Question

我有一个数据框test，其列是因子

class(test)
[1] "data.frame"

sapply(test, class)
  street     city    state 
"factor" "factor" "factor"

如果我尝试将这些列转换为sapply()的字符，则会出错并且我不确定原因

test <- as.data.frame(sapply(test, as.character))

sapply(test, class)
  street     city    state 
"factor" "factor" "factor"

我希望输出是所有字符列。为什么列不转换以及如何将所有因子列转换为字符？

以下是测试数据：

> dput(test)
structure(list(street = structure(c(5L, 1L, 6L, 2L, 3L, 4L), .Label = c("12057 Wilshire Blvd", 
"15300 Sunset Boulevard", "17380 Sunset Blvd", "1898 Westwood Blvd.", 
"3006 Sepulveda Blvd.", "514 Palisades Drive"), class = "factor"), 
    city = structure(c(1L, 1L, 2L, 2L, 2L, 3L), .Label = c("Los Angeles", 
    "Pacific Palisades", "Westwood"), class = "factor"), state = structure(c(1L, 
    1L, 1L, 1L, 1L, 1L), .Label = "CA", class = "factor")), .Names = c("street", 
"city", "state"), row.names = c(NA, -6L), class = "data.frame")

Answer 1

尝试mutate_if，这也应该为您提供更多控制权：

mutate_if(test, is.factor, as.character)

Answer 2

试试这个：

test[] <- lapply(test, as.character)

或者这个：

test <- modifyList(test, lapply(test, as.character))

或者这个：

test <- replace(test, TRUE, lapply(test, as.character))

Answer 3

这应该可以解决问题，它会将apply函数应用于数据帧的每一列。 as.data.frame函数将返回一个矩阵，因此需要通过用test <- as.data.frame(apply(test, 2, as.character), stringsAsFactors = FALSE)

包装它来强制转换为数据帧。

func convertType(from item: Int) -> Float {
    return Float(item)
}

var item: Float = convertType(from: 1)

使用sapply将列转换为字符

3 个答案: