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创建嵌套字典的迭代问题

nested dictionary python-3.x python

Alex_P · 技术社区 · 7 年前

我的数据如下:

                 Application                       WorkflowStep
0                WF:ACAA-CR (auto)                      Manager
1                WF:ACAA-CR (auto)           Access Responsible
2                WF:ACAA-CR (auto)                    Automatic
3                WF:ACAA-CR-AccResp (auto)              Manager
4                WF:ACAA-CR-AccResp (auto)   Access Responsible
5                WF:ACAA-CR-AccResp (auto)            Automatic
6                WF:ACAA-CR-IT-AccResp[AUTO]              Group
7                WF:ACAA-CR-IT-AccResp[AUTO] Access Responsible
8                WF:ACAA-CR-IT-AccResp[AUTO]          Automatic

除了这两列之外,我还想添加第三列,显示所有列的总和 WorkflowStep s。这个 字典应该如下(或相似) :

{'WF:ACAA-CR (auto)': 
             [{'Workflow': ['Manager', 'Access Responsible','Automatic'], 'Summary': 3}], 
 'WF:ACAA-CR-AccResp (auto)': 
             [{'Workflow': ['Manager','Access Responsible','Automatic'], 'Summary': 3}], 
 'WF:ACAA-CR-IT-AccResp[AUTO]': 
             [{'Workflow': ['Group','Access Responsible','Automatic'], 'Summary': 3}]
}

我在上面两列中创建字典的代码工作正常。

for i in range(len(df)):
    currentid = df.iloc[i,0]
    currentvalue = df.iloc[i,1]
    dict.setdefault(currentid, [])
    dict[currentid].append(currentvalue)

创建 工作流程 如下所示,也可以正常工作:

for key, values in dict.items():
    val = values
    match = ["Manager", "Access Responsible", "Automatic", "Group"]
    c = Counter(val)
    sumofvalues = 0
    for m in match:
        if c[m] == 1:
            sumofvalues += 1

我的 初始观念 是调整我的第一个代码,初始键是 Application 和 工作流程 , Summary 将是子字典。

for i in range(len(df)):
    currentid = df.iloc[i,0]
    currentvalue = df.iloc[i,1]
    dict.setdefault(currentid, [])
    dict[currentid].append({"Workflow": [currentvalue], "Summary": []})

但结果并不令人满意,因为它没有增加 currentvalue 到已经存在的 Workflow 但在每次迭代后重新创建它们。

例子

 {'WF:ACAA-CR (auto)': [{'Workflow': ['Manager'], 'Summary': []},
                        {'Workflow': ['Access Responsible'], 'Summary': []}, 
                        {'Workflow': ['Automatic'], 'Summary': []}]
 }

如何创建一个类似于我上面所写内容的字典?

2 回复 | 直到 7 年前

meW 7 年前

IIUC,这是有帮助的-

val = df.groupby('Application')['WorkflowStep'].unique()
{val.index[i]: [{'WorkflowStep':list(val[i]), 'Summary':len(val[i])}] for i in range(len(val))}

导致,

{'WF:ACAA-CR (auto)': [{'WorkflowStep': ['Manager', 'Access Responsible', 'Automatic'], 'Summary': 3}],
 'WF:ACAA-CR-AccResp (auto)': [{'WorkflowStep': ['Manager', 'Access Responsible', 'Automatic'], 'Summary': 3}],
 'WF:ACAA-CR-IT-AccResp[AUTO]': [{'WorkflowStep': ['Group', 'Access Responsible', 'Automatic'], 'Summary': 3}]}

Andrew McDowell 7 年前

我认为MEW的答案是一种更好的方法,它利用了数据框架的简洁功能,但作为参考,如果您想按照您尝试的方式来做,我认为这是可行的:

# Create the data for testing.
d = {'Application': ["WF:ACAA-CR (auto)", "WF:ACAA-CR (auto)", "WF:ACAA-CR (auto)",
                     "WF:ACAA-CR-AccResp (auto)", "WF:ACAA-CR-AccResp (auto)", "WF:ACAA-CR-AccResp (auto)"],
     'WorkflowStep': ["Manager", "Access Responsible","Automatic","Manager","Access Responsible", "Automatic"]}
df = pd.DataFrame(d)

new_dict = dict()
# Iterate through the rows of the data frame. 
for index, row in df.iterrows():
    # Get the values for the current row.
    current_application_id = row['Application']
    current_workflowstep = row['WorkflowStep']

    # Set the default values if not already set.
    new_dict.setdefault(current_application_id, {'Workflow': [], 'Summary' : 0})

    # Add the new values.
    new_dict[current_application_id]['Workflow'].append(current_workflowstep)
    new_dict[current_application_id]['Summary'] += 1

print(new_dict)

其输出为:

{'WF:ACAA-CR (auto)': {'Workflow': ['Manager', 'Access Responsible', 'Automatic'], 'Summary': 3}, 
'WF:ACAA-CR-AccResp (auto)': {'Workflow': ['Manager', 'Access Responsible', 'Automatic'], 'Summary': 3}}