6

虽然我知道如何计算 R 中的平均值，但我似乎无法理解如何在不同的条件下进行计算。我读过的其他帖子没有那么复杂。

>      Target/dictractor    TrialType    Bin0    Bin1    Bin2    Bin3
1          Target              2C           3       0       2       0
1          Target              2C           2       0       3       0
1          Target              2E           0       1       1       2
1          Target              2E           0       0       0       0
1        Distractor            2C           0       3       0       1
1        Distractor            2C           0       0       0       0
1        Distractor            2E           0       0       1       0
1        Distractor            2E           0       0       0       3
2          Target              2C           1       1       0       1
2          Target              2C           2       0       0       2
2        Distractor            2E           0       0       0       0
2        Distractor            2E           0       0       0       0

鉴于上面的数据集，我想分别获得每个主题的每个 bin ( Bin0, Bin1, ) 的平均值。例如，对于主题 1，我想计算每个 bin 的平均值，对于, 在一个条件下，然后是每个 bin在一个条件下的平均值，然后是每个 bin在条件和in的平均值。Bin2TrialTypeTarget/dictractorTrialType 2CTargetTrialType 2ETargetTrialType 2CDistractorTrialType 2EDistractor

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2 回答 2

3

这是我的尝试。您可以使用group_by()和获取id,target和的组合trial。对于每个组合，您都希望得到 bin0-bin3 的平均值。在这种情况下，您可以这样做summarise_each()。

mydf <- data.frame(id = c(1,1,1,1,1,1,1,1,2,2,2,2),
                   target = c("target", "target", "target", "target", "distractor",
                        "distractor", "distractor", "distractor",
                        "target", "target", "distractor", "distractor"),
                   trial = c("2c", "2c", "2e", "2e", "2c", "2c", "2e", "2e",
                             "2c", "2c", "2e", "2e"),
                   bin0 = c(3,2,0,0,0,0,0,0,1,2,0,0),
                   bin1 = c(0,0,1,0,3,0,0,0,1,0,0,0),
                   bin2 = c(2,3,1,0,0,0,1,0,0,0,0,0),
                   bin3 = c(0,0,2,0,1,0,0,3,1,2,0,0),
                   stringsAsFactors = FALSE)

library(dplyr)          
group_by(mydf, id, target, trial) %>%
summarise_each(funs(mean(., na.rm = TRUE)), bin0:bin3)

#  id     target trial bin0 bin1 bin2 bin3
#1  1 distractor    2c  0.0  1.5  0.0  0.5
#2  1 distractor    2e  0.0  0.0  0.5  1.5
#3  1     target    2c  2.5  0.0  2.5  0.0
#4  1     target    2e  0.0  0.5  0.5  1.0
#5  2 distractor    2e  0.0  0.0  0.0  0.0
#6  2     target    2c  1.5  0.5  0.0  1.5

或者，您可以尝试使用该data.table软件包执行相同的操作。

foo <- setDT(mydf)[, lapply(.SD, mean), by = list(id, target, trial)]
print(foo)

#   id     target trial bin0 bin1 bin2 bin3
#1:  1     target    2c  2.5  0.0  2.5  0.0
#2:  1     target    2e  0.0  0.5  0.5  1.0
#3:  1 distractor    2c  0.0  1.5  0.0  0.5
#4:  1 distractor    2e  0.0  0.0  0.5  1.5
#5:  2     target    2c  1.5  0.5  0.0  1.5
#6:  2 distractor    2e  0.0  0.0  0.0  0.0
于 2015-02-02T00:24:06.410 回答
2

使用该dplyr软件包的一种方法是：

数据

df <- read.table(header=T,text='    Subject Target/dictractor    TrialType    Bin0    Bin1    Bin2    Bin3
1          Target              2C           3       0       2       0
1          Target              2C           2       0       3       0
1          Target              2E           0       1       1       2
1          Target              2E           0       0       0       0
1        Distractor            2C           0       3       0       1
1        Distractor            2C           0       0       0       0
1        Distractor            2E           0       0       1       0
1        Distractor            2E           0       0       0       3
2          Target              2C           1       1       0       1
2          Target              2C           2       0       0       2
2        Distractor            2E           0       0       0       0
2        Distractor            2E           0       0       0       0', stringsAsFactors=F)

解决方案

df %>%
  group_by(Subject, Target.dictractor,TrialType) %>%
  summarise(mean_Bin0=mean(Bin0),
            mean_Bin1=mean(Bin1),
            mean_Bin2=mean(Bin2),
            mean_Bin3=mean(Bin3))

输出

Source: local data frame [6 x 7]
Groups: Subject, Target.dictractor

  Subject Target.dictractor TrialType mean_Bin0 mean_Bin1 mean_Bin2 mean_Bin3
1       1        Distractor        2C       0.0       1.5       0.0       0.5
2       1        Distractor        2E       0.0       0.0       0.5       1.5
3       1            Target        2C       2.5       0.0       2.5       0.0
4       1            Target        2E       0.0       0.5       0.5       1.0
5       2        Distractor        2E       0.0       0.0       0.0       0.0
6       2            Target        2C       1.5       0.5       0.0       1.5
于 2015-02-02T00:22:30.793 回答