我的代码中的性能不佳

时间:2012-09-08 05:09:18

标签: python django django-views

我的代码中存在性能问题。我对Python有点新,我想不出更好的方法来执行以下代码。

我有一个带有名为“cdr”的表的外部数据库,它不是django项目的一部分,我需要对行进行一些计算。为了得到变量的值,我正在查询cdr表中的每一行,这使我的代码变得非常慢。

这是我在view.py

中的课程
def cdr_adm(request):
        cursor = connections['cdr'].cursor()
        cursor.execute("SELECT calldate, dst, billsec, accountcode, disposition, userfield FROM cdr where calldate >= '%s' and calldate < '%s' and disposition like '%s' and accountcode like '%s' and dst like '%s' and userfield like '%s'" %(start_date, end_date, status, customer, destino, provider))
        result = cursor.fetchall()
        time = 0
        price = 0
        price_prov = 0
        count = 0
        time_now = 0
        ANS = 0
        asr = 0
        rate_cust = 0
        rate_prov = 0

        for call in result:
                if call[3]:
                        #These 2 lines are the problem - It's very slow to run for all rows.
                        rate_cust = User.objects.get(username = call[3])
                        rate_prov = Provider.objects.get(name = call[5])
                        time_now = call[2] / 60
                        time =  time + time_now
                        count = count + 1
                        price = price + (float(rate_cust.first_name) * time_now)
                        price_prov = price_prov + (float(rate_prov.rate) * time_now)
                        if call[4] == "ANSWERED":
                                ANS = ANS + 1
        time = float(time)
        lucro = price - price_prov
        lucro = float(lucro)
        if count > 0:
                asr = (ANS / count)  * 100
        return render_to_response("cdr_all.html",
                {'calls':result,'price':price,'time':time,'count':count,'customers':customers, 'providers':providers,'price_prov':price_prov, 'lucro':lucro, 'asr':asr }, context_instance=RequestContext(request))

我正在考虑创建一个字典并在其中搜索,但我也不确定。

1 个答案:

答案 0 :(得分:2)

您可以创建所有UserProvider对象的字典,并按照您感兴趣的内容编制索引,如下所示:

users = dict([(u.username, u) for u in User.objects.all()])
providers = dict([(p.name, p) for p in Provider.objects.all()])

(确保在for call in result: for循环之外执行此操作!)然后,您可以将慢查询更改为:

                    rate_cust = users[call[3]]
                    rate_prov = provided[call[5]]

我猜测用户和提供商的数量远远少于调用,这意味着将它们保存在字典中比访问每个调用一个查询要快得多。