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0から6.115053の範囲の周波数の連続変数があります。私はそれを6つのレベルに分割する必要があります、私の分析はこの方法でより読みやすくなります。

私が試してみました：

frequency.new <-  hist(all$frequency, 6, plot = FALSE)
all$frequency <- as.factor(frequency.new)

しかし、私は理解できないエラーが発生します：

Error in sort.list(y) : 
  'x' must be atomic for 'sort.list'
Have you called 'sort' on a list?

誰かが私を助けることができますか？

どうもありがとう！

カテリーナ

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1 に答える 1

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ベースRの関数を確認する必要がありますcut()。さらに冒険する前に、回答の最後の行（太字）にも注意してください。

> set.seed(42)
> cut(runif(50), 6)
 [1] (0.825,0.99]    (0.825,0.99]    (0.167,0.332]   (0.825,0.99]   
 [5] (0.496,0.661]   (0.496,0.661]   (0.661,0.825]   (0.00296,0.167]
 [9] (0.496,0.661]   (0.661,0.825]   (0.332,0.496]   (0.661,0.825]  
[13] (0.825,0.99]    (0.167,0.332]   (0.332,0.496]   (0.825,0.99]   
[17] (0.825,0.99]    (0.00296,0.167] (0.332,0.496]   (0.496,0.661]  
[21] (0.825,0.99]    (0.00296,0.167] (0.825,0.99]    (0.825,0.99]   
[25] (0.00296,0.167] (0.496,0.661]   (0.332,0.496]   (0.825,0.99]   
[29] (0.332,0.496]   (0.825,0.99]    (0.661,0.825]   (0.661,0.825]  
[33] (0.332,0.496]   (0.661,0.825]   (0.00296,0.167] (0.825,0.99]   
[37] (0.00296,0.167] (0.167,0.332]   (0.825,0.99]    (0.496,0.661]  
[41] (0.332,0.496]   (0.332,0.496]   (0.00296,0.167] (0.825,0.99]   
[45] (0.332,0.496]   (0.825,0.99]    (0.825,0.99]    (0.496,0.661]  
[49] (0.825,0.99]    (0.496,0.661]  
6 Levels: (0.00296,0.167] (0.167,0.332] (0.332,0.496] ... (0.825,0.99]

cut()は、観測データが該当する 6 つのグループ (この場合は 6 つのグループ) のどれにインデックスを付ける係数を返します。これは、データの範囲を等間隔の 6 つのグループに単純に分割したものです。?cut間隔の端で何をすべきかについての詳細を読んでください。

コードが失敗する理由は、によって返されるオブジェクトがhist()、グループに分割されたデータよりもはるかに多くを含むリストであるためです。

> foo <- hist(runif(50), breaks = 6, plot = FALSE)
> str(foo)
List of 7
 $ breaks     : num [1:6] 0 0.2 0.4 0.6 0.8 1
 $ counts     : int [1:5] 12 13 7 13 5
 $ intensities: num [1:5] 1.2 1.3 0.7 1.3 0.5
 $ density    : num [1:5] 1.2 1.3 0.7 1.3 0.5
 $ mids       : num [1:5] 0.1 0.3 0.5 0.7 0.9
 $ xname      : chr "runif(50)"
 $ equidist   : logi TRUE
 - attr(*, "class")= chr "histogram"

したがって、これを因数に変換することはできません-Rはそれを行う方法を知りません。hist()また、 6 つのグループに分類されたデータは返されないことに注意してください。これは、ヒストグラムの作成に役立つ他の情報を提供します。とは異なり、かなりのブレークが発生することにも注意してくださいcut()。これらのかなりの休憩が必要な場合は、次の方法で再現できますhist()。

> set.seed(42)
> x <- runif(50)
> brks <- pretty(range(x), n = 6, min.n = 1)
> cut(x, breaks = brks)
 [1] (0.8,1]   (0.8,1]   (0.2,0.4] (0.8,1]   (0.6,0.8] (0.4,0.6] (0.6,0.8]
 [8] (0,0.2]   (0.6,0.8] (0.6,0.8] (0.4,0.6] (0.6,0.8] (0.8,1]   (0.2,0.4]
[15] (0.4,0.6] (0.8,1]   (0.8,1]   (0,0.2]   (0.4,0.6] (0.4,0.6] (0.8,1]  
[22] (0,0.2]   (0.8,1]   (0.8,1]   (0,0.2]   (0.4,0.6] (0.2,0.4] (0.8,1]  
[29] (0.4,0.6] (0.8,1]   (0.6,0.8] (0.8,1]   (0.2,0.4] (0.6,0.8] (0,0.2]  
[36] (0.8,1]   (0,0.2]   (0.2,0.4] (0.8,1]   (0.6,0.8] (0.2,0.4] (0.4,0.6]
[43] (0,0.2]   (0.8,1]   (0.4,0.6] (0.8,1]   (0.8,1]   (0.6,0.8] (0.8,1]  
[50] (0.6,0.8]
Levels: (0,0.2] (0.2,0.4] (0.4,0.6] (0.6,0.8] (0.8,1]

しかし、なぜデータをそのように離散化する必要があるのか​​、それが理にかなっているかどうかを自問する必要があります。

于 2011-09-02T12:06:17.940 に答える