.5 23
8 61.5 23
9 61.5 23
10 61.5 23
11 61.5 23
12 61.5 23
13 31.5 59
14 31.5 59
15 31.5 59
16 31.5 59
17 31.5 59
18 31.5 59
19 82.0 43
20 82.0 43
21 82.0 43
22 82.0 43
23 82.0 43
24 82.0 43
使用 tidyverse：
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