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假设我有一个包含 3 个字段的数据库表：字符串标题、int A、int B。A 和 B 的范围都是 1 到 500。
我想将部分值表示为 5x5 矩阵。这样 (1, 1) 将是 A 和 B 都最低的字符串；(5, 5) 将具有最高的 A 和 B；(1, 5) 将具有最低的 A 和最高的 B。依此类推。
我应该使用哪种算法？

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2 回答 2

1

你有

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

等等...？

减少到一个 3x3 矩阵，它看起来像

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

问题是，(2,2) 指向什么？平均值？好的，在 5x5 矩阵中？您的定义缺少一些信息。

上述矩阵的算法将是：

  1. 为 A 和 B 计算 min、max、avg
  2. 向数据库询问元组 (Amin, Bmin), (Aavg, Bmin), (Amax, Bmin) 等等
  3. 将值填充到矩阵中

附加：如果没有匹配项，请尝试最小值、最大值和平均值的范围。

于 2012-04-21T10:22:00.837 回答
1

我在这里设置了一个模拟，评论将描述这些步骤。

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随机数，分别代表分数 A 和 B。

接下来，我将 A 和 B 的范围划分为五个等距的 bin，每个 bin 代表一个单元格的最小值和最大值。

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
于 2012-04-21T11:16:43.417 回答