创建相似度矩阵

Question

我有一个如下所示的矩阵：

col_1 col_2 value

 A      B     2.1

 A      C     1.3

 B      C     4.6

 A      D     1.4

....

我想得到一个相似度矩阵：

    A    B    C   D

A    X   2.1  1.3  1.4

B    2.1  X   4.6   ...

C    ...  ...  X    ...

D    ...  ... ...    X

所以行名和列名是 A、B、C、D，它从第三列取值并将其添加到矩阵问题还在于原始矩阵的长度约为 10 000 行。

Answer 1

您可以通过以下方式进行。我在 Python 中编写代码，因为没有指定语言

#I assume that your data is in a python pandas dataframe called df

df = ..load your data  
list_of_labels = [ 'A','B','C','D' ]
nb_labels = len(list_of_labels)
similarity = np.zeros( (nb_labels,nb_labels) )

for l1, l2, val in zip( df['col_1'] , df['col_2'] , df['value'] ):
    i = list_of_labels.index( l1 )
    j = list_of_labels.index( l2 )
    similarity[i][j] = val


similarity_df = pd.DataFrame(data=similarity, index=list_of_labels, columns=list_of_labels)

Answer 2

正如 Roland 所建议的，您可以使用 dcast():

library(data.table)
dcast(df, col_1 ~ col_2)
##   col_1   B   C   D
## 1     A 2.1 1.3 1.4
## 2     B  NA 4.6  NA

其中：

df <- data.frame(
  col_1 = c("A", "A", "B", "A"), 
  col_2 = c("B","C", "C", "D"), 
  value = c(2.1, 1.3, 4.6, 1.4)
)

Answer 3

与 xtabs 和 mutate_at。 sparse = TRUE 将输出转换为稀疏矩阵：

library(dplyr)

mat <- df %>%
  mutate_at(1:2, factor, levels = unique(c(levels(.$col_1), levels(.$col_2)))) %>%
  xtabs(value ~ col_1 + col_2, data=., sparse = TRUE)

mat[lower.tri(mat)] <- mat[upper.tri(mat)]

结果：

4 x 4 sparse Matrix of class "dgCMatrix"
     col_2
col_1   A   B   C   D
    A .   2.1 1.3 1.4
    B 2.1 .   4.6 .  
    C 1.3 1.4 .   .  
    D 4.6 .   .   .

创建相似度矩阵

Create Similarity Matrix

r

transformation

matrix