Question

假设我有一个包含3个字段的数据库表：string title，int A，int B. A和B的范围是1到500 我想将值的一部分表示为矩阵5x5。这样（1,1）将是A和B都最低的字符串; （5,5）A和B都最高; （1,5）将具有最低A和最高B.等等我应该使用哪种算法？

Answer 1

你有吗

title  A  B
one    1  1
two    1  2
three  2  1
four   3  3
five   4  4
six    5  5
seven  5  1
eight  1  5

依旧......？

缩小为3x3矩阵，看起来像

a/b  1     2    3  
1   one   two  eight    
2   three four  ?
3   seven  ?   six

问题是，（2,2）指向的是什么？平均？好的，在5x5矩阵中？你的定义缺乏一些信息。

上述矩阵的算法将是：

表示A和B计算min，max，avg
向数据库询问元组（Amin，Bmin），（Aavg，Bmin），（Amax，Bmin）等等
将值填入矩阵

添加：如果没有匹配项，请尝试使用min，max和avg。

Answer 2

我在这里设置了模拟，评论将描述步骤。

首先我生成一些数据：一系列元组，每个元组包含一个字符串和两个代表分数A和B的随机数。

接下来，我将A和B的范围分成五个等距的区间，每个区间代表一个细胞的最小值和最大值。

然后我串行查询数据集以提取每个单元格中的字符串。

根据您使用的实际数据结构和存储，有一百种优化方法。

from random import random

# Generate data and keep record of scores
data = []
a_list = []
b_list = []
for i in range(50):
    a = int(random()*500)+1
    b = int(random()*500)+1
    rec = { 's' : 's%s' % i,
            'a' : a,
            'b' : b
             }
    a_list.append(a)
    b_list.append(b)
    data.append(rec)

# divide A and B ranges into five bins

def make_bins(f_list):
    f_min = min(f_list)
    f_max = max(f_list)
    f_step_size = (f_max - f_min) / 5.0
    f_steps = [ (f_min + i * f_step_size,
                 f_min + (i+1) * f_step_size)
                for i in range(5) ]
    # adjust top bin to be just larger than maximum
    top = f_steps[4]
    f_steps[4] = ( top[0], f_max+1 )
    return f_steps

a_steps = make_bins(a_list)
b_steps = make_bins(b_list)

# collect the strings that fit into any of the bins
# thus all the strings in cell[4,3] of your matrix
# would fit these conditions:
# string would have a Score A that is
# greater than or equal to the first element in a_steps[3]
# AND less than the second element in a_steps[3]
# AND it would have a Score B that is
# greater than or equal to the first element in b_steps[2]
# AND less than the second element in a_steps[2]
# NOTE: there is a need to adjust the pointers due to
#       the way you have numbered the cells of your matrix

def query_matrix(ptr_a, ptr_b):
    ptr_a -= 1
    from_a = a_steps[ptr_a][0]
    to_a = a_steps[ptr_a][1]

    ptr_b -= 1
    from_b = b_steps[ptr_b][0]
    to_b = b_steps[ptr_b][1]

    results = []
    for rec in data:
        s = rec['s']
        a = rec['a']
        b = rec['b']
        if (a >= from_a and
            a < to_a and
            b >= from_b and
            b < to_b):
            results.append(s)
    return results

# Print out the results for a visual check
total = 0
for i in range(5):
    for j in range(5):
        print '=' * 80
        print 'Cell: ', i+1, j+1, ' contains: ',
        hits = query_matrix(i+1,j+1)
        total += len(hits)
        print hits
print '=' * 80
print 'Total number of strings found: ', total

将表格表示为矩阵

2 个答案: