如何在data.table中按名称删除列？ [英] How do you delete a column by name in data.table?

问题描述

要移除 data.frame 中名为foo的列，我可以这样做：

df <-df [-grep（'foo'，colnames（df））]

一旦 df 被转换为 data.table 对象，则无法只删除列。

示例：

  df<  -  data.frame（id = 1： 100，foo = rnorm（100））
 df2 < -  df [-grep（'foo'，colnames（df））]＃works 
 df3<  -  data.table（df）
 df3 [-grep（'foo'，colnames（df3））] 
  

转换为 data.table 对象，这不再工作。

解决方案

以下任何操作都会从数据中删除列 foo 表 df3 ：

 ＃方法1即使在20GB的data.table）
 df3 [，foo：= NULL] 
 
 df3 [，c（foo，bar）：= NULL]＃删除两列
 
 myVar =foo
 df3 [，（myVar）：= NULL]＃lookup myVar contents 
 
＃方法2a  - 可能是多个）
＃列匹配regex 
 df3 [，grep（^ foo $，colnames（df3））：= NULL] 
 
＃方法2b  - 替代2a，在下面的意义上也是安全的
 df3 [，which（grepl（^ foo $，colnames（df3）））：= NULL] 
  
 
 
 
  data.table 也支持以下语法：
  ##方法3（然后可以分配给df3，
 df3 [，！foo，with = FALSE] 
  
虽然如果你实际上想从 df3 删除foo code>（与只打印 df3 减列foo的视图相反） 
 
 
 
（请注意，如果你使用的方法依赖于 grep（）或 grepl（），您需要设置 pattern =^ foo $ foo和buffoon c $ c>（ie包含 foo 作为子字符串的那些）也可以匹配和删除。）
 
 
 
 use：
 
 
 
接下来的两个成语也会起作用 -  如果 df3  c $ c>foo   - 但如果没有，可能会以意外的方式失败。例如，如果你使用它们中的任何一个来搜索不存在的列bar，你将得到一个零行data.table。 / p> 
 
 
因此，它们真的最适合于交互式使用，例如，希望显示一个data.table减去任何包含子字符foo。对于编程目的（或者如果你想从 df3 而不是从它的副本中实际删除列），方法1，2a和2b真的最佳选项。
 ＃方法4a：
 df3 [，-grep（^ foo $，colnames df3）），with = FALSE] 
 
＃方法4b：
 df3 [，！grepl（^ foo $，colnames（df3）），with = FALSE] 
  
 
To get rid of a column named "foo" in a data.frame, I can do:

df <- df[-grep('foo', colnames(df))]

However, once df is converted to a data.table object, there is no way to just remove a column.

Example:
df <- data.frame(id = 1:100, foo = rnorm(100))
df2 <- df[-grep('foo', colnames(df))] # works
df3 <- data.table(df)
df3[-grep('foo', colnames(df3))] 
But once it is converted to a data.table object, this no longer works. 
解决方案
Any of the following will remove column foo from the data.table df3:
# Method 1 (and preferred as it takes 0.00s even on a 20GB data.table)
df3[,foo:=NULL]

df3[, c("foo","bar"):=NULL]  # remove two columns

myVar = "foo"
df3[, (myVar):=NULL]   # lookup myVar contents

# Method 2a -- A safe idiom for excluding (possibly multiple)
# columns matching a regex
df3[, grep("^foo$", colnames(df3)):=NULL]

# Method 2b -- An alternative to 2a, also "safe" in the sense described below
df3[, which(grepl("^foo$", colnames(df3))):=NULL]
data.table also supports the following syntax:
## Method 3 (could then assign to df3, 
df3[, !"foo", with=FALSE]  
though if you were actually wanting to remove column "foo" from df3 (as opposed to just printing a view of df3 minus column "foo") you'd really want to use Method 1 instead.

(Do note that if you use a method relying on grep() or grepl(), you need to set pattern="^foo$" rather than "foo", if you don't want columns with names like "fool" and "buffoon" (i.e. those containing foo as a substring) to also be matched and removed.)

Less safe options, fine for interactive use:

The next two idioms will also work -- if df3 contains a column matching "foo" -- but will fail in a probably-unexpected way if it does not. If, for instance, you use any of them to search for the non-existent column "bar", you'll end up with a zero-row data.table.

As a consequence, they are really best suited for interactive use where one might, e.g., want to display a data.table minus any columns with names containing the substring "foo". For programming purposes (or if you are wanting to actually remove the column(s) from df3 rather than from a copy of it), Methods 1, 2a, and 2b are really the best options.
# Method 4a:
df3[, -grep("^foo$", colnames(df3)), with=FALSE]

# Method 4b: 
df3[, !grepl("^foo$", colnames(df3)), with=FALSE]


                        
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