Python-比其他方式更好的方法

如何解决Python-比其他方式更好的方法

我正在尝试改进旧代码，想知道是否有更好的方法可以处理以下示例：

利用api获取搜索结果。结果包括不同的问题严重性，我想获取单个计数并将每个案例按严重性追加到列表中。

在当前代码中，我可以这样做：

sev = dict()
sev_list = list()
sev_list_2 = list()
sev_list_3 = list()

for i in search_result:
    issue_key = i.key
    issue_severity = i.severity.name

    if issue_severity == 'Sev_1':
        sev_list.append(issue_key)
        sev_count = len(sev_list)
        sev[issue_severity] = (sev_list,sev_count)
    elif issue_severity == 'Sev_2':
        sev_list_2.append(issue_key)
        sev_count = len(sev_list_2)
        sev[issue_severity] = (sev_list_2,sev_count)
    elif issue_severity == 'Sev_3':
        sev_list_3.append(issue_key)
        sev_count = len(sev_list_3)
        sev[issue_severity] = (sev_list_3,sev_count)

结果：

{'Sev_1': (['issue_1','issue_4','issue_5'],3),'Sev_2': (['issue_2','issue_3','issue_6','issue_7'],4),'Sev_3': (['issue_8'],1)}

虽然这很好，但它不是最佳方法，因为当存在多个严重级别时很难扩展。

我正在考虑类似的方法，但是我不确定如果要使用此方法，追加什么正确方法。任何其他建议也将受到赞赏。

sev = dict()
sev_list = list

for i in query_result:
    issue_key = i.key
    issue_severity = i.severity.name

    severity_list = ['Sev_1','Sev_2','Sev_3']

    for x in severity_list:
        if issue_severity == x:
            sev_list.append(bug_key)
            sev_count = len(sev_list)
            sev[x] = (sev_list,sev_count)



      

解决方法

下面的代码将起作用。

sev = dict()
for i in search_result:
    issue_key = i.key
    issue_severity = i.severity.name
    if issue_severity in sev:
        sev[issue_severity][0].append(issue_key)
        sev[issue_severity][1] +=1
    else:
        sev[issue_severity] = ([issue_key],1)

,

因此，由于您已经在考虑添加另一个循环以降低复杂性，所以我建议在此处使用字典理解

# Initialize the empty dict.
sev = {sev_level:([],0) for sev_level in ['Sev_1','Sev_2','Sev_3']}

# Fill the items
for i in search_result:
    sev[i.key][0].append(i.severity.name)
    sev[i.key][1] += 1