2016-08-09 19 views
0

私は同じ人が書いたテキストをたくさん持っており、それぞれのテキストに使用するテンプレートを見積もっています。私はこのことについてつもりな方法は次のとおりです。Rデータを並べ替え

  1. は、すべてのテキスト
  2. ためTermDocumentMatrixは、各ペア
  3. の生のユークリッド距離がためにより大きいX距離(10を任意のペアをカット取る作成します引数)
  4. は森
  5. 復帰いくつかの要約統計情報と各テンプレートの一例を平ら

私は距離のペアを持つ点に到達することができますが、distインスタンスを私が扱えるものに変換することができません。 下部に再現可能な例があります。

distインスタンス内のデータが次のようになります。

dist instance example

行と列の名前は私がステップ5.

を達成行うために使用できるテキストの元のリストのインデックスに対応

これを解決しようとしてきたのは、col name, row name, valueの疎な行列です。私はつまり、この例では3つのテンプレートを返す、私のカットオフよりも大きいすべてのペアを切り出し、森を平坦に快適になります。この時点から

col, row, value 
    1 2 14.966630 
    1 3 12.449900 
    1 4 13.490738 
    1 5 12.688578 
    1 6 12.369317 
    2 3 12.449900 
    2 4 13.564660 
    2 5 12.922848 
    2 6 12.529964 
    3 4 5.385165 
    3 5 5.830952 
    3 6 5.830952 
    4 5 7.416198 
    4 6 7.937254 
    5 6 7.615773 

、グループのみが文書1を含む、グループは文書2及び含みます3番目、4番目、5番目、6番目のグループが含まれています。

distクラス内のベクトルを直接使用することで、これからマトリックスを作成して疎行させようとしましたが、私はそれを理解できないようです。

再現例:

tdm <- matrix(c(1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,1,0,1,0,0,0,0,1,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,2,0,0,0,0,0,1,0,0,0,0,0,3,1,2,2,2,3,2,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,1,1,2,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,4,1,1,1,1,1,0,0,1,1,1,1,0,1,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,2,0,0,1,0,1,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,1,0,0,0,0,0,0,2,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,1,0,1,1,1,1,0,1,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,1,0,1,0,0,1,1,1,1,0,1,0,1,0,0,2,0,0,0,0,0,1,0,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,3,1,1,1,1,1,0,0,0,0,0,0,0,1,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,1,1,1,0,0,1,1,1,1,0,0,0,1,0,0,2,0,0,0,1,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,1,3,1,1,1,1,0,1,0,0,0,0,1,2,0,1,1,0,0,0,0,1,0,0,1,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,1,1,1,1,0,1,0,0,0,0,0,0,0,1,0,0,1,1,1,1,1,1,0,0,0,0,1,0,0,1,0,1,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,1,1,0,1,0,0,0,0,0,1,1,1,2,1,1,1,0,0,0,0,1,2,2,1,1,1,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,1,1,1,1,1,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,1,1,1,0,2,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,2,0,2,2,3,2,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,1,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,2,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,2,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,2,1,1,1,1,1,0,1,0,0,0,0,1,1,0,0,0,0,1,0,0,0,1,0,0,1,1,1,1,1,1,0,0,0,0,0,1,0,0,0,0,1,0,1,1,1,1,1,0,0,1,1,1,0,0,1,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,1,2,1,1,1,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,1,1,0,2,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,3,0,1,1,1,1,0,0,1,0,1,1,1,0,0,0,0,0,1,0,0,0,0,0,4,2,4,6,4,3,1,0,1,2,1,1,0,1,0,0,0,0,2,0,0,0,0,0,0,1,1,1,1,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,2,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,1,1,1,1,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,2,1,2,2,2,2,1,0,1,2,1,1,0,1,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,2,2,2,2,2,2,3,3,4,5,3,1,2,1,1,1,1,1,1,0,0,0,0,3,3,0,0,1,1,0,1,0,0,0,0), nrow=6) 
rownames(tdm) <- 1:6 
colnames(tdm) <- paste("term", 1:229, sep="") 
tdm.dist <- dist(tdm) 
# I'm stuck turning tdm.dist into what I have shown 
+2

直接関係はありませんが、 'distree 'オブジェクトから文書を直接グループ化するには' cutree(hclust(tdm.dist)、h = 10) 'のようなものがあります。 –

+0

@alexis_lazこれははるかに良い方法です。あなたが答えを加えれば、私はそれを受け入れることができます –

答えて

2

古典的なアプローチは、as.data.frame(as.table(.))経路です。具体的には、次のようになります。

しかし、あまりにも多くの強制と大きなオブジェクトの作成だけがすぐにサブセットになることがあります。

distは「より低い」にその値を格納します。DISTトリ「我々が行/ COLインデックスを生成するcombnを使用しcbindができ角形 『』オブジェクト:また

data.frame(do.call(rbind, combn(attr(tdm.dist, "Size"), 2, simplify = FALSE)), c(tdm.dist)) 

、 『マトリックス』パッケージには、そのオブジェクトを作成する際、そのメモリ効率に沿ってある程度の柔軟性を有し、ここで使用することができる。

library(Matrix) 
tmp = combn(attr(tdm.dist, "Size"), 2) 
summary(sparseMatrix(i = tmp[2, ], j = tmp[1, ], x = c(tdm.dist), 
        dims = rep_len(attr(tdm.dist, "Size"), 2), symmetric = TRUE)) 

また、ハンドル「DIST」オブジェクトの異なる機能のうち、

cutree(hclust(tdm.dist), h = 10) 
#1 2 3 4 5 6 
#1 2 3 3 3 3 

基を切断時間を指定して8。

1

私はdplyrtidyrパッケージを使用して、過去に非常によく似たことをやってきた方法です。 連鎖された(%>%)スクリプトを行単位で実行して、データセットが段階的にどのように更新されるかを確認することができます。オブジェクト様「マトリックス」[行、列、値「data.frame」をオンにする

tdm <- matrix(c(1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,1,0,1,0,0,0,0,1,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,2,0,0,0,0,0,1,0,0,0,0,0,3,1,2,2,2,3,2,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,1,1,2,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,4,1,1,1,1,1,0,0,1,1,1,1,0,1,0,0,0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,2,0,0,1,0,1,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,1,0,0,0,0,0,0,2,0,0,0,0,0,1,0,0,0,0,0,1,0,1,0,1,1,0,1,1,1,1,0,1,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,1,1,0,1,0,0,1,1,1,1,0,1,0,1,0,0,2,0,0,0,0,0,1,0,1,1,1,1,1,1,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,3,1,1,1,1,1,0,0,0,0,0,0,0,1,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,1,1,1,1,0,0,1,1,1,1,0,0,0,1,0,0,2,0,0,0,1,0,0,1,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,1,3,1,1,1,1,0,1,0,0,0,0,1,2,0,1,1,0,0,0,0,1,0,0,1,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,1,1,1,1,0,1,0,0,0,0,0,0,0,1,0,0,1,1,1,1,1,1,0,0,0,0,1,0,0,1,0,1,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,1,1,0,1,0,0,0,0,0,1,1,1,2,1,1,1,0,0,0,0,1,2,2,1,1,1,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,1,1,1,1,1,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,0,0,1,1,1,1,0,2,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,2,0,2,2,3,2,1,0,0,0,0,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,1,0,0,0,0,0,0,1,1,1,1,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,2,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,2,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,1,1,0,0,0,0,0,0,1,0,1,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,1,1,0,0,0,0,0,2,1,1,1,1,1,0,1,0,0,0,0,1,1,0,0,0,0,1,0,0,0,1,0,0,1,1,1,1,1,1,0,0,0,0,0,1,0,0,0,0,1,0,1,1,1,1,1,0,0,1,1,1,0,0,1,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,0,1,2,1,1,1,0,0,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,1,1,0,2,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,3,0,1,1,1,1,0,0,1,0,1,1,1,0,0,0,0,0,1,0,0,0,0,0,4,2,4,6,4,3,1,0,1,2,1,1,0,1,0,0,0,0,2,0,0,0,0,0,0,1,1,1,1,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,2,0,0,0,0,1,0,0,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,2,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,1,1,1,1,1,1,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,2,1,2,2,2,2,1,0,1,2,1,1,0,1,0,0,0,0,0,0,0,1,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,1,1,1,1,0,0,0,0,0,0,2,2,2,2,2,2,3,3,4,5,3,1,2,1,1,1,1,1,1,0,0,0,0,3,3,0,0,1,1,0,1,0,0,0,0), nrow=6) 
rownames(tdm) <- 1:6 
colnames(tdm) <- paste("term", 1:229, sep="") 
tdm.dist <- dist(tdm) 

library(dplyr) 
library(tidyr) 


tdm.dist %>% 
    as.matrix() %>%      # update dist object to a matrix 
    data.frame() %>%      # update matrix to a data frame 
    setNames(nm = 1:ncol(.)) %>%   # update column names 
    mutate(names1 = 1:nrow(.)) %>%  # use rownames as a variable 
    gather(names2, value , -names1) %>% # reshape data 
    filter(names1 <= names2)    # keep the values only once 

# names1 names2  value 
# 1  1  1 0.000000 
# 2  1  2 14.966630 
# 3  2  2 0.000000 
# 4  1  3 12.449900 
# 5  2  3 12.449900 
# 6  3  3 0.000000 
# 7  1  4 13.490738 
# 8  2  4 13.564660 
# 9  3  4 5.385165 
# 10  4  4 0.000000 
# 11  1  5 12.688578 
# 12  2  5 12.922848 
# 13  3  5 5.830952 
# 14  4  5 7.416198 
# 15  5  5 0.000000 
# 16  1  6 12.369317 
# 17  2  6 12.529964 
# 18  3  6 5.830952 
# 19  4  6 7.937254 
# 20  5  6 7.615773 
# 21  6  6 0.000000 
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