使用$和列名称向量动态选择数据框架列 [英] Dynamically select data frame columns using $ and a vector of column names

问题描述

我希望根据不同的列排序一个数据框，一个轮到一个。我有一个字符向量，其中包含订单应该基于的相关列名：

I wish to order a data frame based on different columns, one at a turn. I have a character vector with the relevant column names on which the order should be based:

parameter <- c("market_value_LOCAL", "ep", "book_price", "sales_price", "dividend_yield",
               "beta", "TOTAL_RATING_SCORE", "ENVIRONMENT", "SOCIAL", "GOVERNANCE")

我想循环参数中的名称，并动态选择列用于订单我的数据：

I wish to loop over the names in "parameter" and dynamically select the column to be used to order my data:

Q1_R1000_parameter <- Q1_R1000[order(Q1_R1000$parameter[X]), ]

其中 X 是 1:10 （因为我在参数中有10个项目）。

where X is 1:10 (because I have 10 items in "parameter").

为了使我的示例可重现，请考虑数据集mtcars和存储在字符向量cols中的一些变量名。当我尝试使用动态子集cols从mtcars中选择一个变量时，以与上述相似的方式（ Q1_R1000 $ parameter [X] ），列未选择：

To make my example reproducible, consider the data set "mtcars" and some variable names stored in a character vector "cols". When I try to select a variable from "mtcars" using a dynamic subset of "cols", in a similar way as above (Q1_R1000$parameter[X]), the column is not selected:

cols <- c("cyl", "am")
mtcars$cols[1]
# NULL

推荐答案

用 $ 做这种子集。在源代码（ R / src / main / subset.c ）中，它指出：

You can't do that kind of subsetting with $. In the source code (R/src/main/subset.c) it states:

/ * $子集运算符。

我们需要确保只评估第一个参数。

第二个将是需要匹配的符号，而不是评估。

* /

/*The $ subset operator.
We need to be sure to only evaluate the first argument.
The second will be a symbol that needs to be matched, not evaluated.
*/

第二个参数？什么？！你必须意识到 $ 像R中的其他内容一样（包括例如（， + ， ^ 等）是一个函数，它接受参数并进行评估。 df $ V1 可以重写为

Second argument? What?! You have to realise that $, like everything else in R, (including for instance ( , + , ^ etc) is a function, that takes arguments and is evaluated. df$V1 could be rewritten as

`$`(df , V1)

或确实

`$`(df , "V1")

但是...

`$`(df , paste0("V1") )

...例如永远不会工作，也不会在第二个参数中首先被评估的任何东西。您只能传递一个 评估的字符串。

...for instance will never work, nor will anything else that must first be evaluated in the second argument. You may only pass a string which is never evaluated.

而是使用 [（或 [[ if您只需要提取一个列作为向量）。

Instead use [ (or [[ if you want to extract only a single column as a vector).

例如，

var <- "mpg"
#Doesn't work
mtcars$var
#These both work, but note that what they return is different
# the first is a vector, the second is a data.frame
mtcars[[var]]
mtcars[var]

您可以使用 do.call 执行无循环的排序来构造对订单的调用。以下是可重复的示例：

You can perform the ordering without loops, using do.call to construct the call to order. Here is a reproducible example below:

#  set seed for reproducibility
set.seed(123)
df <- data.frame( col1 = sample(5,10,repl=T) , col2 = sample(5,10,repl=T) , col3 = sample(5,10,repl=T) )

#  We want to sort by 'col3' then by 'col1'
sort_list <- c("col3","col1")

#  Use 'do.call' to call order. Seccond argument in do.call is a list of arguments
#  to pass to the first argument, in this case 'order'.
#  Since  a data.frame is really a list, we just subset the data.frame
#  according to the columns we want to sort in, in that order
df[ do.call( order , df[ , match( sort_list , names(df) ) ]  ) , ]

   col1 col2 col3
10    3    5    1
9     3    2    2
7     3    2    3
8     5    1    3
6     1    5    4
3     3    4    4
2     4    3    4
5     5    1    4
1     2    5    5
4     5    3    5

这篇关于使用$和列名称向量动态选择数据框架列的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！