在R中，如何对data.frame的特定子集执行操作？
（我有一种感觉，在我得到答案后，我会觉得自己很傻，但我就是想不出来。） 我有一个data.frame，末尾有一个空列。它将主要填充NAs，但我想用一个值填充其中的一些行。此列表示对data.frame中某列缺少的数据的猜测 我的初始data.frame如下所示：在R中，如何对data.frame的特定子集执行操作？,r,R,（我有一种感觉，在我得到答案后，我会觉得自己很傻，但我就是想不出来。） 我有一个data.frame，末尾有一个空列。它将主要填充NAs，但我想用一个值填充其中的一些行。此列表示对data.frame中某列缺少的数据的猜测 我的初始data.frame如下所示： Game | Rating | MinPlayers | MaxPlayers | MaxPlayersGuess --------------------------------------------------------- A
Game | Rating | MinPlayers | MaxPlayers | MaxPlayersGuess
---------------------------------------------------------
A | 6 | 3 | 6 |
B | 7 | 3 | 7 |
C | 6.5 | 3 | N/A |median(df$MaxPlayers[df$MinPlayers ==3,])
D | 7 | 3 | 6 |
E | 7 | 3 | 5 |
F | 9.5 | 2 | 5 |
G | 6 | 2 | 4 |
H | 7 | 2 | 4 |
I | 6.5 | 2 | N/A |median(df$MaxPlayers[df$MinPlayers ==2,])
J | 7 | 2 | 2 |
K | 7 | 2 | 4 |
Game | Rating | MinPlayers | MaxPlayers | MaxPlayersGuess
---------------------------------------------------------
A | 6 | 3 | 6 |
B | 7 | 3 | 7 |
C | 6.5 | 3 | N/A |6
D | 7 | 3 | 6 |
E | 7 | 3 | 5 |
F | 9.5 | 2 | 5 |
G | 6 | 2 | 4 |
H | 7 | 2 | 4 |
I | 6.5 | 2 | N/A |4
J | 7 | 2 | 2 |
K | 7 | 2 | 4 |
Game | Rating | MinPlayers | MaxPlayers | MaxPlayersGuess
---------------------------------------------------------
A | 6 | 3 | 6 |
B | 7 | 3 | 7 |
C | 6.5 | 3 | N/A |median(df$MaxPlayers[df$MinPlayers ==3,])
D | 7 | 3 | 6 |
E | 7 | 3 | 5 |
F | 9.5 | 2 | 5 |
G | 6 | 2 | 4 |
H | 7 | 2 | 4 |
I | 6.5 | 2 | N/A |median(df$MaxPlayers[df$MinPlayers ==2,])
J | 7 | 2 | 2 |
K | 7 | 2 | 4 |
Game | Rating | MinPlayers | MaxPlayers | MaxPlayersGuess
---------------------------------------------------------
A | 6 | 3 | 6 |
B | 7 | 3 | 7 |
C | 6.5 | 3 | N/A |6
D | 7 | 3 | 6 |
E | 7 | 3 | 5 |
F | 9.5 | 2 | 5 |
G | 6 | 2 | 4 |
H | 7 | 2 | 4 |
I | 6.5 | 2 | N/A |4
J | 7 | 2 | 2 |
K | 7 | 2 | 4 |
gld$MaxPlayersGuess <- ifelse(is.na(gld$MaxPlayers), median(gld$MaxPlayers[gld$MinPlayers,]), NA)
Error in gld$MaxPlayers[gld$MinPlayers, ] :
incorrect number of dimensions
gld$MaxPlayersGuess相对于发布的示例进行更新
这是我一天的小贴士，有时候计算你想要什么，然后在需要的时候抓住它比使用所有这些逻辑连接更容易。你试图想出一种方法，一次计算出所有的数据，这会让人困惑，把它分成几个步骤。您需要知道每个可能的“MinPlayer”组的“MaxPlayer”的中值。然后，您希望在缺少MaxPlayer时使用该值。所以这里有一个简单的方法
#generate fake data
MinPlayer <- rep(3:2, each = 4)
MaxPlayer <- rep(2:5, each = 2, times = 2)
df <- data.frame(MinPlayer, MaxPlayer)
#replace some values of MaxPlayer with NA
df$MaxPlayer <- ifelse(df$MaxPlayer == 3, NA, df$MaxPlayer)
####STARTING DATA
# > df
# MinPlayer MaxPlayer
# 1 3 2
# 2 3 2
# 3 3 NA
# 4 3 NA
# 5 2 4
# 6 2 4
# 7 2 5
# 8 2 5
# 9 3 2
# 10 3 2
# 11 3 NA
# 12 3 NA
# 13 2 4
# 14 2 4
# 15 2 5
# 16 2 5
####STEP 1
#find the median of MaxPlayer for each group of MinPlayer (e.g., when MinPlayer == 1, 2 or whatever)
#just add a column to the data frame that has the right median value for each subset of MinPlayer in it and grab that value to use later.
library(plyr) #plyr is a great way to compute things across data subsets
df <- ddply(df, c("MinPlayer"), transform,
median.minp = median(MaxPlayer, na.rm = TRUE)) #ignore NAs in the median
####STEP 2
#anytime that MaxPlayer == NA, grab the median value to replace the NA, otherwise keep the MaxPlayer value
df$MaxPlayer <- ifelse(is.na(df$MaxPlayer), df$median.minp, df$MaxPlayer)
####STEP 3
#you had to compute an extra column you don't really want, so drop it now that you're done with it
df <- df[ , !(names(df) %in% "median.minp")]
####RESULT
# > df
# MinPlayer MaxPlayer
# 1 2 4
# 2 2 4
# 3 2 5
# 4 2 5
# 5 2 4
# 6 2 4
# 7 2 5
# 8 2 5
# 9 3 2
# 10 3 2
# 11 3 2
# 12 3 2
# 13 3 2
# 14 3 2
# 15 3 2
# 16 3 2
#生成假数据
MinPlayer相对于发布的示例更新
这是我一天的小贴士，有时候计算你想要什么，然后在需要的时候抓住它比使用所有这些逻辑连接更容易。你试图想出一种方法，一次计算出所有的数据，这会让人困惑，把它分成几个步骤。您需要知道每个可能的“MinPlayer”组的“MaxPlayer”的中值。然后，您希望在缺少MaxPlayer时使用该值。所以这里有一个简单的方法
#generate fake data
MinPlayer <- rep(3:2, each = 4)
MaxPlayer <- rep(2:5, each = 2, times = 2)
df <- data.frame(MinPlayer, MaxPlayer)
#replace some values of MaxPlayer with NA
df$MaxPlayer <- ifelse(df$MaxPlayer == 3, NA, df$MaxPlayer)
####STARTING DATA
# > df
# MinPlayer MaxPlayer
# 1 3 2
# 2 3 2
# 3 3 NA
# 4 3 NA
# 5 2 4
# 6 2 4
# 7 2 5
# 8 2 5
# 9 3 2
# 10 3 2
# 11 3 NA
# 12 3 NA
# 13 2 4
# 14 2 4
# 15 2 5
# 16 2 5
####STEP 1
#find the median of MaxPlayer for each group of MinPlayer (e.g., when MinPlayer == 1, 2 or whatever)
#just add a column to the data frame that has the right median value for each subset of MinPlayer in it and grab that value to use later.
library(plyr) #plyr is a great way to compute things across data subsets
df <- ddply(df, c("MinPlayer"), transform,
median.minp = median(MaxPlayer, na.rm = TRUE)) #ignore NAs in the median
####STEP 2
#anytime that MaxPlayer == NA, grab the median value to replace the NA, otherwise keep the MaxPlayer value
df$MaxPlayer <- ifelse(is.na(df$MaxPlayer), df$median.minp, df$MaxPlayer)
####STEP 3
#you had to compute an extra column you don't really want, so drop it now that you're done with it
df <- df[ , !(names(df) %in% "median.minp")]
####RESULT
# > df
# MinPlayer MaxPlayer
# 1 2 4
# 2 2 4
# 3 2 5
# 4 2 5
# 5 2 4
# 6 2 4
# 7 2 5
# 8 2 5
# 9 3 2
# 10 3 2
# 11 3 2
# 12 3 2
# 13 3 2
# 14 3 2
# 15 3 2
# 16 3 2
#生成假数据
MinPlayer我想你已经在@griffmer的答案中找到了你所需要的一切。但一种不那么优雅但可能更直观的方式可能是循环：
## Your data:
df <- data.frame(
Game = LETTERS[1:11],
Rating = c(6,7,6.5,7,7,9.5,6,7,6.5,7,7),
MinPlayers = c(rep(3,5), rep(2,6)),
MaxPlayers = c(6,7,NA,6,5,5,4,4,NA,2,4)
)
## Loop over rows:
df$MaxPlayersGuess <- vapply(1:nrow(df), function(ii){
if (is.na(df$MaxPlayers[ii])){
median(df$MaxPlayers[df$MinPlayers == df$MinPlayers[ii]],
na.rm = TRUE)
} else {
df$MaxPlayers[ii]
}
}, numeric(1))
我想你已经在@griffmer的答案中找到了你所需要的一切。但一种不那么优雅但可能更直观的方式可能是循环：
## Your data:
df <- data.frame(
Game = LETTERS[1:11],
Rating = c(6,7,6.5,7,7,9.5,6,7,6.5,7,7),
MinPlayers = c(rep(3,5), rep(2,6)),
MaxPlayers = c(6,7,NA,6,5,5,4,4,NA,2,4)
)
## Loop over rows:
df$MaxPlayersGuess <- vapply(1:nrow(df), function(ii){
if (is.na(df$MaxPlayers[ii])){
median(df$MaxPlayers[df$MinPlayers == df$MinPlayers[ii]],
na.rm = TRUE)
} else {
df$MaxPlayers[ii]
}
}, numeric(1))
如果要使用dplyr
，可以尝试：
输入：
df <- data.frame(
Game = LETTERS[1:11],
Rating = c(6,7,6.5,7,7,9.5,6,7,6.5,7,7),
MinPlayers = c(rep(3,5), rep(2,6)),
MaxPlayers = c(6,7,NA,6,5,5,4,4,NA,2,4)
)
这将对数据基础MinPlayers
进行分组，然后将MaxPlayers
的中值分配给缺少数据的行
输出：
Source: local data frame [11 x 4]
Groups: MinPlayers [2]
Game Rating MinPlayers MaxPlayers
<fctr> <dbl> <dbl> <dbl>
1 A 6.0 3 6
2 B 7.0 3 7
3 C 6.5 3 6
4 D 7.0 3 6
5 E 7.0 3 5
6 F 9.5 2 5
7 G 6.0 2 4
8 H 7.0 2 4
9 I 6.5 2 4
10 J 7.0 2 2
11 K 7.0 2 4
来源：本地数据帧[11 x 4]
组别：MinPlayers[2]
游戏评级MinPlayers MaxPlayers
1A 6.03 6
2 B 7.0 3 7
3 C 6.5 3 6
4 D 7.0 3 6
5 E 7.0 3 5
6 F 9.5 2 5
7 G 6.0 2 4
8小时7.0 2 4
9 I 6.5 2 4
10 J 7.0 2
11 K 7.0 2 4
如果要使用dplyr
，可以尝试：
输入：
df <- data.frame(
Game = LETTERS[1:11],
Rating = c(6,7,6.5,7,7,9.5,6,7,6.5,7,7),
MinPlayers = c(rep(3,5), rep(2,6)),
MaxPlayers = c(6,7,NA,6,5,5,4,4,NA,2,4)
)
这将对数据基础MinPlayers
进行分组，然后将MaxPlayers
的中值分配给缺少数据的行
输出：
Source: local data frame [11 x 4]
Groups: MinPlayers [2]
Game Rating MinPlayers MaxPlayers
<fctr> <dbl> <dbl> <dbl>
1 A 6.0 3 6
2 B 7.0 3 7
3 C 6.5 3 6
4 D 7.0 3 6
5 E 7.0 3 5
6 F 9.5 2 5
7 G 6.0 2 4
8 H 7.0 2 4
9 I 6.5 2 4
10 J 7.0 2 2
11 K 7.0 2 4
来源：本地数据帧[11 x 4]
组别：MinPlayers[2]
游戏评级MinPlayers MaxPlayers
1A 6.03 6
2 B 7.0 3 7
3 C 6.5 3 6
4 D 7.0 3 6
5 E 7.0 3 5
6 F 9.5 2 5
7 G 6.0 2 4
8小时7.0 2 4
9 I 6.5 2 4
10 J 7.0 2
11 K 7.0 2 4
抱歉，因为我甚至不知道如何开始编写程序，我不知道如何提供一个可复制的示例。感谢您尝试回答。通过尝试你的一些建议，我能够更好地看到这个问题，并找出如何发布一个示例。@Zelbinian，所以通常你会将griffmer的作为answer@Chris但我仍然不知道如何解决这个问题。。。我所学到的只是如何更准确地陈述它。关键的区别在于计算所基于的值取决于当前正在执行的同一行中的MinPlayers值，我不知道如何访问该值。@Zelbinian，is.na（）是您要确定在何处输入最大玩家猜测的测试。所以：max_play_idx表示歉意，因为我甚至不知道如何开始编写程序，我不知道如何提供一个可复制的示例。感谢您尝试回答。通过尝试你的一些建议，我能够更好地看到这个问题，并找出如何发布一个示例。@Zelbinian，所以通常你会将griffmer的作为answer@Chris但我仍然不知道如何解决这个问题。。。我所学到的只是如何更准确地陈述它。关键的区别在于计算所基于的值取决于当前正在执行的同一行中的MinPlayers值，我不知道如何访问该值。@Zelbinian，is.na（）是您要确定在何处输入最大玩家猜测的测试。所以：max_play_idx