R函数(lapply {ifelse})？根据索引df返回数据文件中的值

Question

我有一个相当大的位置和关联点的数据集。所有的点和位置都是索引值。我试图以简洁的方式将索引值从另一个df转换为位置代码，同时保留"na“值。这是我尝试过的。

a <- c(1,2,3,5, "na", "na")
b <- c(2,1,5,7,3,6)
c <- c(3,6,2,4,1,"na")

df <- rbind(a,b,c) 
df <- as.data.frame(df)

d <- c(1,2,3,4,5,6,7,8)
e <- c(0115,0116,1117,1119,1237,1456,1901,2135)
df2 <- cbind(d,e)
df2 <- as.data.frame(df2)

test <- lapply(df[2:6], function(x){ifelse(x %in% df2$d, df2$e, "na")})

结果是在第一列关联值的行之间重复值。有什么办法解决这个问题吗。显然，我不明白我是如何应用函数的。

Answer 1

在数据中使用NA而不是"na“。

df3 <- melt(df, id.var = "V1", value.name = "d")
merge(df3, df2, by = "d", all = TRUE)

#     d V1 variable    e
# 1   1  3       V5  115
# 2   1  2       V2  115
# 3   2  1       V2  116
# 4   2  3       V3  116
# 5   3  1       V3 1117
# 6   3  2       V5 1117
# 7   4  3       V4 1119
# 8   5  2       V3 1237
# 9   5  1       V4 1237
# 10  6  3       V2 1456
# 11  6  2       V6 1456
# 12  7  2       V4 1901
# 13  8 NA     <NA> 2135
# 14 NA  1       V6   NA
# 15 NA  1       V5   NA
# 16 NA  3       V6   NA

数据

df <- structure(list(V1 = c(1, 2, 3), V2 = c(2, 1, 6), V3 = c(3, 5, 
2), V4 = c(5, 7, 4), V5 = c(NA, 3, 1), V6 = c(NA, 6, NA)), .Names = c("V1", 
"V2", "V3", "V4", "V5", "V6"), row.names = c("a", "b", "c"), class = "data.frame")

df2 <- structure(list(d = c(1, 2, 3, 4, 5, 6, 7, 8), e = c(115, 116, 
1117, 1119, 1237, 1456, 1901, 2135)), .Names = c("d", "e"), row.names = c(NA, 
-8L), class = "data.frame")

使用lapply的

lapply(df[2:6], function(x){
  y <- which(!is.na(x))
  x[y] <- df2$e[na.omit(x)]
  # x[y] <- df2$e[df2$d[na.omit(x)]]  # using d
  return(x)
})
# $V2
# [1]  116  115 1456
# 
# $V3
# [1] 1117 1237  116
# 
# $V4
# [1] 1237 1901 1119
# 
# $V5
# [1]   NA 1117  115
# 
# $V6
# [1]   NA 1456   NA