如何在R中grep ?

I would like to choose rows based on the subsets of their names, for example

例如，我想根据它们的名称的子集来选择行。

If I have the following data:

如果我有以下数据:

data <- structure(c(91, 92, 108, 104, 87, 91, 91, 97, 81, 98), 
.Names = c("fee-", "fi", "fo-", "fum-", "foo-", "foo1234-", "123foo-", 
"fum-", "fum-", "fum-"))

how do I select the rows matching 'foo'?

如何选择匹配“foo”的行?

using grep() doesn't work:

使用grep()不工作:

 grep('foo', data)

returns:

integer(0)

what am I doing wrong? or, is there a better way?

我做错了什么?或者，有没有更好的方法?

Thanks!

谢谢!

3 个解决方案

#1

You need to grep the names property of data, not the values property.

您需要grep命名数据的属性，而不是值属性。

For your example, use

在你的例子中,使用

> grep("foo",names(data))
[1] 5 6 7
> data[grep("foo",names(data))]
  foo- foo1234-  123foo- 
  87       91       91

One other clean way to do this is using data frames.

另一种干净的方法是使用数据帧。

> data <- data.frame(values=c(91, 92, 108, 104, 87, 91, 91, 97, 81, 98), 
                   names = c("fee-", "fi", "fo-", "fum-", "foo-", "foo1234-", "123foo-", 
                   "fum-", "fum-", "fum-"))

> data$values[grep("foo",data$names)]
[1] 87 91 91

#2

Use subset in combination with regular expressions:

与正则表达式结合使用子集:

subset(your_data, regexpr("foo", your_data$your_column_to_match) > 0))

If you just care about a dataset with one column I guess you do not need to specify a column name...

如果您只关心一个包含一个列的数据集，那么您不需要指定一个列名……

Philip

菲利普

#3

> grep("foo",names(data), value=T)
[1] "foo-"     "foo1234-" "123foo-"

if value is true, it returns the content instead of the index

如果值为true，它将返回内容而不是索引。

#1