使用管道时无需输入第一个参数 [英] Use pipe without feeding first argument

问题描述

%>%管道运算符是否总是将左侧(LHS)馈入右侧(RHS)的第一个自变量?即使在RHS调用中再次指定了第一个参数?

Is the %>% pipe operator always feeding the left-hand side (LHS) to the first argument of the right-hand side (RHS)? Even if the first argument is specified again in the RHS call?

说我想指定在cor()中使用哪个变量:

Say I want to specify which variable to use in cor():

library(magrittr)
iris  %>%
  cor(x=.$Sepal.Length, y=.$Sepal.Width)

但是这失败了，看起来像是调用了cor(., x=.$Sepal.Length, y=.$Sepal.Width)之类的东西?

But this fails, it looks like it call something like cor(., x=.$Sepal.Length, y=.$Sepal.Width) ?

我知道我可以代替

iris  %$%
  cor(x=Sepal.Length, y=Sepal.Width)

但想用%>% ...

推荐答案

%>%管道运算符是否总是将左侧(LHS)馈入右侧(RHS)的第一个自变量?即使在RHS调用中再次指定了第一个参数?

Is the %>% pipe operator always feeding the left-hand side (LHS) to the first argument of the right-hand side (RHS)? Even if the first argument is specified again in the RHS call?

不.您自己已经注意到了该异常:如果右侧使用.，则左侧的第一个参数是 not 输入的.您需要手动传递它.

No. You’ve noticed the exception yourself: if the right-hand side uses ., the first argument of the left-hand side is not fed in. You need to pass it manually.

但是，由于您不是单独使用.而是在表达式中使用它，因此在您的情况下不会发生 .为了避免将左侧作为第一个参数，您还需要使用花括号:

However, this is not happening in your case because you’re not using . by itself, you’re using it inside an expression. To avoid the left-hand side being fed as the first argument, you additionally need to use braces:

iris %>% {cor(x = .$Sepal.Length, y = .$Sepal.Width)}

或者:

iris %$% cor(x = Sepal.Length, y = Sepal.Width)

-毕竟，这就是%$%的用途，而不是%>%.

— after all, that’s what %$% is there for, as opposed to %>%.

但是比较一下:

iris %>% lm(Sepal.Width ~ Sepal.Length, data = .)

在这里，我们将左侧表达式显式传递为lm的data参数.这样，我们可以防止将它作为第一个参数传递给lm.

Here, we’re passing the left-hand side expression explicitly as the data argument to lm. By doing so, we prevent it being passed as the first argument to lm.

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