Question

mysql服务器版本5.7.21 mysql workbench版本：6.3

我做了以下查询

import numpy as np
import numpy.random as npr
import scipy.sparse as sparse
from scipy.sparse import coo_matrix
import numba
from numba import jit


@jit(nopython=True) # comment this if want just numpy
def countSketch(input_rows, input_data,
                      input_nnz,
                      sketch_size, seed=None):
   '''
   input_rows: row indices for data (can be repeats)
   input_data: values seen in row location,
   input_nnz : number of nonzers in the data (can replace with
   len(input_data) but avoided here for speed)
   sketch_size: int
   seed=None : random seed
   '''
   hashed_rows = np.empty(input_rows.shape,dtype=np.int32)
   current_row = 0
   hash_val = npr.choice(sketch_size)
   sign_val = npr.choice(np.array([-1.0,1.0]))
   #print(hash_val)
   hashed_rows[0] = hash_val
   #print(hash_val)
   for idx in np.arange(input_nnz):
       print(idx)
       row_id = input_rows[idx]
       data_val = input_data[idx]
       if row_id == current_row:
           hashed_rows[idx] = hash_val
           input_data[idx] = sign_val*data_val
       else:
           # make new hashes
           hash_val = npr.choice(sketch_size)
           sign_val = npr.choice(np.array([-1.0,1.0]))
           hashed_rows[idx] = hash_val
           input_data[idx] = sign_val*data_val
   return hashed_rows, input_data

def sort_row_order(input_data):
   sorted_row_column = np.array((input_data.row,
                                 input_data.col,
                                 input_data.data))

   idx  = np.argsort(sorted_row_column[0])
   sorted_rows = np.array(sorted_row_column[0,idx], dtype=np.int32)
   sorted_cols = np.array(sorted_row_column[1,idx], dtype=np.int32)
   sorted_data = np.array(sorted_row_column[2,idx], dtype=np.float64)
   return sorted_rows, sorted_cols, sorted_data


if __name__=="__main__":
   import time
   from tabulate import tabulate

   matrix = sparse.random(1000, 50, 0.1)
   x = np.random.randn(matrix.shape[1])
   true_norm = np.linalg.norm(matrix@x,ord=2)**2
   tidy_data =  sort_row_order(matrix)

   sketch_size = 300
   start = time.time()
   hashed_rows, sketched_data = countSketch(tidy_data[0],\
                                            tidy_data[2], matrix.nnz,sketch_size)
   duration_slow = time.time() - start
   S_A = sparse.coo_matrix((sketched_data, (hashed_rows,matrix.col)))
   approx_norm_slow = np.linalg.norm(S_A@x,ord=2)**2
   rel_error_slow = approx_norm_slow/true_norm
   #print("Sketch time: {}".format(duration_slow))
   start = time.time()
   hashed_rows, sketched_data = countSketch(tidy_data[0],\
                                           tidy_data[2], matrix.nnz,sketch_size)
   duration = time.time() - start
   #print("Sketch time: {}".format(duration))
   S_A = sparse.coo_matrix((sketched_data, (hashed_rows,matrix.col)))
   approx_norm = np.linalg.norm(S_A@x,ord=2)**2
   rel_error = approx_norm/true_norm
   #print("Relative norms: {}".format(approx_norm/true_norm))
   print(tabulate([[duration_slow, rel_error_slow, 'Yes'],
                   [duration, rel_error, 'No']],
                   headers=['Sketch Time', 'Relative Error', 'Dry Run'],
                   tablefmt='orgtbl'))

这让我觉得我购买的产品最多。以这种方式

SELECT articles.name as 'Product Name', article_count
FROM
    (select articel_id, count(articel_id) as article_count
    FROM sales_records_row
    GROUP BY  articel_id
    ORDER BY count(articel_id) DESC) 
AS 
    overview,
    articles
WHERE 
    articles.articel_id = overview.articel_id

我现在想从这里获得前三名。包括重复。所以在这种情况下，我会在我的前3名列表中获得5个项目，因为product4,2和7都计算了5次。最简单的方法是什么？

如何获得前3名（重复）

0 个答案: