使用grepl重置列

Question

我正在尝试重写一些东西，比如：

数据$ProgramData$Program%在% c('A1'，'1A1'，'A'，'AA'，'11A') <- 'A‘

我想使用SQL中类似于%A%的东西来重写这行代码。我相信有一种方法可以使用grepl函数来实现这一点，但我不知道该如何实现。

Answer 1

你差不多就在那儿了。一个简单的grepl应用程序就可以做到这一点。

Data$Program[grepl(Data$Program,'A')] <- 'A'

例如，对于虹膜数据集：

myIris<-iris
myIris$Species <- as.character(myIris$Species)
myIris$Species[grepl('i',myIris$Species)] <- "I"
myIris$Species
  [1] "setosa" "setosa" "setosa" "setosa" "setosa" "setosa" "setosa" "setosa"
  [9] "setosa" "setosa" "setosa" "setosa" "setosa" "setosa" "setosa" "setosa"
 [17] "setosa" "setosa" "setosa" "setosa" "setosa" "setosa" "setosa" "setosa"
 [25] "setosa" "setosa" "setosa" "setosa" "setosa" "setosa" "setosa" "setosa"
 [33] "setosa" "setosa" "setosa" "setosa" "setosa" "setosa" "setosa" "setosa"
 [41] "setosa" "setosa" "setosa" "setosa" "setosa" "setosa" "setosa" "setosa"
 [49] "setosa" "setosa" "I"      "I"      "I"      "I"      "I"      "I"     
 [57] "I"      "I"      "I"      "I"      "I"      "I"      "I"      "I"     
 [65] "I"      "I"      "I"      "I"      "I"      "I"      "I"      "I"     
 [73] "I"      "I"      "I"      "I"      "I"      "I"      "I"      "I"     
 [81] "I"      "I"      "I"      "I"      "I"      "I"      "I"      "I"     
 [89] "I"      "I"      "I"      "I"      "I"      "I"      "I"      "I"     
 [97] "I"      "I"      "I"      "I"      "I"      "I"      "I"      "I"     
[105] "I"      "I"      "I"      "I"      "I"      "I"      "I"      "I"     
[113] "I"      "I"      "I"      "I"      "I"      "I"      "I"      "I"     
[121] "I"      "I"      "I"      "I"      "I"      "I"      "I"      "I"     
[129] "I"      "I"      "I"      "I"      "I"      "I"      "I"      "I"     
[137] "I"      "I"      "I"      "I"      "I"      "I"      "I"      "I"     
[145] "I"      "I"      "I"      "I"      "I"      "I"