Question

我正在尝试使用列表从2列以上（下面给出的2列）中提取信息，并创建另一列，其中包含列表中的字符串中的字符串，该列指定要查找的列中的哪一列第一。我有下面的示例以及所需的输出。希望这有助于我正在寻找的东西。

A<-c("This contains NYU", "This has NYU", "This has XT", "This has FIT", 
"Something something UNH","I got into UCLA","Hello XT")
B<-c("NYU","UT","USC","FIT","UNA","UCLA", "CA")
data<-data.frame(A,B)

list <- c("NYU","FIT","UCLA","CA","UT","USC")

                        A    B
1       This contains NYU  NYU
2            This has NYU   UT
3             This has XT  USC
4            This has FIT  FIT
5 Something something UNH  UNA
6         I got into UCLA UCLA
7                Hello XT   CA

我希望代码从列表中搜索并首先查看列A，如果找不到字符串，则查看列B，如果不是，则给出null。通过查看列表，我希望所需的输出看起来如下所示。

                        A    B    C
1       This contains NYU  NYU  NYU
2            This has NYU   UT  NYU
3             This has XT  USC  USC
4            This has FIT  FIT  FIT
5 Something something UNH  UNA <NA>
6         I got into UCLA UCLA UCLA
7                Hello XT   CA   CA

Answer 1

您可以将列表转换为regexpr，然后应用R regexpr函数：

expr <- paste0(list,collapse = "|")
# expr = "NYU|FIT|UCLA|CA|UT|USC" -> Reg expr means NYU or FIT or ......

data[,"C"] <- ""
cols <- rev(names(data)[-(which(names(data)=="C"))])

for(c in cols) {
 index <- regexpr(expr,data[,c])
 data[,"C"] <- ifelse(index != -1,substr(data[,c],index,index + attr(index,"match.length")-1),data[,"C"])     
}

希望这会有所帮助

Gottavianoni

Answer 2

使用来自tokenizers包的库（tokenizers）。

合并两列并使用合并的A和B

创建一个新列

ConstraintLayout

然后，按照下面的循环，它将在向量中提取值，然后你可以在现有数据帧中绑定向量。

data$newC <- paste(data$A, data$B, sep = " " )

希望它有所帮助。我正在按照你的预期获得上面的输出。

解决方案图片：

Answer 3

另一种方法可能是

#common between column A & vector l
C_tempA <- sapply(df$A, function(x) intersect(strsplit(as.character(x), split = " ")[[1]], l))
#common between column B & vector l
C_tempB <- sapply(df$B, function(x) intersect(as.character(x), l))

#column C calculation
df$C <- ifelse(C_tempA=="character(0)", C_tempB, C_tempA)
df$C[df$C=="character(0)"] <- NA

#final dataframe
df

输出是：

                        A    B    C
1       This contains NYU  NYU  NYU
2            This has NYU   UT  NYU
3             This has XT  USC  USC
4            This has FIT  FIT  FIT
5 Something something UNH  UNA   NA
6         I got into UCLA UCLA UCLA
7                Hello XT   CA   CA

示例数据：

df <- structure(list(A = structure(c(4L, 6L, 7L, 5L, 3L, 2L, 1L), .Label = c("Hello XT", 
"I got into UCLA", "Something something UNH", "This contains NYU", 
"This has FIT", "This has NYU", "This has XT"), class = "factor"), 
    B = structure(c(3L, 7L, 6L, 2L, 5L, 4L, 1L), .Label = c("CA", 
    "FIT", "NYU", "UCLA", "UNA", "USC", "UT"), class = "factor")), .Names = c("A", 
"B"), row.names = c(NA, -7L), class = "data.frame")

l <- c("NYU","FIT","UCLA","CA","UT","USC")

使用R中的列表从多个列中提取字符串

3 个答案: