dplyr :: mutate取消引用RHS [英] dplyr::mutate unquote RHS

问题描述

我想知道如何正确地UQ在dplyr方法(如mutate)中在RHS上字符串创建变量名.在此MWE的wilcox.test部分中查看我在注释中得到的错误消息:

I am wondering how to properly UQ string created variable names on the RHS in dplyr methods like mutate. See the error messages I got in comments in the wilcox.test part of this MWE:

require(dplyr)

dfMain <- data.frame(
    base = c(rep('A', 5), rep('B', 5)),
    id   = letters[1:10],
    q0   = rnorm(10)
)

backgs <- list(
    A = rnorm(13),
    B = rnorm(11)
)

fun <- function(dfMain, i = 0){

    pcol <- sprintf('p%i', i)
    qcol <- sprintf('q%i', i)

    (
        dfMain %>%
        group_by(id) %>%
        mutate(
            !!pcol := ifelse(
                !is.nan(!!qcol) &
                length(backgs[[base]]),
                wilcox.test(
                    # !!(qcol) - backgs[[base]] 
                    # object 'base' not found
                    # (!!qcol) - backgs[[base]]
                    #  non-numeric argument to binary operator
                    (!!qcol) - backgs[[base]]
                )$p.value,
                NaN
            )
        )
    )

}

dfMain <- dfMain %>% fun()

我猜在!!(qcol) ...上它被解释为我想取消整个表达式的引用，不仅是变量名，这就是为什么它找不到base的原因?我还发现(!!qcol)返回字符串本身，因此-运算符无法处理它也就不足为奇了.

I guess at !!(qcol) ... it is interpreted as I would like to unquote the whole expression not only the variable name that's why it does not find base? I also found out that (!!qcol) returns the string itself so no surprise the - operator is unable to handle it.

推荐答案

通过将定义qcol的行更改为:

Your code should work as you expect by changing the line where you define qcol to:

qcol <- as.symbol(sprintf('q%i', i))

也就是说，由于qcol是字符串，因此您需要先将其转换为符号，然后才能取消引用，以便在mutate中对其进行正确的评估.另外，我假设您要引用的列是您在数据中定义的q0列，而不是不存在的名为qval0的列.

That is, since qcol was a string, you needed to turn it into a symbol before unquoting for it to be evaluated correctly in your mutate. Also I presume the column you wanted to refer to was the q0 column you defined in your data, not a non-existent column named qval0.

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