blocks|key|655349|text|我只是简单地在数据框中这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|655350|DT$col+=+NULL|code-block|syntax|javascript|655351|运行速度很快，据我所知不会造成任何问题。|655352|更新:如果您的DT非常大，则不是最好的方法，因为使用$<-操作符将导致对象复制。所以更好地使用：|offset|length|style|CODE|655353|DT[,+col:=NULL]|655354|entityMap^0|0|0|0|Q|3|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|V|8|@$K|W|L|X|M|N]]|9|@]|A|$]]|$1|O|3|P|5|D|7|Y|8|@]|9|@]|A|$E|F]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

I simply do it in the data frame kind of way:

<pre><code>DT$col = NULL
</code></pre>

Works fast and as far as I could see doesn't cause any problems.

UPDATE: not the best method if your DT is very large, as using the <code>$&lt;-</code> operator will lead to object copying. So better use:

<pre><code>DT[, col:=NULL]
</code></pre>

blocks|key|655429|text|您也可以使用set来实现这一点，这样可以避免循环中使用[.data.table的开销：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|655430|dt+<-+data.table(+a=letters,+b=LETTERS,+c=seq(26),+d=letters,+e=letters+)
set(+dt,+j=c(1L,3L,5L),+value=NULL+)
>+dt[1:5]
+++b+d
1:+A+a
2:+B+b
3:+C+c
4:+D+d
5:+E+e|code-block|syntax|javascript|655431|如果你想通过列名来做这件事，那么which(colnames(dt)+%25in%25+c("a","c","e"))应该适用于j。|655432|entityMap^0|6|3|R|C|0|0|G|13|1O|1|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|N|$]]

You can also use <code>set</code> for this, which avoids the overhead of <code>[.data.table</code> in loops:

<pre><code>dt &lt;- data.table( a=letters, b=LETTERS, c=seq(26), d=letters, e=letters )
set( dt, j=c(1L,3L,5L), value=NULL )
&gt; dt[1:5]
 b d
1: A a
2: B b
3: C c
4: D d
5: E e
</code></pre>

If you want to do it by column name, <code>which(colnames(dt) %in% c("a","c","e"))</code> should work for <code>j</code>.

blocks|key|441052|text|对于data.table，将该列指定为NULL会将其删除：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|441053|DT[,c("col1",+"col1",+"col2",+"col2")]+<-+NULL
%5E
%7C----+Notice+the+extra+comma+if+DT+is+a+data.table|code-block|syntax|javascript|441054|..。这相当于：|441055|DT$col1+<-+NULL
DT$col2+<-+NULL
DT$col3+<-+NULL
DT$col4+<-+NULL|441056|data.frame的等价物是：|441057|DF[c("col1",+"col1",+"col2",+"col2")]+<-+NULL
++++++%5E
++++++%7C----+Notice+the+missing+comma+if+DF+is+a+data.frame|441058|问:为什么在data.table版本中有逗号，而在data.frame版本中没有逗号？|441059|答:由于data.frames是以列列表的形式存储的，您可以跳过逗号。您也可以将其添加到中，但是需要将它们分配给一个NULL列表，即DF[,+c("col1",+"col2",+"col3")]+<-+list(NULL)。|offset|length|style|CODE|441060|entityMap^0|0|0|0|0|0|0|0|1M|4|1U|19|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|10|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|12|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|14|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|15|8|@$S|16|T|17|U|V]|$S|18|T|19|U|V]]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|X|$]]

For a data.table, assigning the column to NULL removes it:

<pre><code>DT[,c("col1", "col1", "col2", "col2")] &lt;- NULL
^
|---- Notice the extra comma if DT is a data.table
</code></pre>

... which is the equivalent of:

<pre><code>DT$col1 &lt;- NULL
DT$col2 &lt;- NULL
DT$col3 &lt;- NULL
DT$col4 &lt;- NULL
</code></pre>

The equivalent for a data.frame is:

<pre><code>DF[c("col1", "col1", "col2", "col2")] &lt;- NULL
 ^
 |---- Notice the missing comma if DF is a data.frame
</code></pre>

Q. Why is there a comma in the version for data.table, and no comma in the version for data.frame?

A. As data.frames are stored as a list of columns, you can skip the comma. You could also add it in, however then you will need to assign them to a list of <code>NULL</code>s, <code>DF[, c("col1", "col2", "col3")] &lt;- list(NULL)</code>.

blocks|key|441103|text|下面是一种方法，当您想要将列的#+of列设置为NULL时，给定它们的列名，您可以使用一个函数:)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|441104|deleteColsFromDataTable+<-+function+(train,+toDeleteColNames)+{

+++++++for+(myNm+in+toDeleteColNames)

+++++++train+<-+train+[,(myNm):=NULL]

+++++++return+(train)
}|code-block|syntax|javascript|441105|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here is a way when you want to set a # of columns to NULL given their column names
a function for your usage :)

<pre><code>deleteColsFromDataTable &lt;- function (train, toDeleteColNames) {

 for (myNm in toDeleteColNames)

 train &lt;- train [,(myNm):=NULL]

 return (train)
}
</code></pre>

blocks|key|655599|text|非常简单的选项，如果您有许多单独的列要在数据表中删除，并且您希望避免键入所有列名#careadviced|type|unstyled|depth|inlineStyleRanges|entityRanges|data|655600|dt+<-+dt[,+-c(1,4,6,17,83,104)]|code-block|syntax|javascript|655601|这将改为根据列号删除列。|655602|它的效率显然不是很高，因为它忽略了data.table的优点，但是如果您处理的行数少于500,000行，它就可以很好地工作|655603|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Very simple option in case you have many individual columns to delete in a data table and you want to avoid typing in all column names #careadviced

<pre><code>dt &lt;- dt[, -c(1,4,6,17,83,104)]
</code></pre>

This will remove columns based on column number instead.

It's obviously not as efficient because it bypasses data.table advantages but if you're working with less than say 500,000 rows it works fine

blocks|key|441150|text|DT[,c:=NULL]+#+remove+column+c|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|441151|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>DT[,c:=NULL] # remove column c
</code></pre>

blocks|key|655624|text|假设您的dt具有列col1、col2、col3、col4、col5和coln。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|655625|要删除其中的子集，请执行以下操作：|655626|vx+<-+as.character(bquote(c(col1,+col2,+col3,+coln)))[-1]
DT[,+paste0(vx):=NULL]|code-block|syntax|javascript|655627|entityMap^0|9|4|E|4|J|4|O|4|T|4|Y|4|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|N|$]]

Suppose your dt has columns <code>col1</code>, <code>col2</code>, <code>col3</code>, <code>col4</code>, <code>col5</code>, <code>coln</code>.

To delete a subset of them:

<pre><code>vx &lt;- as.character(bquote(c(col1, col2, col3, coln)))[-1]
DT[, paste0(vx):=NULL]
</code></pre>

To get rid of a column named "foo" in a <code>data.frame</code>, I can do:

<code>df &lt;- df[-grep('foo', colnames(df))]</code>

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

Example:

<pre><code>df &lt;- data.frame(id = 1:100, foo = rnorm(100))
df2 &lt;- df[-grep('foo', colnames(df))] # works
df3 &lt;- data.table(df)
df3[-grep('foo', colnames(df3))] 
</code></pre>

But once it is converted to a <code>data.table</code> object, this no longer works.

How do you delete a column by name in data.table?

要在data.frame中删除名为"foo“的列，我可以这样做：df <- df[-grep('foo', colnames(df))]但是，一旦将df转换为data.table对象，就不可能只删除一列。示例：df <- data.frame(id = 1:100, foo = rnorm(100))df2 <- df...

问如何在data.table中按名称删除列？
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问如何在data.table中按名称删除列？
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