R：计算我的数据集子集的相关性？

时间：2011-03-27 13:36:30

标签： r correlation

我有一个包含20个变量v1 - v20的数据集。现在我想使用cor（...）计算v2和v10之间的相关性，直到v15和v3，v10到v15。最好的方法是什么？我是否必须使用

为每个变量对执行此操作 
cor(v2, v10)
cor(v2, v11)
cor(v2, v12)
and so on?

以下是实际数据集：

   > dput(dataset)
structure(list(Number = 1:15, Question.1.1 = c(3L, 4L, 5L, 5L, 
4L, 5L, 5L, 4L, 4L, 5L, 5L, 5L, 5L, 5L, 5L), Question.1.2 = c(1L, 
2L, 1L, 1L, 4L, 1L, 1L, 2L, 3L, 1L, 1L, 1L, 1L, 1L, 1L), Question.2.1 = c(5L, 
3L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L), Question.2.2 = c(2L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), Question.3.1 = c(2L, 
NA, 4L, 5L, 4L, 3L, 5L, 3L, 5L, 5L, 5L, 5L, 4L, 4L, 4L), Question.3.2 = c(2L, 
NA, 1L, 1L, 2L, 2L, 1L, 4L, 3L, 1L, 1L, 1L, 2L, 2L, 1L), Question.4.1 = c(3L, 
2L, 5L, 2L, 5L, 5L, 5L, 3L, 5L, 5L, 5L, 5L, 4L, 5L, 2L), Question.4.2 = c(2L, 
2L, 1L, 2L, 2L, 1L, 2L, 3L, 3L, 1L, 1L, 1L, 1L, 1L, 4L), Question.5.1 = c(5L, 
3L, 5L, 3L, 4L, 4L, 5L, 3L, 5L, 5L, 5L, 5L, 5L, 4L, 4L), Question.5.2 = c(2L, 
2L, 1L, 1L, 3L, 2L, 1L, 3L, 4L, 1L, 1L, 1L, 1L, 1L, 1L), Question.6.1 = c(5L, 
2L, 2L, 2L, 3L, 2L, 3L, 1L, 3L, 3L, 5L, 4L, 3L, 3L, 1L), Question.6.2 = c(2L, 
3L, 2L, 1L, 2L, 3L, 3L, 3L, 3L, 2L, 1L, 1L, 2L, 2L, 1L), Question.7.1 = c(5L, 
2L, 5L, 5L, 5L, 3L, 5L, 5L, 2L, 4L, 5L, 5L, 5L, 4L, 5L), Question.7.2 = c(1L, 
4L, 1L, 1L, 2L, 2L, 1L, 1L, 3L, 1L, 1L, 1L, 1L, 1L, 1L), Question.8.1 = c(4L, 
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L), Question.8.2 = c(1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), Question.9.1 = c(5L, 
3L, 5L, 4L, 4L, 5L, 5L, 5L, 4L, 5L, 5L, 5L, 5L, 4L, 3L), Question.9.2 = c(1L, 
1L, 1L, 2L, 2L, 1L, 2L, 1L, 4L, 2L, 1L, 2L, 2L, 1L, 2L), AQ.1 = c(5L, 
5L, 5L, 1L, 3L, 5L, 5L, 5L, 5L, 2L, 2L, 2L, 5L, 5L, 3L), AQ.2 = c(2L, 
5L, 2L, 1L, 2L, 5L, 2L, 1L, 5L, 1L, 1L, 4L, 2L, 3L, 3L), Task.1 = c(5L, 
2L, 5L, 1L, 4L, 5L, 5L, 4L, 4L, 5L, 5L, 4L, 5L, 5L, 5L), Task.2 = c(4L, 
3L, 5L, 4L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L), Task.3 = c(4L, 
3L, 4L, 1L, 3L, 5L, 4L, 5L, 5L, 5L, 5L, 5L, 4L, 4L, 4L), Task.4 = c(5L, 
5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L, 5L), Task.5 = c(5L, 
4L, 5L, 4L, 4L, 5L, 4L, 5L, 4L, 5L, 5L, 5L, 5L, 5L, 4L), GQ.1 = c(4L, 
2L, 2L, 5L, 4L, 4L, 5L, 4L, 5L, 5L, 5L, 4L, 4L, 5L, 4L), GQ.2 = c(4L, 
4L, 4L, 5L, 5L, 4L, 4L, 3L, 3L, 3L, 5L, 5L, 5L, 4L, 3L), GQ.3 = c(5L, 
3L, 2L, 5L, 3L, 5L, 5L, 5L, 4L, 5L, 5L, 5L, 4L, 4L, 4L), GQ.4 = c(5L, 
2L, 1L, 4L, 4L, 4L, 4L, 3L, 3L, 3L, 5L, 5L, 4L, 4L, 1L), GQ.5 = c(4L, 
3L, 4L, 5L, 5L, 4L, 4L, 4L, 4L, 5L, 5L, 5L, 5L, 4L, 3L), GQ.6 = c(2L, 
2L, 1L, 1L, 2L, 1L, 4L, 1L, 4L, 5L, 5L, 1L, 5L, 1L, 5L), GQ.7 = c(4L, 
5L, 5L, 5L, 4L, 2L, 3L, 5L, 3L, 5L, 5L, 2L, 5L, 3L, 2L), GQ.8 = c(2L, 
4L, 3L, 2L, 3L, 3L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 2L), GQ.9 = c(3L, 
5L, 2L, 3L, 4L, 4L, 5L, 3L, 4L, 4L, 3L, 3L, 4L, 2L, 2L), GQ.10 = c(3L, 
4L, 1L, 2L, 3L, 4L, 5L, 5L, 5L, 5L, 4L, 4L, 5L, 5L, 2L), Feature.1 = c(4L, 
4L, 2L, 3L, 4L, 4L, 5L, 5L, 4L, 5L, 5L, 4L, 5L, 3L, 4L), Feature.2 = c(4L, 
4L, 2L, 1L, 5L, 4L, 5L, 5L, 5L, 4L, 4L, 3L, 5L, 3L, 2L), Feature.3 = c(3L, 
2L, 1L, 2L, 5L, 5L, 2L, 4L, 2L, 4L, 4L, 5L, 2L, 4L, 2L), Feature.4 = c(3L, 
3L, 3L, 4L, 3L, 4L, 5L, 5L, 4L, 4L, 4L, 3L, 4L, 3L, 3L), Feature.5 = c(2L, 
2L, 3L, 3L, 4L, 3L, 4L, 4L, 2L, 4L, 3L, 4L, 5L, 3L, 1L), Feature.6 = c(5L, 
5L, 1L, 1L, 5L, 5L, 5L, 4L, 4L, 5L, 5L, 5L, 5L, 4L, 4L), Feature.7 = c(5L, 
3L, 2L, 5L, 4L, 5L, 3L, 5L, 4L, 5L, 5L, 5L, 5L, 4L, 4L)), .Names = c("Number", 
"Question.1.1", "Question.1.2", "Question.2.1", "Question.2.2", 
"Question.3.1", "Question.3.2", "Question.4.1", "Question.4.2", 
"Question.5.1", "Question.5.2", "Question.6.1", "Question.6.2", 
"Question.7.1", "Question.7.2", "Question.8.1", "Question.8.2", 
"Question.9.1", "Question.9.2", "AQ.1", "AQ.2", "Task.1", "Task.2", 
"Task.3", "Task.4", "Task.5", "GQ.1", "GQ.2", "GQ.3", "GQ.4", 
"GQ.5", "GQ.6", "GQ.7", "GQ.8", "GQ.9", "GQ.10", "Feature.1", 
"Feature.2", "Feature.3", "Feature.4", "Feature.5", "Feature.6", 
"Feature.7"), class = "data.frame", row.names = c(NA, -15L))

4 个答案:

答案 0 :(得分：4)

我可能误解了这个问题......但为什么不只是在数据框上运行cor？

例如：

data <- data.frame(q1=sample(1:5, 15, rep=1), 
          q2=sample(1:5, 15, rep=1), 
          q3=sample(1:5, 15, rep=1), 
          q4=sample(1:5, 15, rep=1), 
          q5=sample(1:5, 15, rep=1), 
          q6=sample(1:5, 15, rep=1), 
          q7=sample(1:5, 15, rep=1), 
          q8=sample(1:5, 15, rep=1), 
          q9=sample(1:5, 15, rep=1), 
          q10=sample(1:5, 15, rep=1))

print(cor(data))

你甚至可以

image(cor(data), x=1:10, y=1:10, zlim=c(-1,1))

如果你只需要某些相关值，只需将corr的结果放在一个变量中，然后取出你需要的结果。

例如，我们希望第2列与第5列到第10列的相关性：

corrs <- cor(data)
print(corrs[2, 5:10]) # or corrs[5:10, 2], the correlation matrix is symmetric

答案 1 :(得分：1)

显式对数据集进行子集并在该数据集上运行关联命令。假设您的变量排序正确，请尝试以下方法：

cor(dat[,c(2, 10:15)][,1]
cor(dat[,c(3, 10:15)][,1]

如果没有订购，您只需要订购它们或者用引号命名变量。例如：

cor(dat[,c('v3', 'v10', 'v11', 'v12', 'v13', 'v14', 'v15')][,1]

答案 2 :(得分：1)

使用subset命令：

dtf <- subset(mtcars, select = c(mpg, hp, wt))
cor(dtf)
         mpg         hp         wt
mpg  1.0000000 -0.7761684 -0.8676594
hp  -0.7761684  1.0000000  0.6587479
wt  -0.8676594  0.6587479  1.0000000

或使用psych包和corr.test功能：

library(psych)
corr.test(dtf)
Call:corr.test(x = dtf)
Correlation matrix 
      mpg    hp    wt
mpg  1.00 -0.78 -0.87
hp  -0.78  1.00  0.66
wt  -0.87  0.66  1.00
Sample Size 
    mpg hp wt
mpg  32 32 32
hp   32 32 32
wt   32 32 32
Probability value 
    mpg hp wt
mpg   0  0  0
hp    0  0  0
wt    0  0  0

答案 3 :(得分：1)

问题似乎是由于在整个数据帧上运行corr导致的信息过载。我没有太多使用它，但是ggplot成名的Hadley Wickham的plyr package似乎为分组和管理输出提供了一些优雅的解决方案。

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