R：如何在lapply（）中应用grep（）

HNSKD 发表于 Dev

HNSKD

我想在R中应用grep（），但是在lapply（）中我并不是很好。我了解lapply可以列出一个列表，将功能应用于每个成员并输出一个列表。例如，让x一个由2个成员组成的列表。

> x<-strsplit(docs$Text," ")
> 
> x
[[1]]
 [1] "I"         "lovehttp"  "my"        "mum."      "I"         "love"     
 [7] "my"        "dad."      "I"         "love"      "my"        "brothers."

[[2]]
 [1] "I"         "live"      "in"        "Eastcoast" "now."      "Job.I"    
 [7] "used"      "to"        "live"      "in"        "WestCoast."

我想应用grep（）函数删除由http组成的单词。因此，我将申请：

> lapply(x,grep(pattern="http",invert=TRUE, value=TRUE))

但它不起作用，它说

Error in grep(pattern = "http", invert = TRUE, value = TRUE) : 
argument "x" is missing, with no default

所以，我尝试

> lapply(x,grep(pattern="http",invert=TRUE, value=TRUE,x))

但是它说

Error in match.fun(FUN) : 
'grep(pattern = "http", invert = TRUE, value = TRUE, x)' is not a 
function, character or symbol

请帮助，谢谢！

雷特尔

可以一行完成：

lst <- lapply(lst, grep, pattern="http", value=TRUE, invert=TRUE)

#lst
#[[1]]
# [1] "I"         "my"        "mum."      "I"         "love"      "my"        "dad."      "I"         "love"      "my"        "brothers."
#
#[[2]]
# [1] "I"          "live"       "in"         "Eastcoast"  "now."       "Job.I"      "used"       "to"         "live"       "in"         "WestCoast."

如果您不想删除包含模式的整个单词，而只删除模式本身，而保留其余单词（如注释中所述），则可以使用gsub代替grep：

lapply(lst, gsub, pattern="http", replacement="")
#[[1]]
# [1] "I"         "love"      "my"        "mum."      "I"         "love"      "my"        "dad."      "I"         "love"      "my"        "brothers."
#
#[[2]]
# [1] "I"          "live"       "in"         "Eastcoast"  "now."       "Job.I"      "used"       "to"         "live"       "in"         "WestCoast."

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