如何在R中摺疊/重新編碼一個變量

我使用Outlook on Life數據集並對收入感興趣。受訪者選擇了以下19個選項之一：

Less than $5,000  
$5,000 to $7,499  
$7,500 to $9,999  
$10,000 to $12,499 
$12,500 to $14,999 
$15,000 to $19,999 
$20,000to $24,999  
$25,000 to $29,999 
$30,000 to $34,999 
$35,000 to $39,999 
$40,000 to $49,999 
$50,000 to $59,999 
$60,000 to $74,999 
$75,000 to $84,999 
$85,000 to $99,999 
$100,000 to $124,999 
$125,000 to $149,999 
$150,000 to $174,999 
$175,000 or more

我要崩潰了，並簡化該下列只是爲了讓情節更加理解：

在貧困線（$ 0 - 24,999）
工人階級（$ 25 000 - 34,999），
下階層（$ 35,000個 - 60,000），
中產階級（$ 60,000 - 100,000），
上層中產階級（$ 100,000 - 150,000），
最高5％（$ 150,000+）。

我該如何去重新編碼呢？

謝謝！

來源

2015-09-26 Katherine

嘗試切功能 – Chris

你的間隔是有問題的。如果有人賺了22,000，他們會選擇第7組（20k - 24,999）。你會希望他們在貧困線下。但是24k的人也會選擇組7，但他們在工作班。你怎麼知道它的區別？ –

是的，這是有問題的。我可以按摩我想要的分組，使它們更適合預先建立的間隔。所以我可以讓貧困線下去24,999。然後工作班34,999。 – Katherine

重新編碼因子的最簡單方法是認識到levels函數可以接受可用於重新映射因子水平的值列表。

我假設你的數據已經是一個因素（正如你所說的「受訪者必須選擇以下19個選項之一」），這意味着使用cut函數並沒有什麼意義。

這裏是它在動作的簡單示例：

z <- gl(3, 2, 12) # [1] 1 1 2 2 3 3 1 1 2 2 3 3, Levels: 1 2 3 
levels(z) <- list(A = c(1,3), B = 2) 
z # [1] A A B B A A A A B B A A, Levels: A B

正如可以從上面的例子中看到的，我們已重新編碼的水平1和3爲A組和電平2爲基B.所以你的問題可以用類似的方式完成：

groups <- as.factor(sample(c("Less than $5,000", 
"$5,000 to $7,499", 
"$7,500 to $9,999", 
"$10,000 to $12,499", 
"$12,500 to $14,999", 
"$15,000 to $19,999", 
"$20,000to $24,999", 
"$25,000 to $29,999", 
"$30,000 to $34,999", 
"$35,000 to $39,999", 
"$40,000 to $49,999", 
"$50,000 to $59,999", 
"$60,000 to $74,999", 
"$75,000 to $84,999", 
"$85,000 to $99,999", 
"$100,000 to $124,999", 
"$125,000 to $149,999", 
"$150,000 to $174,999", 
"$175,000 or more"), size=100, replace=T)) 

levels(groups) <- list(
    "Under poverty line"=c("Less than $5,000", 
     "$5,000 to $7,499", 
     "$7,500 to $9,999", 
     "$10,000 to $12,499", 
     "$12,500 to $14,999", 
     "$15,000 to $19,999", 
     "$20,000to $24,999"), 
    "Working class"=c("$25,000 to $29,999", 
        "$30,000 to $34,999"), 
    "Lower middle class"=c("$35,000 to $39,999", 
         "$40,000 to $49,999", 
         "$50,000 to $59,999"), 
    "Middle class"=c("$60,000 to $74,999", 
        "$75,000 to $84,999", 
        "$85,000 to $99,999"), 
    "Upper middle class"=c("$100,000 to $124,999", 
         "$125,000 to $149,999"), 
    "Top 5 percent"=c("$150,000 to $174,999", 
        "$175,000 or more") 
)

來源

2015-09-27 11:26:45 chappers

如何在R中摺疊/重新編碼一個變量

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