R：为什么是 mean(NA, na.rm = TRUE) return NaN

Question

当使用所有 NA 的向量估计平均值时，如果 na.rm = TRUE，我们得到 NaN。这是为什么，这是有缺陷的逻辑还是我遗漏了什么？使用 NA 肯定比 NaN?

更有意义

下面的快速示例

mean(NA, na.rm = TRUE)
#[1] NaN

mean(rep(NA, 10), na.rm = TRUE)
#[1] NaN

Answer 1

来自 mean 文档：

na.rm a logical value indicating whether NA values should be stripped before the computation proceeds.

使用此逻辑，在应用函数均值之前删除所有 NA。在您的情况下，您对任何事情都没有任何意义（所有 NA 都被删除），因此返回 NaN。

Answer 2

有点可惜?mean对此没有说什么。 only told you that applying mean on an empty "numeric" results in NaN without more reasoning. tried to reason this but was not accurate, as division by 0 is not always NaN, it can be Inf or -Inf. I once discussed about this in 。然而，我们越来越接近了。虽然mean(x)没有被sum(x) / length(x)编码，但是这个数学事实确实解释了这个NaN.

来自?sum:

 *NB:* the sum of an empty set is zero, by definition.

所以 sum(numeric(0)) 是 0。由于 length(numeric(0)) 是 0，mean(numeric(0)) 是 0 / 0 即 NaN.

R：为什么是 mean(NA, na.rm = TRUE) return NaN

R: Why does mean(NA, na.rm = TRUE) return NaN

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