imdb_success_predictor.py 文件源码

def create_output(movie_info):
    Y = scipy.zeros(len(movie_info))
    for i in range(0, len(movie_info)):
        gross = movie_info[i][15]
        if gross > 1000000:
            Y[i] = 1
    print 'Number of successful movies', sum(Y)
    return Y