list output truncated - 如何在R中使用str()扩展列出的变量

Question

我有一个df包含600多个变量的data.frame .我正在编写一个自动创建列的函数,需要对它们进行一次视觉检查.

该str功能提供了一个很好的总结:

str(df)
'data.frame':   29 obs. of  602 variables:
 $ uniqueSessionsIni: POSIXct, format: "2015-01-05 15:00:00" "2015-01-05 16:00:00" "2015-01-05 17:00:00" ...
 $ uniqueSessionsEnd: POSIXct, format: "2015-01-05 15:59:00" "2015-01-05 16:59:00" "2015-01-05 17:59:00" ...
 $ m0p0             : POSIXct, format: "2015-01-05 15:00:00" "2015-01-05 15:00:00" "2015-01-05 15:00:00" ...
 $ m1p0             : POSIXct, format: "2015-01-05 15:01:00" "2015-01-05 15:01:00" "2015-01-05 15:01:00" ...
 $ m2p0             : POSIXct, format: "2015-01-05 15:02:00" "2015-01-05 15:02:00" "2015-01-05 15:02:00" ...    

它继续...
但截断输出,如下所示:

$ m33p1            : POSIXct, format: "2015-01-05 15:34:00" "2015-01-05 15:34:00" "2015-01-05 15:34:00" ...
$ m34p1            : POSIXct, format: "2015-01-05 15:35:00" "2015-01-05 15:35:00" "2015-01-05 15:35:00" ...
$ m35p1            : POSIXct, format: "2015-01-05 15:36:00" "2015-01-05 15:36:00" "2015-01-05 15:36:00" ...
$ m36p1            : POSIXct, format: "2015-01-05 15:37:00" "2015-01-05 15:37:00" "2015-01-05 15:37:00" ...
[list output truncated]

如何显示602个变量的完整列表？

Answer 1

您可以使用参数list.len:

str(df, list.len=ncol(df))

如果你想打印更多的观察结果你可以设置参数vec.len,也可以查看?str所有参数的文档.

Answer 2

通过使用参数list.len,可以选择要列出的数据框中的变量数.有两种选择:

a)您可以选择要列出的变量数量;

str(df, list.len = 602) # in this case I'm listing 602 variables.

b)您选择列出数据帧的变量总数(如user1981275所述);

str(df, list.len = ncol(df))

查看R帮助以获取更多信息

> ?str