Pythonic方式存储前10名结果

有明显的做法：

if calc > highest_calc:
    second_calc = highest_calc
    highest_calc = calc
if calc < highest_calc and calc > second_calc:
    third_calc =  second_calc
    second_calc = calc
if calc < second_calc and calc > third_calc:
    fourth_calc = third_calc
    third_calc = calc
etc.

但是,有更好,更有活力和更pythonic的方式吗？

奖金

对于我的项目,每个calcultion都有三个相应的名称：name_a name_b name_c.我不想要的是拥有相同三个名字的前10个值中的一个.但是,如果最后一个计算具有相同的名称,我想保持两者中最高的.最好的方法是什么？

例如,假设2.3是calc的值,使用MCD SBUX和CAT来计算calc.但是,如果我已经使用MCD SBUX和CAT进行了一次计算并且它已经达到了顶峰呢？如何找到此计算的值,以便我可以看到它是否小于或大于新计算结果.如果它大于,则删除旧的calc并添加新的calc.如果小于,则传递新的计算结果.希望这是有道理的：

If name_a in top10 and name_b in top10 and name_c in top10:
   if calc > old_calc_with_same_names:
       add_calc = calc,name_a,name_b,name_c
       top10.insert(bisect.bisect(calc,top10[0]),add_calc)
else:
   add to top10

完成的代码

csc = []
top_ports = []
add_sharpe = [sharpe,weight_a,exchange_a,weight_b,exchange_b,name_c,weight_c,exchange_c]
    if init__calc == 0:
            csc.append(add_sharpe)
    if init__calc > 1:
        if name_a == prev_name_a and name_b == prev_name_b and name_c == prev_name_c:
            csc.append(add_sharpe)
        if name_a != prev_name_a or name_b != prev_name_b or name_c != prev_name_c:
            if csc:
                hs = max(csc,key=lambda x: x[0])
                if top_ports:
                    ls = min(top_ports,key=lambda x: x[0])
                    if hs[0] > ls[0]:
                        hsi = csc.index(hs)
                        top_ports.append(csc[hsi])
                else:
                    hsi = csc.index(hs)
                    top_ports.append(csc[hsi])
            csc = []
            csc.append(add_sharpe)

后来在剧本中……

top_ports = sorted(top_ports,key=itemgetter(0),reverse=True)
print "The highest sharpe is: {0}".format(top_ports[0])
print " ==============================================="
print " ==============================================="
print datetime.now() - startTime
print "Second: {0}".format(top_ports[1])
print "Third: {0}".format(top_ports[2])
print "Fourth: {0}".format(top_ports[3])
print "Fifth: {0}".format(top_ports[4])

等等