确定最终列中的值是否存在于各个行中

Determine if Value in Final Column exists in respective rows

我有一个数据框如下:

df1

ColA     ColB    ColC    ColD     ColE     COlF    ColG      Recs
   1      A-1   A - 3       B        B       NA                 C
   1              B-1     C R        D        E      NA         B
   1       NA       A       B        A                          B

如何确定 Recs 列中的最后一个是否在其各自的行中找到?

我在下面尝试过,但它不起作用,因为我的普通数据集中有重复项:

df1$Exist <- apply(df1, 1, FUN = function(x) 
            c("No", "Yes")[(anyDuplicated(x[!is.na(x) & x != "" ])!=0) +1])

还有空格、NA 和包含空格和破折号的字符值。

最终输出应该是:

ColA     ColB    ColC    ColD     ColE     COlF    ColG      Recs    Exist?
   1      A-1   A - 3       B        B       NA                 C        No
   1              B-1     C R        D        E      NA         B        No
   1       NA       A       B        A                          B       Yes

谢谢

exist <- rep(NA, nrow(df1))
for (i in 1:nrow(df1)) {
exist[i] <- df1$Recs[i] %in% df1[i, 1:7]
}
df1 <- cbind(df1, exist)

如果我理解正确的话,这应该有效:

# Compute column index of reference variable
col_ind <- which(colnames(df1) == "Recs")

# Compute boolean vector of presence
present_bool <- apply(df1, 1, function(row) {
  any(row[col_ind] == row[-col_ind], na.rm = TRUE)
})

# Create the desired column
df1$Exist <- ifelse(present_bool, "Yes", "No")

为了效率,你可以在这里使用data.table

library(data.table)
setDT(df)[, Exist := Recs %chin% unlist(.SD), .SDcols=-"Recs", by=1:nrow(df)]

这给出了

   ColA ColB ColC ColD ColE COlF ColG Recs  Exist
1:    1  A-1  A-3    B    B   NA   NA    C  FALSE
2:    1       B-1  C R    D    E   NA    B  FALSE
3:    1   NA    A    B    A        NA    B   TRUE

原始数据:

df <-structure(list(ColA = c(1L, 1L, 1L), ColB = c("A-1", "", NA), 
    ColC = c("A-3", "B-1", "A"), ColD = c("B", "C R", "B"), ColE = c("B", 
    "D", "A"), COlF = c(NA, "E", ""), ColG = c(NA, NA, NA), Recs = c("C", 
    "B", "B")), .Names = c("ColA", "ColB", "ColC", "ColD", "ColE", 
"COlF", "ColG", "Recs"), row.names = c(NA, -3L), class = "data.frame")

这应该是另一种获得所需结果的方法:

f.checkExist <- function(x) {
 grepl(df[x, 8], df[x, 1:7])
}

df$exists <- grepl(T, lapply(1:nrow(df), f.checkExist))