Question

我已生成以下列表，我只想提取cor值（最后一个值）。有没有办法做到这一点？我尝试使用$ estimate来做到这一点，但无济于事。在此先感谢您的帮助。

[[6]]

Pearson's product-moment correlation

data:  cer[, x] and par[, y] 
t = 109.0053, df = 1898, p-value < 2.2e-16
alternative hypothesis: true correlation is not equal to 0 
95 percent confidence interval:
0.9221188 0.9345283 
sample estimates:
cor 
0.9285826 #this value

str（list）的输出：

$ :List of 9
..$ statistic  : Named num 68.5
.. ..- attr(*, "names")= chr "t"
..$ parameter  : Named int 1898
.. ..- attr(*, "names")= chr "df"
..$ p.value    : num 0
..$ estimate   : Named num 0.844
.. ..- attr(*, "names")= chr "cor"
..$ null.value : Named num 0
.. ..- attr(*, "names")= chr "correlation"
..$ alternative: chr "two.sided"
..$ method     : chr "Pearson's product-moment correlation"
..$ data.name  : chr "cer[, x] and par[, y]"
..$ conf.int   : atomic [1:2] 0.83 0.856
.. ..- attr(*, "conf.level")= num 0.95
..- attr(*, "class")= chr "htest"

Answer 1

＆LT; mind.read＆gt;

您似乎提供了一个包含结果的列表   cor.test(..., method = 'pearson')

＆LT; \ mind.read＆gt;

如果此列表名为z

然后

 sapply(z, '[[', 'estimate')

将从estimate

的每个元素中提取z元素

记住财富（312）

很有用

fortune(312)

The problem here is that the $ notation is a magical shortcut and like any other
magic if used incorrectly is likely to do the programmatic equivalent of turning
yourself into a toad.
   -- Greg Snow (in response to a user that wanted to access a column whose name is
      stored in y via x$y rather than x[[y]])
      R-help (February 2012)

从列表中提取项目

1 个答案: