convert dictionary to sparse matrix

问题

I have a dictionary with keys as user_ids and values as list of movie_ids liked by that user with #unique_users = 573000 and # unique_movies =16000.

{1: [51, 379, 552, 2333, 2335, 4089, 4484], 2: [51, 379, 552, 1674, 1688, 2333, 3650, 4089, 4296, 4484], 5: [783, 909, 1052, 1138, 1147, 2676], 7: [171, 321, 959], 9: [3193], 10: [959], 11: [131,567,897,923],..........}

Now i want to convert this into into a matrix with rows as user_ids and columns as movies_id with values 1 for the movies which user has liked i.e it will be 573000*16000

Ultimately i have to multiply this matrix with it's transpose to have co-occurrence matrix with dim (#unique_movies,#unique_movies).

Also, what will be the time complexity of X'*X operation where X is like (500000,12000).

回答1:

I think you can construct an empty dok_matrix and fill the values. Then transpose it and convert it to csr_matrix for efficient matrix multiplications.

import numpy as np
import scipy.sparse as sp
d = {1: [51, 379, 552, 2333, 2335, 4089, 4484], 2: [51, 379, 552, 1674, 1688, 2333, 3650, 4089, 4296, 4484], 5: [783, 909, 1052, 1138, 1147, 2676], 7: [171, 321, 959], 9: [3193], 10: [959], 11: [131,567,897,923]}

mat = sp.dok_matrix((573000,16000), dtype=np.int8)

for user_id, movie_ids in d.items():
    mat[user_id, movie_ids] = 1

mat = mat.transpose().tocsr()
print mat.shape

回答2:

df = {1: [51, 379, 552, 2333, 2335, 4089, 4484], 2: [51, 379, 552, 1674, 1688, 2333, 3650, 4089, 4296, 4484], 5: [783, 909, 1052, 1138, 1147, 2676], 7: [171, 321, 959], 9: [3193], 10: [959], 11: [131,567,897,923],..........}
df2 = pd.DataFrame.from_dict(df, orient='index')
df2 = df2.stack().reset_index()
df2.level_1=1
df2.pivot(index='level_0',columns=0,values='level_1').fillna(0)

This converts the dict into a dataframe, followed by stacking to get userIDs and movieIDs in separate columns, then all the values of unused column level_1 is set to 1. Last statement creates a pivot table filling non-existant combinations with zeros.

回答3:

You can create csr_matrix at once (like this format: csr_matrix((data, (row_ind, col_ind))). Here is a snippet on how to do that.

import scipy.sparse as sp
d = {0: [0,1], 1: [1,2,3], 
     2: [3,4,5], 3: [4,5,6], 
     4: [5,6,7], 5: [7], 
     6: [7,8,9]}
row_ind = [k for k, v in d.items() for _ in range(len(v))]
col_ind = [i for ids in d.values() for i in ids]
X = sp.csr_matrix(([1]*len(row_ind), (row_ind, col_ind))) # sparse csr matrix

You can use matrix X to find cooccurrence matrix later (i.e. X.T * X) (credit github @daniel-acuna). I guess there is a faster way to convert dictionary of list to row_ind, col_ind.

来源：https://stackoverflow.com/questions/37862139/convert-dictionary-to-sparse-matrix

标签

python

dictionary

matrix