Question

拥有这样的数据框：

   structure(list(price = structure(1:4, .Label = c("price1", "price2", 
"price3", "price4"), class = "factor"), col1 = structure(c(1L, 
2L, NA, 3L), .Label = c("text1", "text2", "text3"), class = "factor"), 
    col2 = structure(c(NA, 1L, NA, NA), .Label = "text1", class = "factor"), 
    col3 = structure(c(NA, 1L, NA, NA), .Label = "text3", class = "factor"), 
    col4 = structure(c(NA, 1L, NA, NA), .Label = "text4", class = "factor")), .Names = c("price", 
"col1", "col2", "col3", "col4"), class = "data.frame", row.names = c(NA, 
-4L))

如何将每行的值更改为列名并且是否存在（1和0）值？

示例输出：

price text1 text2 text3 text4
price1 1      0    0      0
price2 1      1    1      1
price3 0      0    0      0
price4 0      0    1      0

Answer 1

我们使用LOWER()为1以外的列创建逻辑矩阵，将其强制转换为二进制（is.na）并将输出分配回子集数据

或另一个选项是df1[-1] <- +(!is.na(df1[-1])) df1 # price col1 col2 #1 price1 1 0 #2 price2 1 1 #3 price3 0 0

lapply

对于新数据集

df1[-1] <- lapply(df1[-1], function(x) as.integer(!is.na(x)))

从多个值到分类值

1 个答案: