如何遍歷字典列表

Question

我的代碼是

index = 0
for key in dataList[index]:
    print(dataList[index][key])

似乎可以很好地打印index = 0的字典鍵的值。 但是，我不知道如何遍歷dataList中未知數量的字典。

Answer 1

您可以迭代list的len range的索引：

dataList = [{'a': 1}, {'b': 3}, {'c': 5}]
for index in range(len(dataList)):
    for key in dataList[index]:
        print(dataList[index][key])

或者您可以使用帶有index計數器的 while 循環：

dataList = [{'a': 1}, {'b': 3}, {'c': 5}]
index = 0
while index < len(dataList):
    for key in dataList[index]:
        print(dataList[index][key])
    index += 1

你甚至可以直接遍歷列表中的元素：

dataList = [{'a': 1}, {'b': 3}, {'c': 5}]
for dic in dataList:
    for key in dic:
        print(dic[key])

只需遍歷字典的值，甚至可以不進行任何查找：

dataList = [{'a': 1}, {'b': 3}, {'c': 5}]
for dic in dataList:
    for val in dic.values():
        print(val)

或者將迭代包裝在列表理解或生成器中，稍后解包：

dataList = [{'a': 1}, {'b': 3}, {'c': 5}]
print(*[val for dic in dataList for val in dic.values()], sep='\n')

可能性是無止境。 這是一個選擇你喜歡什么的問題。

Answer 2

您可以輕松地做到這一點：

for dict_item in dataList:
  for key in dict_item:
    print dict_item[key]

它將遍歷列表，並且對於列表中的每個字典，它將遍歷鍵並打印其值。

Answer 3

use=[{'id': 29207858, 'isbn': '1632168146', 'isbn13': '9781632168146', 'ratings_count': 0}]
for dic in use:
    for val,cal in dic.items():
        print(f'{val} is {cal}')

Answer 4

def extract_fullnames_as_string(list_of_dictionaries): return list(map(lambda e : "{} {}".format(e['first'],e['last']),list_of_dictionaries)) names = [{'first': 'Zhibekchach', 'last': 'Myrzaeva'}, {'first': 'Gulbara', 'last': 'Zholdoshova'}] print(extract_fullnames_as_string(names)) #Well...the shortest way (1 line only) in Python to extract data from the list of dictionaries is using lambda form and map together.

Answer 5

"""提供最大靈活性並且對我來說似乎更動態的方法如下："""

在名為.....的函數中循環列表

def extract_fullnames_as_string(list_of_dictionaries):

    result = ([val for dic in list_of_dictionaries for val in 
    dic.values()])

    return ('My Dictionary List is ='result)


    dataList = [{'first': 3, 'last': 4}, {'first': 5, 'last': 7},{'first': 
    15, 'last': 9},{'first': 51, 'last': 71},{'first': 53, 'last': 79}]
    
    print(extract_fullnames_as_string(dataList))

"""這樣一來，Datalist 可以是任何格式的 Dictionary 格式，我發現，否則你最終會處理格式問題。試試下面的方法，它仍然可以工作......"" "

    dataList1 = [{'a': 1}, {'b': 3}, {'c': 5}]
    dataList2 = [{'first': 'Zhibekchach', 'last': 'Myrzaeva'}, {'first': 
    'Gulbara', 'last': 'Zholdoshova'}]

    print(extract_fullnames_as_string(dataList1))
    print(extract_fullnames_as_string(dataList2))

Answer 6

另一個 pythonic 解決方案是使用集合模塊。

這是一個示例，我想生成一個僅包含 'Name' 和 'Last Name' 值的字典：

from collections import defaultdict

test_dict = [{'Name': 'Maria', 'Last Name': 'Bezerra', 'Age': 31},
             {'Name': 'Ana', 'Last Name': 'Mota', 'Age': 31},
             {'Name': 'Gabi', 'Last Name': 'Santana', 'Age': 31}]

collect = defaultdict(dict)

# at this moment, 'key' becomes every dict of your list of dict
for key in test_dict:
    collect[key['Name']] = key['Last Name']

print(dict(collect))

輸出應該是：

{'Name': 'Maria', 'Last Name': 'Bezerra'}, {'Name': 'Ana', 'Last Name': 'Mota'}, {'Name': 'Gabi', 'Last Name': 'Santana'}

Answer 7

有多種方法可以遍歷字典列表。 但是，如果您喜歡Pythonic代碼，請考慮以下方法，但首先，讓我們使用data_list而不是dataList因為在 Python 中， snake_case優於camelCase 。

方式#1：遍歷字典的鍵

# let's assume that data_list is the following dictionary
data_list = [{'Alice': 10}, {'Bob': 7}, {'Charlie': 5}]

for element in data_list:
    for key in element:
        print(key, element[key])

Output

Alice 10
Bob 7
Charlie 5

解釋：

• for element in data_list: -> element將在每次迭代時成為data_list中的字典，即第一次迭代中為{'Alice': 10} ，第二次迭代中為{'Bob': 7} ，以及{'Charlie': 5} ，在第三次迭代中。
• for key in element: -> key將是每次迭代時element的鍵，因此當element為{'Alice': 10}時， key的值將是'Alice' 。 請記住， element可以包含更多鍵，但在這個特定示例中它只有一個。
• print(key, element[key]) -> 它打印key key key和element的值，即它訪問 `element.

方式#2：迭代字典的鍵和值

# let's assume that data_list is the following dictionary
data_list = [{'Alice': 10}, {'Bob': 7}, {'Charlie': 5}]

for element in data_list:
    for key, value in element.items():
        print(key, value)

此代碼段的 output 與上一個相同。

解釋：

• for element in data_list: -> 它與之前代碼中的解釋相同。
• for key, value in element.items(): -> 在每次迭代中， element.items()將返回一個包含兩個元素的元組。 前者元素是鍵，后者是與該鍵關聯的值，因此當element為{'Alice': 10}時， key的值為'Alice' ，值的value 10 。 請記住，該字典只有一個鍵值對。
• print(key, value) -> 它打印key和value 。

如前所述，有多種方法可以遍歷字典列表，但要使您的代碼更加 Pythonic，請避免使用索引或 while 循環。

Answer 8

有一個類似的問題，通過使用單個 for 循環遍歷列表來修復我的問題，請參閱代碼片段

de = {"file_name":"jon","creation_date":"12/05/2022","location":"phc","device":"s3","day":"1","time":"44692.5708703703","year":"1900","amount":"3000","entity":"male"}
se = {"file_name":"bone","creation_date":"13/05/2022","location":"gar","device":"iphone","day":"2","time":"44693.5708703703","year":"2022","amount":"3000","entity":"female"}
re = {"file_name":"cel","creation_date":"12/05/2022","location":"ben car","device":"galaxy","day":"1","time":"44695.5708703703","year":"2022","amount":"3000","entity":"male"}
te = {"file_name":"teiei","creation_date":"13/05/2022","location":"alcon","device":"BB","day":"2","time":"44697.5708703703","year":"2022","amount":"3000","entity":"female"}
ye = {"file_name":"js","creation_date":"12/05/2022","location":"woji","device":"Nokia","day":"1","time":"44699.5708703703","year":"2022","amount":"3000","entity":"male"}
ue = {"file_name":"jsdjd","creation_date":"13/05/2022","location":"town","device":"M4","day":"5","time":"44700.5708703703","year":"2022","amount":"3000","entity":"female"}


d_list = [de,se,re,te,ye,ue]


for dic in d_list:
    print (dic['file_name'],dic['creation_date'])