如何用轻敲来对因子的每个水平执行t检验 [英] How to perform t-tests for each level of a factor with tapply

问题描述

我的数据和代码如下:

my_vector <- rnorm(150)
my_factor1 <- gl(3,50)
my_factor2 <- gl(2,75)

tapply(my_vector, my_factor1, function(x)
  t.test(my_vector~my_factor2, paired=T))

我想对my_factor1的每个级别进行单独的t检验，以针对my_factor2的两个级别都测试my_vector.

I want to do a separate t-test for each level of my_factor1, to test my_vector for both levels of my_factor2.

但是，对于我的代码，t检验不会拆分my_factor1的级别，并且每个级别的结果均相等，因为my_vector完全包含在每个t.test中.

However, with my code the t-test is not splitting the levels of my_factor1, and the results are equal for each level because my_vector is entirely included in each t.test.

这是我的代码的输出:

$`1`

Paired t-test

data:  my_vector by my_factor2
t = 0.2448, df = 74, p-value = 0.8073
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.2866512  0.3669667
sample estimates:
mean of the differences 
         0.04015775 


$`2`

Paired t-test

data:  my_vector by my_factor2
t = 0.2448, df = 74, p-value = 0.8073
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.2866512  0.3669667
sample estimates:
mean of the differences 
         0.04015775 


$`3`

Paired t-test

data:  my_vector by my_factor2
t = 0.2448, df = 74, p-value = 0.8073
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -0.2866512  0.3669667
sample estimates:
mean of the differences 
         0.04015775 

我想念什么或做错什么了?

What am I missing or doing wrong?

推荐答案

您的示例存在一些问题，因为如果您进行了设置，

Your example is slightly problematic, since if you set:

df <- data.frame(my_vector = rnorm(150),
                 my_factor1 = gl(3,50),
                 my_factor2 = gl(2,75)
                )

当my_factor1 = 1或3时，由于重复的重叠方式，对于my_factor2只有一个唯一值.请参阅?gl.也是这样:

You will have only one unique value for my_factor2 when my_factor1 = 1 or 3 because of how your repetitions overlap. See ?gl. So do:

df <- data.frame(my_vector = rnorm(150),
                 my_factor1 = gl(3,1,150),
                 my_factor2 = gl(2,1,150)
                )
with(df,
       by(df, my_factor1,
          function(x) t.test(my_vector ~ my_factor2, data=x)
       )
     )

似乎会产生所需的输出.

Which appears to produce your desired output.

作为旁注-考虑对多个比较进行更正: https://stats.stackexchange.com/questions/16779/when-is-multiple-comparison-correction-necessary

As a side note-- consider correction for multiple comparisons: https://stats.stackexchange.com/questions/16779/when-is-multiple-comparison-correction-necessary

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