Question

我有一个2005年至2016年的数据框架列表。除了年份的数字外，它们的编写方式相同：

m  =list(X2016_kvish_1_10t = X2016_kvish_1_10t, X2015_kvish_1_10t = X2015_kvish_1_10t, X2014_kvish_1_10t = X2014_kvish_1_10t,
     X2013_kvish_1_10t = X2013_kvish_1_10t, X2012_kvish_1_10t = X2012_kvish_1_10t, X2011_kvish_1_10t = X2011_kvish_1_10t,
     X2010_kvish_1_10t = X2010_kvish_1_10t, X2009_kvish_1_10t = X2009_kvish_1_10t, X2008_kvish_1_10t = X2008_kvish_1_10t, 
     X2007_kvish_1_10t = X2007_kvish_1_10t, X2006_kvish_1_10t = X2006_kvish_1_10t, X2005_kvish_1_10t = X2005_kvish_1_10t)

是否有更短的编写方式，而无需单独编写所有内容？

Answer 1

尝试mget：

df_names = paste0("X", 2005:2016, "_kvish_1_10t")
m = mget(df_names)

修改

正如@ d.b指出的那样，你甚至不需要创建df_names

m = mget(ls(pattern="_kvish_1_10t$"))

Answer 2

您可以使用mget函数在工作区中提供对象名称的字符向量。

我制作了一个可复制的例子，目的是展示如何做到这一点。

df_name <- paste0("x", 2005:2016, "_kvish_1_10t")
df_name
#>  [1] "x2005_kvish_1_10t" "x2006_kvish_1_10t" "x2007_kvish_1_10t"
#>  [4] "x2008_kvish_1_10t" "x2009_kvish_1_10t" "x2010_kvish_1_10t"
#>  [7] "x2011_kvish_1_10t" "x2012_kvish_1_10t" "x2013_kvish_1_10t"
#> [10] "x2014_kvish_1_10t" "x2015_kvish_1_10t" "x2016_kvish_1_10t"
# juste create some dummy table for example
l <- lapply(df_name, assign, value = mtcars[1:2], envir= .GlobalEnv)
# Use mget to get a list of all the object
m <- mget(df_name, envir = .GlobalEnv)
str(m)
#> List of 12
#>  $ x2005_kvish_1_10t:'data.frame':   32 obs. of  2 variables:
#>   ..$ mpg: num [1:32] 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
#>   ..$ cyl: num [1:32] 6 6 4 6 8 6 8 4 4 6 ...
#>  $ x2006_kvish_1_10t:'data.frame':   32 obs. of  2 variables:
#>   ..$ mpg: num [1:32] 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
#>   ..$ cyl: num [1:32] 6 6 4 6 8 6 8 4 4 6 ...
#>  $ x2007_kvish_1_10t:'data.frame':   32 obs. of  2 variables:
#>   ..$ mpg: num [1:32] 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
#>   ..$ cyl: num [1:32] 6 6 4 6 8 6 8 4 4 6 ...
#>  $ x2008_kvish_1_10t:'data.frame':   32 obs. of  2 variables:
#>   ..$ mpg: num [1:32] 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
#>   ..$ cyl: num [1:32] 6 6 4 6 8 6 8 4 4 6 ...
#>  $ x2009_kvish_1_10t:'data.frame':   32 obs. of  2 variables:
#>   ..$ mpg: num [1:32] 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
#>   ..$ cyl: num [1:32] 6 6 4 6 8 6 8 4 4 6 ...
#>  $ x2010_kvish_1_10t:'data.frame':   32 obs. of  2 variables:
#>   ..$ mpg: num [1:32] 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
#>   ..$ cyl: num [1:32] 6 6 4 6 8 6 8 4 4 6 ...
#>  $ x2011_kvish_1_10t:'data.frame':   32 obs. of  2 variables:
#>   ..$ mpg: num [1:32] 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
#>   ..$ cyl: num [1:32] 6 6 4 6 8 6 8 4 4 6 ...
#>  $ x2012_kvish_1_10t:'data.frame':   32 obs. of  2 variables:
#>   ..$ mpg: num [1:32] 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
#>   ..$ cyl: num [1:32] 6 6 4 6 8 6 8 4 4 6 ...
#>  $ x2013_kvish_1_10t:'data.frame':   32 obs. of  2 variables:
#>   ..$ mpg: num [1:32] 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
#>   ..$ cyl: num [1:32] 6 6 4 6 8 6 8 4 4 6 ...
#>  $ x2014_kvish_1_10t:'data.frame':   32 obs. of  2 variables:
#>   ..$ mpg: num [1:32] 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
#>   ..$ cyl: num [1:32] 6 6 4 6 8 6 8 4 4 6 ...
#>  $ x2015_kvish_1_10t:'data.frame':   32 obs. of  2 variables:
#>   ..$ mpg: num [1:32] 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
#>   ..$ cyl: num [1:32] 6 6 4 6 8 6 8 4 4 6 ...
#>  $ x2016_kvish_1_10t:'data.frame':   32 obs. of  2 variables:
#>   ..$ mpg: num [1:32] 21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
#>   ..$ cyl: num [1:32] 6 6 4 6 8 6 8 4 4 6 ...

在r中以短的方式写入多个变量

2 个答案: