<p>我已经在这里设置了一个模拟程序,注释将描述这些步骤。在</p>
<p>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a和B分数的随机数</p>
<p>接下来,我将A和B的范围分成五个等距的箱子,每个箱子代表一个单元格的最小值和最大值。在</p>
<p>然后我连续查询数据集以提取每个单元格中的字符串。在</p>
<p>根据您使用的实际数据结构和存储,有上百种优化方法。在</p>
<pre><code>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
</code></pre>