Question

假设我有一个在数字和字符串之间混合的值向量，如下所示：

df<-structure(c("Location", "SKU", "Manufacturer", "Size", 
            "State", "43488", "43489", "43490", 
            "43491"), .Names = c("col1","col2","col3","col4","col5","col6","col7","col8","col9"))

我只想将数字值（最初是Excel日期）转换为R日期格式，而其余变量保持原样。我正在使用的向量在较大的data.frame中只是一行，所以我宁愿避免将日期从字符串中拆分并随后合并。到目前为止，我尝试了以下方法：

as.Date(as.numeric(df), origin = "1899-12-30")
[1] NA           NA           NA           NA           NA           "2019-01-23" "2019-01-24" "2019-01-25" "2019-01-26"

但是这会将字符串转换为NA。

Answer 1

我们可以找出其中包含数字的索引，然后仅将这些值转换为日期。

DB::select("SELECT * FROM my_table WHERE vendor_id IN (" . implode(array_keys($vendors), ',') . ")", 
    $vendors);

使用inds <- grepl("\\d+", df) df[inds] <- format(as.Date(as.numeric(df[inds]), origin = "1899-12-30")) df # col1 col2 col3 col4 col5 #"Location" "SKU" "Manufacturer" "Size" "State" # col6 col7 col8 col9 #"2019-01-23" "2019-01-24" "2019-01-25" "2019-01-26"对上述答案的一种变体，而没有创建其他变量（ifelse）

inds

Answer 2

尝试一下：
-创建一个函数来检查值是否可以强制为numeric类型
-如果为true，则将其转换为numeric并将其格式化为date
-如果为false，则按原样返回值

df<-structure(c("Location", "SKU", "Manufacturer", "Size", 
                "State", "43488", "43489", "43490", 
                "43491"), .Names = c("col1","col2","col3","col4","col5","col6","col7","col8","col9"))

convert_num_to_date = function(x){
  if (is.na(as.numeric(x))) {
    return(x)
  } else {
    x = format(as.Date(as.numeric(x), origin = "1899-12-30"))
    return(x)
  }
}

df = sapply(df, convert_num_to_date)

仅将R中的数值转换为日期字符串

2 个答案: