R:gsub,pattern = vector和replacement = vector

时间:2013-10-17 10:52:09

标签: r

正如标题所述,我正在尝试使用gsub,其中我使用向量作为“模式”和“替换”。目前,我有一个代码如下:

  names(x1) <- gsub("2110027599", "Inv1", names(x1)) #x1 is a data frame
  names(x1) <- gsub("2110025622", "Inv2", names(x1))
  names(x1) <- gsub("2110028045", "Inv3", names(x1))
  names(x1) <- gsub("2110034716", "Inv4", names(x1))
  names(x1) <- gsub("2110069349", "Inv5", names(x1))
  names(x1) <- gsub("2110023264", "Inv6", names(x1))

我希望做的是这样的事情:

  a <- c("2110027599","2110025622","2110028045","2110034716", "2110069349", "2110023264")
  b <- c("Inv1","Inv2","Inv3","Inv4","Inv5","Inv6")
  names(x1) <- gsub(a,b,names(x1))

我猜测某个地方有一个可以做到这一点的应用功能,但我不太确定使用哪一个!

编辑:名称(x1)看起来像这样(还有更多列,但我将它们排除在外):

> names(x1)
  [1] "2110023264A.Ms.Amp"        "2110023264A.Ms.Vol"        "2110023264A.Ms.Watt"       "2110023264A1.Ms.Amp"      
  [5] "2110023264A2.Ms.Amp"       "2110023264A3.Ms.Amp"       "2110023264A4.Ms.Amp"       "2110023264A5.Ms.Amp"      
  [9] "2110023264B.Ms.Amp"        "2110023264B.Ms.Vol"        "2110023264B.Ms.Watt"       "2110023264B1.Ms.Amp"      
 [13] "2110023264Error"           "2110023264E-Total"         "2110023264GridMs.Hz"       "2110023264GridMs.PhV.phsA"
 [17] "2110023264GridMs.PhV.phsB" "2110023264GridMs.PhV.phsC" "2110023264GridMs.TotPFPrc" "2110023264Inv.TmpLimStt"  
 [21] "2110023264InvCtl.Stt"      "2110023264Mode"            "2110023264Mt.TotOpTmh"     "2110023264Mt.TotTmh"      
 [25] "2110023264Op.EvtCntUsr"    "2110023264Op.EvtNo"        "2110023264Op.GriSwStt"     "2110023264Op.TmsRmg"      
 [29] "2110023264Pac"             "2110023264PlntCtl.Stt"     "2110023264Serial Number"   "2110025622A.Ms.Amp"       
 [33] "2110025622A.Ms.Vol"        "2110025622A.Ms.Watt"       "2110025622A1.Ms.Amp"       "2110025622A2.Ms.Amp"      
 [37] "2110025622A3.Ms.Amp"       "2110025622A4.Ms.Amp"       "2110025622A5.Ms.Amp"       "2110025622B.Ms.Amp"       
 [41] "2110025622B.Ms.Vol"        "2110025622B.Ms.Watt"       "2110025622B1.Ms.Amp"       "2110025622Error"          
 [45] "2110025622E-Total"         "2110025622GridMs.Hz"       "2110025622GridMs.PhV.phsA" "2110025622GridMs.PhV.phsB"

我希望得到的是:

> names(x1)
  [1] "Inv6A.Ms.Amp"        "Inv6A.Ms.Vol"        "Inv6A.Ms.Watt"       "Inv6A1.Ms.Amp"       "Inv6A2.Ms.Amp"      
  [6] "Inv6A3.Ms.Amp"       "Inv6A4.Ms.Amp"       "Inv6A5.Ms.Amp"       "Inv6B.Ms.Amp"        "Inv6B.Ms.Vol"       
 [11] "Inv6B.Ms.Watt"       "Inv6B1.Ms.Amp"       "Inv6Error"           "Inv6E-Total"         "Inv6GridMs.Hz"      
 [16] "Inv6GridMs.PhV.phsA" "Inv6GridMs.PhV.phsB" "Inv6GridMs.PhV.phsC" "Inv6GridMs.TotPFPrc" "Inv6Inv.TmpLimStt"  
 [21] "Inv6InvCtl.Stt"      "Inv6Mode"            "Inv6Mt.TotOpTmh"     "Inv6Mt.TotTmh"       "Inv6Op.EvtCntUsr"   
 [26] "Inv6Op.EvtNo"        "Inv6Op.GriSwStt"     "Inv6Op.TmsRmg"       "Inv6Pac"             "Inv6PlntCtl.Stt"    
 [31] "Inv6Serial Number"   "Inv2A.Ms.Amp"        "Inv2A.Ms.Vol"        "Inv2A.Ms.Watt"       "Inv2A1.Ms.Amp"      
 [36] "Inv2A2.Ms.Amp"       "Inv2A3.Ms.Amp"       "Inv2A4.Ms.Amp"       "Inv2A5.Ms.Amp"       "Inv2B.Ms.Amp"       
 [41] "Inv2B.Ms.Vol"        "Inv2B.Ms.Watt"       "Inv2B1.Ms.Amp"       "Inv2Error"           "Inv2E-Total"        
 [46] "Inv2GridMs.Hz"       "Inv2GridMs.PhV.phsA" "Inv2GridMs.PhV.phsB" 

6 个答案:

答案 0 :(得分:24)

已有很多解决方案,这里还有一个:

qdap包:

library(qdap)
names(x1) <- mgsub(a,b,names(x1))

答案 1 :(得分:16)

来自str_replace_all的字符串文档,&#34;如果要将多个模式和替换应用于同一个字符串,请将命名的版本传递给模式。&#34; < / p>

因此使用上面的a,b和名称(x1)

library(stringr)
names(b) <- a
str_replace_all(names(x1), b)

答案 2 :(得分:9)

新答案

如果我们可以做出另一个假设,以下内容应该有效。这次的假设是你真的有兴趣替换names(x1)中每个值的前10个字符。

在这里,我将names(x1)存储为名为“X1”的字符向量。该解决方案主要使用substr将X1中的值分为2个部分,match以找出正确的替换选项,并paste将所有内容重新组合在一起。

a <- c("2110027599", "2110025622", "2110028045",
       "2110034716", "2110069349", "2110023264")
b <- c("Inv1","Inv2","Inv3","Inv4","Inv5","Inv6")

X1pre <- substr(X1, 1, 10)
X1post <- substr(X1, 11, max(nchar(X1)))

paste0(b[match(X1pre, a)], X1post)
#  [1] "Inv6A.Ms.Amp"        "Inv6A.Ms.Vol"        "Inv6A.Ms.Watt"      
#  [4] "Inv6A1.Ms.Amp"       "Inv6A2.Ms.Amp"       "Inv6A3.Ms.Amp"      
#  [7] "Inv6A4.Ms.Amp"       "Inv6A5.Ms.Amp"       "Inv6B.Ms.Amp"       
# [10] "Inv6B.Ms.Vol"        "Inv6B.Ms.Watt"       "Inv6B1.Ms.Amp"      
# [13] "Inv6Error"           "Inv6E-Total"         "Inv6GridMs.Hz"      
# [16] "Inv6GridMs.PhV.phsA" "Inv6GridMs.PhV.phsB" "Inv6GridMs.PhV.phsC"
# [19] "Inv6GridMs.TotPFPrc" "Inv6Inv.TmpLimStt"   "Inv6InvCtl.Stt"     
# [22] "Inv6Mode"            "Inv6Mt.TotOpTmh"     "Inv6Mt.TotTmh"      
# [25] "Inv6Op.EvtCntUsr"    "Inv6Op.EvtNo"        "Inv6Op.GriSwStt"    
# [28] "Inv6Op.TmsRmg"       "Inv6Pac"             "Inv6PlntCtl.Stt"    
# [31] "Inv6Serial Number"   "Inv2A.Ms.Amp"        "Inv2A.Ms.Vol"       
# [34] "Inv2A.Ms.Watt"       "Inv2A1.Ms.Amp"       "Inv2A2.Ms.Amp"      
# [37] "Inv2A3.Ms.Amp"       "Inv2A4.Ms.Amp"       "Inv2A5.Ms.Amp"      
# [40] "Inv2B.Ms.Amp"        "Inv2B.Ms.Vol"        "Inv2B.Ms.Watt"      
# [43] "Inv2B1.Ms.Amp"       "Inv2Error"           "Inv2E-Total"        
# [46] "Inv2GridMs.Hz"       "Inv2GridMs.PhV.phsA" "Inv2GridMs.PhV.phsB"

旧答案

如果我们可以假设names(x1)与模式和替换的顺序相同,并且它基本上是一对一替换,那么您可能只能使用sapply

以下是该特定情况的一个示例:

想象一下“names(x)”看起来像这样:

X1 <- paste0("A2", a, sequence(length(a)))
X1
# [1] "A221100275991" "A221100256222" "A221100280453" 
# [4] "A221100347164" "A221100693495" "A221100232646"

这是我们的patternreplacement向量:

a <- c("2110027599", "2110025622", "2110028045", 
       "2110034716", "2110069349", "2110023264")
b <- c("Inv1","Inv2","Inv3","Inv4","Inv5","Inv6")

如果这些假设有效,我们可以使用sapply

sapply(seq_along(a), function(x) gsub(a[x], b[x], X1[x]))
# [1] "A2Inv11" "A2Inv22" "A2Inv33" "A2Inv44" "A2Inv55" "A2Inv66"

答案 3 :(得分:2)

不知何故,names<-match似乎更适合......

names( x1 ) <- b[ match( names( x1 ) , a ) ]

但我假设向量a的元素是names的实际data.frame

如果a确实是names x1 grepl中找到的模式,那么此names<-方法会new <- sapply( a , grepl , x = names( x1 ) ) names( x1 ) <- b[ apply( new , 1 , which.max ) ] 可能有用......

{{1}}

答案 4 :(得分:1)

尝试mapply

names(x1) <- mapply(gsub, a, b, names(x1), USE.NAMES = FALSE)

或者更容易str_replace来自stringr

library(stringr)
names(x1) <- str_replace(names(x1), a, b)

答案 5 :(得分:0)

我需要做类似的事情,但必须使用基数R。只要您的向量长度相同,我认为这就会起作用

for (i in seq_along(a)){
  names(x1) <- gsub(a[i], b[i], names(x1))
}