R中具有特定ID的條目數量的平均值？

Question

因此，假設我有一個數據集，看起來像這一點，我與R工作組：

player  at_bat opponent_name  game result 
Torri_Hunter 1 Pittsburgh Pirates 1 home run 
Torri_Hunter 2 Pittsburgh Pirates 1 triple 
Torri_Hunter 3 Pittsburgh Pirates 1 strikeout 
Torri_Hunter 4 Pittsburgh Pirates 1 strikeout 
Torri_Hunter 1 Pittsburgh Pirates 2 groundout 
Torri_Hunter 2 Pittsburgh Pirates 2 home run 
Torri_Hunter 3 Pittsburgh Pirates 2 flyout 
Torri_Hunter 1 Pittsburgh Pirates 2 home run 
Torri_Hunter 2 Pittsburgh Pirates 3 triple 
Torri_Hunter 3 Pittsburgh Pirates 3 strikeout 
Torri_Hunter 4 Pittsburgh Pirates 3 strikeout 
Torri_Hunter 1 Detroit Tigers  1 home run 
Torri_Hunter 2 Detroit Tigers  1 home run 
Torri_Hunter 3 Detroit Tigers  1 home run 
Torri_Hunter 4 Detroit Tigers  1 strikeout 

（我知道鳥居的名字被拼寫錯了，多包涵在這裏）。

而我最終想通過遊戲來計算本壘打的百分比在一個系列，有一些看起來像這樣結束了：

   opponent_name  game_1s game_2s game_3s 
Torri Hunter Pittsburgh Pirates 25%  50%  0% 
Torri Hunter Detroit Tigers  75%  --  -- 

我可以dplyr ::過濾下來的結果，理貨（）各遊戲的ID統計，然後導出到.csv，我可以在excel中獲得平均值（這就是我一直在做的），但是在R中完全可以做到這一點。任何想法？

Answer 1

你可以這樣做：

library(dplyr) 
df %>% 
    group_by(player, opponent_name, game) %>% 
    summarise(p = sum(result == "home run")/n()) 

其中給出：

#Source: local data frame [4 x 4] 
#Groups: player, opponent_name 
# 
#  player  opponent_name game p 
#1 Torri_Hunter  Detroit Tigers 1 0.75 
#2 Torri_Hunter Pittsburgh Pirates 1 0.25 
#3 Torri_Hunter Pittsburgh Pirates 2 0.50 
#4 Torri_Hunter Pittsburgh Pirates 3 0.00 

以匹配所需的輸出，你也可以這樣做：

df %>% 
    group_by(player, opponent_name, game) %>% 
    summarise(p = mean(result == "home run")) %>% 
    tidyr::spread(game, p) %>% 
    arrange(desc(opponent_name)) %>% 
    setNames(c(names(.)[1:2], paste0("game_", names(.)[3:5], "s"))) %>% 
    mutate_each(funs(ifelse(is.na(.), "--", paste0(. * 100, "%"))), -(player:opponent_name)) 

其中給出：

#Source: local data frame [2 x 5] 
# 
#  player  opponent_name game_1s game_2s game_3s 
#1 Torri_Hunter Pittsburgh Pirates  25%  50%  0% 
#2 Torri_Hunter  Detroit Tigers  75%  --  -- 

Answer 2

如何編寫兩個函數來幫助你？假設你的數據框是調用df。

perc_res <- function(opponent, game="1" player="Torri_Hunter", result="home run"){ 
    return(
    dim(df[df$player==player & df$opponent==opponent & df$result==result & df$game==game,])[1]/ 
     dim(df[df$player==player & df$opponent==opponent & df$game==game,])[1] 
) 
} 

然後可以使輸出數據框，看起來像

out.df <- data.frame(Opponent=levels(factor(df$opponent)), Player="Torri_Hunter") 
out.df$game1s <- lapply(out.df$Opponent, perc_res, game=1) 

等 如果以後希望有更多的玩家，你可以使用mapply。

ps：實際上還沒有運行代碼，所以可能仍然存在一些常見的錯誤。但我認爲這至少應該讓你開始！

Answer 3

甲data.table溶液澆鑄將是

require(data.table) 
setDT(dat) 
percentage <- dat[,mean(result == "home run"), by = c("player", "opponent_name", "game")] 

結果：

> percentage 

     player  opponent_name game V1 
1: Torri_Hunter Pittsburgh Pirates 1 0.25 
2: Torri_Hunter Pittsburgh Pirates 2 0.50 
3: Torri_Hunter Pittsburgh Pirates 3 0.00 
4: Torri_Hunter  Detroit Tigers 1 0.75 

它投射到如問題

require(reshape2) 
dcast(percentage, player + opponent_name ~ game , value.var = "V1") 

結果所需的輸出：

 player  opponent_name 1 2 3 
1 Torri_Hunter  Detroit Tigers 0.75 NA NA 
2 Torri_Hunter Pittsburgh Pirates 0.25 0.5 0