Question

我正在使用具有结构的df

df = data.frame(customer = c(1,2)
                , destination_1 = c("c", "b")
                , destination_2 = c("a", NA)
                )
+----------+----------------+---------------+
| customer |  destination_1 | destination_2 | 
+----------+----------------+---------------+
| 1        | c              | a             |
+----------+----------------+---------------+
| 2        | b              | NA            | 
+----------+----------------+---------------+

没有行（无列）目标的排序。例如。我们看到，对于客户1，他的目的地布置为c，a

我希望将n列添加到df（其中n =整个df中唯一目标的数量）作为标记字段，以显示每个客户是否都曾去过每个目标。即

df$destination_a <- c(1,0)
df$destination_b <- c(0,1)
df$destination_c <- c(1,0)

+----------+----------------+---------------+---------------+---------------+---------------+
| customer |  destination_1 | destination_2 | destination_a | destination_b | destination_c |
+----------+----------------+---------------+---------------+---------------+---------------+
| 1        | c              | a             | 1             | 0             | 1             |
+----------+----------------+---------------+---------------+---------------+---------------+
| 2        | b              | NA            | 0             | 1             | 0             |
+----------+----------------+---------------+---------------+---------------+---------------+

我对如何实现这一目标只有丝毫想法。可能在循环中使用Apply？如果可能的话，我将更喜欢基本的R解决方案。我期待听到任何想法。谢谢。

Answer 1

您可以

# 1) get the number and destination names of required columns by extracting unique values from your dataframe:
df2 <- df[,c(2,3,4,5)]  #subsetting variables containing the destinations (modify column numbers if needed)
all <- c()
for (i in 1:ncol(df2)){for (j in 1:nrow(df2)){all <- c(all, as.character((df2[j,i])))}}
all <- sort(unique(all))

# 2) from these values make column names and columns and fill them with NAs
all_names <- paste("destination_", all, sep="")
df[,all_names] <- NA

# At this point you have the dataframe with required columns, and you can 
# 3) now you can fill these columns with values (0 or 1):

for (j in 1:nrow(df)){
  rowvalues <- c()
  for (k in 1:ncol(df)){rowvalues <- c(rowvalues, as.character((df[j,k])))}

  for (i in 1:length(all)){
    x <- (ncol(df)-length(all))
    if (all[i] %in% rowvalues){val <- 1}
    else {val <- 0}
    df[j, (i+x)] <- val  
  }
}

从宽格式标记字段

1 个答案: