抑制函数结果中attr()的输出

问题描述

在data.frame的源代码中，最后三行代码设置属性并返回结果.

In the source code for data.frame, the last three lines of code set the attributes and return the result.

    ...
    attr(value, "row.names") <- row.names
    attr(value, "class") <- "data.frame"
    value
}

在我编写的函数中，结果是由lapply创建的命名列表.在函数主体中设置任何属性之前，结果如下.

In a function I wrote, the result is a named list created by lapply. Before I set any attributes in the function body, result is as follows.

> x <- data.frame(a = 1:5, b = letters[1:5])
> (g <- grep.dataframe("a|c", x))
# ...
# $b
#   value row
# 1     a   1
# 2     c   3
> attributes(g)  # I want "list" in here...
# $names
# [1] "a" "b"

我希望"class"包含在属性列表中，所以我在res之前添加attr(res, "class") <- "list"(res是最终结果).现在，类别"显示在属性列表中.但是，它也可以打印出我不想要的功能的结果.我尝试用invisible包装它，但是没有用.

I'd like "class" to be included in the attributes list, so I add attr(res, "class") <- "list" (res is the final result) just before res. "class" now shows up in the attributes list. However,it also prints out with the result of the function, which I don't want. I tried wrapping it with invisible, but that didn't work.

为什么手动分配的属性随函数结果一起打印，但在我创建的新数据框中却被抑制?

Why do the manually assigned attributes print with the function result, but are suppressed in a new data frame I create?

> (h <- grep.dataframe("a|c", x))
# ...
# $b
#   value row
# 1     a   1
# 2     c   3

# attr(,"class")  # ...This prints with the result. I don't want that.
# [1] "list"
> attributes(h)   # ...But I want these attributes
# $names
# [1] "a" "b"

# $class
# [1] "list"

推荐答案

?class文档提供了一些指针:

The ?class documentation offers some pointers:

当将通用函数fun应用于具有类属性c("first"，"second")的对象时，系统搜索名为fun.first的函数，如果找到该函数，则将其应用于目的.如果找不到这样的函数，则尝试一个名为fun.second的函数.如果没有任何类名会产生合适的函数，则使用fun.default函数(如果存在).如果没有class属性，则尝试使用隐式类，然后使用默认方法.

When a generic function fun is applied to an object with class attribute c("first", "second"), the system searches for a function called fun.first and, if it finds it, applies it to the object. If no such function is found, a function called fun.second is tried. If no class name produces a suitable function, the function fun.default is used (if it exists). If there is no class attribute, the implicit class is tried, then the default method.

从那开始并运行一些简单的测试，我得到了这一点:

From that and running a few simple tests, I gather that:

• 列表是这些隐式类之一:请参见attributes(list(1))，typeof(list(1))
• 在列表上调用print时，它使用的是print.default
• print.default打印对象的属性

• a list is one of these implicit classes: see attributes(list(1)), typeof(list(1))
• when print is called on a list, it is using print.default
• print.default prints the attributes of an object

因此您可以定义一个print.list来处理您的特殊情况:

So you could define a print.list that will handle your special case:

 print.list <- function(x, ...) {
    if (is.list(x)) attr(x, "class") <- NULL
    print.default(x, ...)
 }

res <- list(1)
attr(res, "class") <- "list"
res
# [[1]]
# [1] 1

attributes(res)
# $class
# [1] "list"

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