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

....

我想要一个相似度矩阵：

    A    B    C   D

A    X   2.1  1.3  1.4

B    2.1  X   4.6   ...

C    ...  ...  X    ...

D    ...  ... ...    X

所以行和列的名称分别是A，B，C，D，它从第三列获取值并将其添加到矩阵中问题还在于原始矩阵的长度约为10000行。

Answer 1

罗兰（Roland）建议，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

位置：

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 2

使用xtabs和mutate_at。 sparse = TRUE将输出转换为sparseMatrix：

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 .   .   .

Answer 3

您可以通过以下方式进行操作。因为没有指定语言，所以我用Python编写了代码

#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)

创建相似度矩阵

3 个答案: