如何從嵌套字典創建逗號分隔值？

Question

我需要創建一個 CSV 文件，其中填充了來自嵌套字典的數據。

我指的是這個問題： Convert Nested Dictionary to CSV Table 。

我檢查了這個問題： 從嵌套字典到 CSV 文件，並找到了一種創建逗號分隔值的方法。

我的嵌套字典如下所示：

users_item = {
    "Angelica": {
    "Blues Traveler": 3.5,
    "Broken Bells": 2.0,
    "Norah Jones": 4.5,
    "Phoenix": 5.0,
    "Slightly Stoopid": 1.5,
    "The Strokes": 2.5,
    "Vampire Weekend": 2.0
},
"Bill":{
    "Blues Traveler": 2.0,
    "Broken Bells": 3.5,
    "Deadmau5": 4.0,
    "Phoenix": 2.0,
    "Slightly Stoopid": 3.5,
    "Vampire Weekend": 3.0}
}

with open('output1.csv', 'w') as csv_file:
    csvwriter = csv.writer(csv_file, delimiter=',')
    for session in users_item:
        for item in users_item[session]:
            csvwriter.writerow([session, item, users_item[session][item]])

我生成了這個 output。

Angelica,Blues Traveler,3.5
Angelica,Broken Bells,2.0
Angelica,Norah Jones,4.5
Angelica,Phoenix,5.0
Angelica,Slightly Stoopid,1.5
Angelica,The Strokes,2.5
Angelica,Vampire Weekend,2.0
Bill,Blues Traveler,2.0
Bill,Broken Bells,3.5
Bill,Deadmau5,4.0
Bill,Phoenix,2.0
Bill,Slightly Stoopid,3.5
Bill,Vampire Weekend,3.0

但是，我需要按以下格式生成 output：

Angelica,Blues Traveler,3.5,Broken Bells,2.0,Norah Jones,4.5,Phoenix,5.0,Slightly Stoopid,1.5,The Strokes,2.5,Vampire Weekend,2.0
Bill,Blues Traveler,2.0,Broken Bells,3.5,Deadmau5,4.0,Phoenix,2.0,Slightly Stoopid,3.5,Vampire Weekend,3.0

你能告訴我如何修改代碼以獲得我上面顯示的 output 嗎？

Answer 1

你的很接近。 我將修改您的方法，使用itertools.chain.from_iterable將內部 dict 項展平為列表，然后將其與用戶名結合起來，並將列表寫入 CSV 文件。

修改后的代碼：

from csv import writer
from itertools import chain

users_item = {
    "Angelica": {
        "Blues Traveler": 3.5,
        "Broken Bells": 2.0,
        "Norah Jones": 4.5,
        "Phoenix": 5.0,
        "Slightly Stoopid": 1.5,
        "The Strokes": 2.5,
        "Vampire Weekend": 2.0,
    },
    "Bill": {
        "Blues Traveler": 2.0,
        "Broken Bells": 3.5,
        "Deadmau5": 4.0,
        "Phoenix": 2.0,
        "Slightly Stoopid": 3.5,
        "Vampire Weekend": 3.0,
    },
}  

with open("output.csv", mode="w", newline='') as f:
    writer = writer(f)
    for user, items in users_item.items():
        flattened_items = list(chain.from_iterable(items.items()))
        writer.writerow([user, *flattened_items])

或使用列表推導展平：

with open("output.csv", mode="w", newline="") as f:
    writer = writer(f)
    for user, items in users_item.items():
        row = [item for kvp in items.items() for item in kvp]
        writer.writerow([user, *row])

或者一種更簡單的方法，只需將鍵值對一次擴展為一個列表：

with open("output.csv", mode="w", newline="") as f:
    writer = writer(f)
    for user, items in users_item.items():
        row = [user]
        for k, v in items.items():
            row.extend([k, v])
        writer.writerow(row)

output.csv

Angelica,Blues Traveler,3.5,Broken Bells,2.0,Norah Jones,4.5,Phoenix,5.0,Slightly Stoopid,1.5,The Strokes,2.5,Vampire Weekend,2.0
Bill,Blues Traveler,2.0,Broken Bells,3.5,Deadmau5,4.0,Phoenix,2.0,Slightly Stoopid,3.5,Vampire Weekend,3.0