主页 / coding / 问题 / 78312055
Accepted
Saeed
Asked: 2024-04-12 01:19:58 +0800 CST

如何根据字典列表的值写入不同的文件并连接 for 循环和三个字典列表

  • 772

首先，如果问题太长，我很抱歉。我试图包含最少的数据，但那是我可以添加的。

这些是我的清单：

products_income = [
    {'client_group': 'A', 'count': 2, 'product': 'A', 'monthly_income': 370},
    {'client_group': 'B', 'count': 1, 'product': 'B', 'monthly_income': 215},
    {'client_group': 'C', 'count': 3, 'product': 'C', 'monthly_income': 495},
    {'client_group': 'A', 'count': 2, 'product': 'D', 'monthly_income': 304},
    {'client_group': 'B', 'count': 1, 'product': 'E', 'monthly_income': 110},
    {'client_group': 'B', 'count': 2, 'product': 'F', 'monthly_income': 560},
    {'client_group': 'A', 'count': 1, 'product': 'G', 'monthly_income': 196},
    ]

client_package = [
    {'client_group': 'A', 'total_income': 870},
    {'client_group': 'B', 'total_income': 885},
    {'client_group': 'C', 'total_income': 495}
]

client_group_user_counts = {
    'A': ['user1', 'user2', 'user3', 'user4', 'user5'], # 5 users
    'B': ['user21', 'user22', 'user23', 'user24'], # 4 users
    'C': ['user41', 'user42', 'user43'], # 3 users
}

这些是我商店的输出，这是写入“ client_group.txt”调用的不同文件的预期输出：

A.txt:

group A has 2 product A, monthly income is 370.
group A has 2 product D, monthly income is 304.
group A has 1 product G, monthly income is 196.

group A total income is 870.

group A has total 5 users, and this is its users:
user1
user2
user3
user4
user5

B.txt:

group B has 1 product B, monthly income is 215.
group B has 1 product E, monthly income is 110.
group B has 2 product F, monthly income is 560.

group B total income is 885.

group B has total 4 users, and this is its users

user21
user22
user23
user24

C.txt:

group C has 3 product C, monthly income is 495.

group C total income is 495.

group C has total 3 users, and this is its users:
user41
user42
user43

这是我当前的代码，我尚未达到预期的输出（实际上，除了我的代码的其他问题之外，目前我不知道如何在这种情况下写入单独的文件）：

# I added this function, because I had error in the last line of next code block inside the `f-string`, so I thought that could be a workaround for this
def to_join():
    return '\n'.join(client_group_user_counts[user])

for product in products_income:
    for client in client_package:
        for user in client_group_user_counts:
            if product['client_group'] == client['client_group'] == user:
                print(f"""group {product['client_group']} has {product['count']} product {product['product']}, monthly income is {product['monthly_income']}.
total income is {client['total_income']}.
group {product['client_group']} has total {len(client_group_user_counts[user])} users, and this is its users:
{to_join()}
""")

这是当前输出：

group A has 2 product A, monthly income is 370.
group A total income is 870.
group A has total 5 users, and this is its users:
user1
user2
user3
user4
user5

group B has 1 product B, monthly income is 215.
group B total income is 885.
group B has total 4 users, and this is its users:
user21
user22
user23
user24

group C has 3 product C, monthly income is 495.
group C total income is 495.
group C has total 3 users, and this is its users:
user41
user42
user43

group A has 2 product D, monthly income is 304.
group A total income is 870.
group A has total 5 users, and this is its users:
user1
user2
user3
user4
user5

group B has 1 product E, monthly income is 110.
group B total income is 885.
group B has total 4 users, and this is its users:
user21
user22
user23
user24

group B has 2 product F, monthly income is 560.
group B total income is 885.
group B has total 4 users, and this is its users:
user21
user22
user23
user24

group A has 1 product G, monthly income is 196.
group A total income is 870.
group A has total 5 users, and this is its users:
user1
user2
user3
user4
user5
python

4 个回答

  1. 逻辑如下：

    您已经拥有 中的所有组名称client_group_user_counts。我们将使用它来生成一堆文件。我们还将把组名映射到文件名（这个程序并不完全需要，但这是一个很好的选择，以防万一您的逻辑在以后变得更加复杂）。

    然后，我们将迭代products_income. 我们使用值查找输出文件名group_name并写入格式化的记录。

    一旦我们写完所有记录，我们将迭代client_package写出总收入。但是等一下，我们将重新格式化该字典，以便键是组名称，值是总收入 - 这只会节省我们一些查找步骤。

    最后，我们将迭代client_group_user_counts。由于键已经是组名称，因此我们可以使用它们来查找输出文件。然后我们只需要将用户名本身写入输出文件

    这是代码：

    # create all the output files
outfiles = {k: open(f"{k}.txt", 'w') for k in client_group_user_counts}

# turn client_package into a dict for easier access
client_package = {row['client_group']: row['total_income'] for row in client_package}

# write everything to the correct files
for row in products_income:
    cgroup = row['client_group']
    prodCount = row['count']
    prodName = row['product']
    income = row['monthly_income']

    outfile = outfiles[cgroup]
    outfile.write(f"group {cgroup} has {prodCount} product {prodName}, monthly income is {income}.\n")

# add the additional space
for outfile in outfiles.values():
    outfile.write('\n')

# write the total income summary
for group, income in client_package.items():
    outfiles[group].write(f"group {group} total income is {income}.\n\n")

# add the users
for group,users in client_group_user_counts.items():
    outfiles[group].write(f"group {group} has total {len(users)} users, and this is its users:\n")
    for user in users:
        outfiles[group].write(f"{user}\n")

# close the files
for outfile in outfiles.values():
    outfile.close()
    • 2

  2. 我会首先重新编写字典，以避免以后忽略它们。

    然后对每个键仅循环一次并使用以下语句打开文件with：

    products = {}
for d in products_income:
    (products
     .setdefault(d['client_group'], []) 
     .append('group {client_group} has {count} product {product}, monthly income is {monthly_income}.'.format(**d))
    )
packages = {}
for d in client_package:
    (packages
     .setdefault(d['client_group'], []) 
     .append('group {client_group} total income is {total_income}.'.format(**d))
    )

for group, users in client_group_user_counts.items():
    with open(f'{group}.txt', 'w') as f:
        f.write('\n'.join(products[group])+'\n\n')
        f.write('\n'.join(packages[group])+'\n\n')
        f.write(f'group {group} has total {len(users)} users, and this is its users:\n')
        f.write('\n'.join(users))

    例子A.txt：

    group A has 2 product A, monthly income is 370.
group A has 2 product D, monthly income is 304.
group A has 1 product G, monthly income is 196.

group A total income is 870.

group A has total 5 users, and this is its users:
user1
user2
user3
user4
user5
    • 2
  3. Best Answer

    由于结果是按“组 ID”聚合的，因此您需要client_group_user_counts首先进行迭代。这还允许您获取open()特定于当前“组 ID”的文件

    这可能需要一些重构，但它应该给你一个想法

    鉴于：

    products_income = [
    {'client_group': 'A', 'count': 2, 'product': 'A', 'monthly_income': 370},
    {'client_group': 'B', 'count': 1, 'product': 'B', 'monthly_income': 215},
    {'client_group': 'C', 'count': 3, 'product': 'C', 'monthly_income': 495},
    {'client_group': 'A', 'count': 2, 'product': 'D', 'monthly_income': 304},
    {'client_group': 'B', 'count': 1, 'product': 'E', 'monthly_income': 110},
    {'client_group': 'B', 'count': 2, 'product': 'F', 'monthly_income': 560},
    {'client_group': 'A', 'count': 1, 'product': 'G', 'monthly_income': 196},
    ]

client_package = [
    {'client_group': 'A', 'total_income': 870},
    {'client_group': 'B', 'total_income': 885},
    {'client_group': 'C', 'total_income': 495}
]

client_group_user_counts = {
    'A': ['user1', 'user2', 'user3', 'user4', 'user5'], # 5 users
    'B': ['user21', 'user22', 'user23', 'user24'], # 4 users
    'C': ['user41', 'user42', 'user43'], # 3 users
}

    尝试：

    for group_key, group_users in client_group_user_counts.items():
    packages = [row for row in client_package if row["client_group"] == group_key]
    products = [row for row in products_income if row["client_group"] == group_key]
    with open(f"{group_key}.txt", "w", encoding="utf-8", newline="") as file_out:
        for product in products:
            file_out.write(f"group {group_key} has {product['count']} product {product['product']}, monthly income is {product['monthly_income']}.\n")
        file_out.write("\n")
        file_out.write(f"group {group_key} total income is {packages[0]['total_income']}.\n")
        file_out.write("\n")
        file_out.write(f"group {group_key} has total {len(group_users)} users, and this is its users:\n")
        for user in group_users:
            file_out.write(f"{user}\n")
    • 1

  4. 不知道我是否理解你的程序结构，如果我错了，请纠正我的。由于您拥有组，因此client_group_user_counts您可以从其键中获取所有组。

    groups = client_group_user_counts.keys()

    现在您已经拥有了所有组，您可以迭代组并打印上面提到的内容。

    for group in groups :
    for product in products_income:
        if product['client_group']==group:
            print(f"group {group} has {product['count']} product {product['product']}, monthly income is {product['monthly_income']}.")
    for package in client_package:
        if package['client_group']==group:
            print(f"group {group} total income is {package['total_income']}.")
    print(f"group {group} has total {len(client_group_user_counts[group])} users, and this is its users:")
    print('\n'.join(client_group_user_counts[group]))
    print()

    但通常建议使用 pandas 等库来完成更复杂的任务。

    • 1

相关问题