用户定义的函数在R中产生警告

I would like to write a function which takes 3 arguments and calculates the mean based on the value of the arguments. The dataframe I'm using has 3 variables out of which the mean uses one of them and it holds numeric values. When I do mean(df$v1, na.rm=TRUE) I get a warning saying:

我想写一个函数，它接受3个参数，并根据参数的值计算平均值。我正在使用的数据帧有3个变量，其中均值使用其中一个变量并保存数值。当我的意思是（df $ v1，na.rm = TRUE）时，我收到一条警告：

Warning message:
In mean.default(df$v1, na.rm = TRUE) :
  argument is not numeric or logical: returning NA

I tried to do mean(df$as.numeric(as.character(v1)), na.rm = TRUE)

我试着做法（df $ as.numeric（as.character（v1）），na.rm = TRUE）

Error in mean(df$as.numeric(as.character(v1)), na.rm = TRUE) : 
  attempt to apply non-function

Not sure how to deal with this.

不知道如何处理这个问题。

3 个解决方案

#1

If you get an warning for mean(df$v1, na.rm=TRUE) then it probably means you have made an error in data entry that resulted in numbers getting coerced to characters and then to factors. Try:

如果你得到一个mean的警告（df $ v1，na.rm = TRUE），那么这可能意味着你在数据输入中犯了一个错误，导致数字被强制转换为字符，然后是强制因素。尝试：

str(df$v)

I predict you will find that to be a factor column. You can change it back to numeric with:

我预测你会发现这是一个因素列。您可以使用以下命令将其更改回数字：

df$v1 <- as.numeric(as.character(df$v1))

#2

You have to use

你必须使用

mean(as.numeric(as.character(df$v1)), na.rm = TRUE)

#3

The suggested way to convert factors to numeric is to use levels. From the as.factor help file:

将因子转换为数字的建议方法是使用级别。从as.factor帮助文件：

In particular, as.numeric applied to a factor is meaningless, and may happen by implicit coercion. To transform a factor f to approximately its original numeric values, as.numeric(levels(f))[f] is recommended and slightly more efficient than as.numeric(as.character(f)).

特别是，as.numeric应用于一个因子是没有意义的，并且可能通过隐式强制发生。要将因子f转换为大约其原始数值，建议使用as.numeric（levels（f））[f]，并且比as.numeric（as.character（f））稍微更有效。

Example:

例：

> x=factor(10:20)
> as.numeric(levels(x))[x]
 [1] 10 11 12 13 14 15 16 17 18 19 20

Note this is logically the same as:

请注意，这在逻辑上与以下相同：

> x=factor(10:20)
> as.numeric(levels(x))[as.numeric(x)]
 [1] 10 11 12 13 14 15 16 17 18 19 20

As factors are integers internally.

因为因素是内部整数。

#1