在Lapply中迭代时打印列表名称在3.2中不起作用 [英] print list names while iterating in lapply not working in 3.2

问题描述

我试图每次通过lapply运行函数时输出列表名称.我较早前发布了这个问题，并由@提供了答案在我将R升级到版本3.2.0之前，Ananda Mahto正常工作.它不再工作，并且出现以下错误消息:Error in eval.parent(quote(names(X)))[substitute(x)[[3]]] : invalid subscript type 'symbol'

x <- ts(rnorm(40,5), start = c(1961, 1), frequency = 12)
y <- ts(rnorm(50,20), start = c(1971, 1), frequency = 12)
z <- ts(rnorm(50,39), start = c(1981, 1), frequency = 12)
a <- ts(rnorm(50,59), start = c(1991, 1), frequency = 12)

dat.list <- list(x=x,y=y,z=z,a=a)

abc <- function(x) {
  r <- mean(x)
  print(eval.parent(quote(names(X)))[substitute(x)[[3]]])
  return(r)
}

forl <- lapply(dat.list, abc)

我不确定问题是什么，但是我检查了新版本R中的所有语法，没有任何改变.任何帮助是极大的赞赏.我也欢迎任何新想法.

解决方案

如果稍微对其进行重组，则可以通过简单的外观获得相同的效果:

set.seed(42)
## using your dat.list construction code from above
abc <- function(x) { r <- mean(x); return(r); }

forl <- mapply(function(n, x) {
        message(n)
        abc(x)
    }, names(dat.list), dat.list, SIMPLIFY=FALSE)
## x
## y
## z
## a

forl
## $x
## [1] 4.960464
## $y
## [1] 20.1141
## $z
## [1] 38.87175
## $a
## [1] 58.89825

我想您想让list中的输出与vector中的ergo SIMPLIFY=FALSE相似，以模仿lapply.否则，它是一个更简单的向量.

虽然您可以创建三明治函数，但必须重新实现lapply功能，但这并不是通用的，因为您必须显式地传递名称和dat.

我在这里使用的个人编码偏好是使用message而不是print.理由是用户可以选择使用suppressMessages(在mapply之外，假设它可以/将被埋在函数中)，而抑制print调用的输出则需要更多的工作. /p>

I'm trying o output the list names every time I run the function thru lapply. I posted this question earlier that I posted earlier, and the answer provided by @Ananda Mahto worked fine until I upgraded my R to version 3.2.0. It is no longer working and I get the following error message: Error in eval.parent(quote(names(X)))[substitute(x)[[3]]] : invalid subscript type 'symbol'

x <- ts(rnorm(40,5), start = c(1961, 1), frequency = 12)
y <- ts(rnorm(50,20), start = c(1971, 1), frequency = 12)
z <- ts(rnorm(50,39), start = c(1981, 1), frequency = 12)
a <- ts(rnorm(50,59), start = c(1991, 1), frequency = 12)

dat.list <- list(x=x,y=y,z=z,a=a)

abc <- function(x) {
  r <- mean(x)
  print(eval.parent(quote(names(X)))[substitute(x)[[3]]])
  return(r)
}

forl <- lapply(dat.list, abc)

I'm not sure what the issues is, but I checked all the syntax in the new version of R nothing has changed. Any help is greatly appreciated. I'm open to any new ideas as well.

解决方案

If you restructure it a little, you can get the same effect with a simpler appearance:

set.seed(42)
## using your dat.list construction code from above
abc <- function(x) { r <- mean(x); return(r); }

forl <- mapply(function(n, x) {
        message(n)
        abc(x)
    }, names(dat.list), dat.list, SIMPLIFY=FALSE)
## x
## y
## z
## a

forl
## $x
## [1] 4.960464
## $y
## [1] 20.1141
## $z
## [1] 38.87175
## $a
## [1] 58.89825

I'm presuming you wanted the output in a list vice a vector, ergo SIMPLIFY=FALSE to mimic lapply. Otherwise, it's a simpler vector.

It's not as generic in that you have to explicitly pass the names as well as the dat, though you can create a sandwich function at the cost of having to reimplement lapply functionality.

A personal coding preference I'm using here is the use of message over print. The rationale is that the user has the option to use suppressMessages (outside the mapply, assuming it could/would be buried in a function), whereas suppressing the output of a print call is a bit more work.

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