在R的替代命令中取消引用字符串 [英] Unquote string in R's substitute command

问题描述

我想知道是否可以取消引用通过替代命令传递给表达式的字符串。

I'd like to know whether it is possible to unquote a string passed to an expression via the substitute command.

具体来说，我正在使用dplyr进行过滤并从数据框中选择：

Specifically, I am using dplyr to filter and select from a data frame:

    > w
       subject sex response
    1        1   M    19.08
    2        2   M    16.46
    ...     ...  ...  ...
    6        6   M    23.60
    7        7   M    23.96
    8        8   F    22.48
    9        9   F    25.79
    ...     ...  ...  ...
    16      16   F    26.66

以下将产生所需的结果：

The following produces the desired result:

    > w %.% filter(sex == "M") %.% select(response)        
      response
    1    19.08
    2    16.46
    3    22.81
    4    18.62
    5    18.75
    6    23.60
    7    23.96

但我想以更一般的方式执行此操作。由于字符串 sex用引号引起来，因此以下内容不会产生所需的结果。

But I want to do this in a more general way. Th following does not produce the required result since the string "sex" is enclosed in quotation marks.

substitute（w％。％filter（ y == M），list（y = paste（names（w）[2]）））

substitute(w %.% filter(y == "M"), list(y = paste(names(w)[2])))

    w %.% filter("sex" == "M")
    > eval(substitute(w %.% filter(y == "M"), list(y = paste(names(w)[2]))))
    [1] subject  sex      response
    <0 rows> (or 0-length row.names)

我总是可以执行以下操作：

I can always do the following:

    eval(parse(text = paste("w %.% filter(", names(w)[2], " == 'M')")))

但这确实有点笨拙。

是否有更优雅的方法？最终，我想将其包装到一个函数中，并使其变得更通用。

Are there more elegant ways of doing this? Eventually, I'd like to wrap this up in a function and make it even more general.

任何帮助/建议都将不胜感激。

Any help / suggestion would be much appreciated.

亲切的问候，

Stefan

推荐答案

也许您可以尝试：

w <- structure(list(subject = c(1L, 2L, 6L, 7L, 8L, 9L, 16L), sex = structure(c(2L, 
2L, 2L, 2L, 1L, 1L, 1L), .Label = c("F", "M"), class = "factor"), 
response = c(19.08, 16.46, 23.6, 23.96, 22.48, 25.79, 26.66
)), .Names = c("subject", "sex", "response"), class = "data.frame", row.names = c("1", 
"2", "6", "7", "8", "9", "16"))

基于@hadley的评论

Based on @hadley's comments

 eval(substitute(w%>% filter(y=="M"), list(y=as.name(names(w)[2]))))

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