如何在列表中获取元组的第一个元素和嵌套字典的值

Question

如何选择元组的第一个元素和列表中嵌套字典的值？

organization = [('research', {'size': 0, 'values': [], 'start': 0}), ('marketing', {'size': 2, 'values': [{'user': {'name': 'anna', 'id': 10, 'displayName': 'Anna'}, 'category': 'Secretary'}, {'user': {'name': 'bob', 'id': 11, 'displayName': 'Bobby'}, 'category': 'Manager'}], 'start': 0}), ('sales', {'size': 1, 'values': [{'user': {'name': 'claire', 'id': 13, 'displayName': 'Clarissa Claire'}, 'category': 'Secretary'}], 'start': 0})]

department = [[x['user']['name'], x['category']] for nest_list in organization for x in nest_list["values"]]
print(department)

预期：

department = [[marketing, anna, Secretary], [marketing, bob, manager], [sales, claire, Secretary]]

Answer 1

本部分讨论的问题是：

for nest_list in organization for x in nest_list["values"]

nest_list是一个元组(organization中的所有元组)，您希望x遍历values列表中的所有对象(这是每个元组的第二个成员)，但是直接跳到values。最后应该是nest_list[1]["values"]--在进入values之前，您需要进入第二个元组元素。

然后，在列表定义中，还需要将nest_list[0]作为每个元素列表的第一个元素，以便您可以根据需要拥有部门。因此，department =行将如下所示：

department = [[nest_list[0], x['user']['name'], x['category']] for nest_list in organization for x in nest_list[1]["values"]]

Answer 2

以下是我想出的：

# Define your list
organization = [('research', {'size': 0, 'values': [], 'start': 0}), ('marketing', {'size': 2, 'values': [{'user': {'name': 'anna', 'id': 10, 'displayName': 'Anna'}, 'category': 'Secretary'}, {'user': {'name': 'bob', 'id': 11, 'displayName': 'Bobby'}, 'category': 'Manager'}], 'start': 0}), ('sales', {'size': 1, 'values': [{'user': {'name': 'claire', 'id': 13, 'displayName': 'Clarissa Claire'}, 'category': 'Secretary'}], 'start': 0})]

# Create empty lists to be used by the loop
employee = []
employee_list = []

# Loop through each department
for counter in range(0,len(organization)):
    # Check if there are any entries by looking at the size key
    if organization[counter][1]['size'] > 0:
        # loop through each user
        for user in range(0,len(organization[counter][1]['values'])):
            # for each user, pull out the Name and Category, and add it to the employee list
            employee = [organization[counter][0],organization[counter][1]['values'][user]['user']['name'],organization[counter][1]['values'][user]['category']]
            # Add each employee as a nested list in the Employee list
            employee_list.append(employee)

print(employee_list)

可能会有一些改进，但这是我能够拼凑在一起的。

编辑:在我说了这句话之后，我对它进行了一些改进：

# Define your list
organization = [('research', {'size': 0, 'values': [], 'start': 0}), ('marketing', {'size': 2, 'values': [{'user': {'name': 'anna', 'id': 10, 'displayName': 'Anna'}, 'category': 'Secretary'}, {'user': {'name': 'bob', 'id': 11, 'displayName': 'Bobby'}, 'category': 'Manager'}], 'start': 0}), ('sales', {'size': 1, 'values': [{'user': {'name': 'claire', 'id': 13, 'displayName': 'Clarissa Claire'}, 'category': 'Secretary'}], 'start': 0})]

# Create empty lists to be used by the loop
employee = []
employee_list = []

# Loop through each department
for counter in range(0,len(organization)):
    # Loop through each user based on the size key
    for user in range(0,organization[counter][1]['size']):
        # for each user, pull out the Name and Category, and add it to the employee list
        employee = [organization[counter][0],organization[counter][1]['values'][user]['user']['name'],organization[counter][1]['values'][user]['category']]
        # Add each employee as a nested list in the Employee list
        employee_list.append(employee)

print(employee_list)