如何計算重復的物件屬性？-有解無憂

我有下一個代碼，我為每個入口寵物創建物件：

from itertools import groupby
class Petshop():
    def __init__(self, entryDate, name, ownerName):
        self.entryDate = entryDate
        self.name = name
        self.ownerName = ownerName

class Pets():
    def __init__(self):
        self.petsList = []
    
    def addPets(self, entryDate, name, ownerName):
        entry_pet = Petshop(entryDate, name, ownerName)
        self.petsList.append(entry_pet)
    
    def printPets(self):
        self.petsList.sort(key=lambda p: p.entryDate)
        counter = 0
        for group in groupby(self.petsList, key=lambda p: p.entryDate):
            ls = list(group)
            print("---------------", ls[0], '------------------')
            for pet in list(ls[1]):
                print("Name:", pet.name)
                print("Owner name:", pet.ownerName)
                if pet.name in list(ls[1]):
                    counter  = 1
                print('There are ',counter,'pets with the same name')

pet = Pets()

pet.addPets('04/13/2021','Pinky', 'David Smith')
pet.addPets('07/10/2020', 'Charlie', 'Joe Davis')
pet.addPets('04/13/2021','Pinky', 'Daniel Trincot')
pet.addPets('07/10/2020', 'Kenny', 'Susan Jones')
pet.addPets('12/22/2018', 'Teddy', 'Carl Johnson')
pet.addPets('07/10/2020', 'Kenny', 'Richard Campbell')
pet.addPets('04/13/2021','Max', 'Bryan Miller')
pet.addPets('07/10/2020', 'Buddy', 'Kathy Brown')
pet.addPets('07/10/2020', 'Kenny', 'John Brown')
pet.printPets()

使用該代碼，我想按日期計算有多少重復的寵物條目，例如我希望在控制臺中：

--------------- 04/13/2021 ------------------
Name: Pinky
Owner name: David Smith
Name: Pinky
Owner name: Daniel Trincot
Name: Max
Owner name: Bryan Miller
There are 2 pets with the same name
--------------- 07/10/2020 ------------------
Name: Charlie
Owner name: Joe Davis
Name: Kenny
Owner name: Susan Jones
Name: Kenny
Owner name: Richard Campbell
Name: Buddy
Owner name: Kathy Brown
Name: Kenny
Owner name: John Brown
There are 3 pets with the same name
--------------- 12/22/2018 ------------------
Name: Teddy
Owner name: Carl Johnson
There are 0 pets with the same name

我試圖這樣做：

if pet.name in list(ls[1]):
     counter  = 1

但這不起作用，因為我認為我的代碼沒有進入 if 而我只是進入控制臺：

--------------- 04/13/2021 ------------------
Name: Pinky
Owner name: David Smith
There are  0 pets with the same name
Name: Pinky
Owner name: Daniel Trincot
There are  0 pets with the same name
Name: Max
Owner name: Bryan Miller
There are  0 pets with the same name
--------------- 07/10/2020 ------------------
Name: Charlie
Owner name: Joe Davis
There are  0 pets with the same name
Name: Kenny
Owner name: Susan Jones
There are  0 pets with the same name
Name: Kenny
Owner name: Richard Campbell
There are  0 pets with the same name
Name: Buddy
Owner name: Kathy Brown
There are  0 pets with the same name
Name: Kenny
Owner name: John Brown
There are  0 pets with the same name
--------------- 12/22/2018 ------------------
Name: Teddy
Owner name: Carl Johnson
There are  0 pets with the same name

所以我來這里尋求幫助。

uj5u.com熱心網友回復：

對printPets方法的更改（針對 PEP 8 進行了調整，但仍應清楚每個名稱所指的內容）（以下完整代碼示例中有更詳細的說明）：

from collections import Counter

...

    def print_all_pets(self):
        for date, group in groupby(sorted(self.pets_list, key=lambda x: x.entry_date),
                                   key=lambda x: x.entry_date):
            print(f'{"-" * 20} {date} {"-" * 20}')
            counter = Counter()
            for pet in group:
                print(f"Name: {pet.name}")
                print(f"Owner name: {pet.owner_name}")
                counter[pet.name]  = 1
            print(f'There are {max(counter.values())} '
                  f'pets with the same name')

完整的代碼示例（它也遵循 PEP 8，我建議你也遵循）；大部分解釋都在代碼注釋中：

from itertools import groupby
from dataclasses import dataclass, field
from collections import Counter


# using dataclasses because they make it easier
# to create more data driven objects and they can pre-build
# all the comparison methods so that sorting is way easier
# and you can exclude what not to sort
@dataclass(order=True)
class PetShop:
    entry_date: str
    name: field(default_factory=str, compare=False)
    owner_name: field(default_factory=str, compare=False)


class Pets:
    def __init__(self):
        self.pets_list = []

    def add_pet(self, entry_date, name, owner_name):
        entry_pet = PetShop(entry_date, name, owner_name)
        self.pets_list.append(entry_pet)

    def print_all_pets(self):
        # as you can see no key is needed for sorting the `PetShop` instances
        # because dataclass allows to do so, then just group them by their date
        # which seems to require some function, also unpack both values:
        # the date and the group
        for date, group in groupby(sorted(self.pets_list), key=lambda x: x.entry_date):
            # use f-string for formatting and also you can multiply strings
            print(f'{"-" * 20} {date} {"-" * 20}')
            # use counter for counting and update it with the pet name
            # each iteration
            counter = Counter()
            for pet in group:
                print(f"Name: {pet.name}")
                print(f"Owner name: {pet.owner_name}")
                # update the counter
                counter[pet.name]  = 1
            # then get the highest count
            print(f'There are {max(counter.values())} '
                  f'pets with the same name')


pets = Pets()

# use a list to make the code less repetitive, this
# allows to loop over the list and add each item to the
# pets.pets_list less repetitively than calling the `add_pet`
# method multiple times manually
pet_shops = [
    ('04/13/2021', 'Pinky', 'David Smith'),
    ('07/10/2020', 'Charlie', 'Joe Davis'),
    ('04/13/2021', 'Pinky', 'Daniel Trincot'),
    ('07/10/2020', 'Kenny', 'Susan Jones'),
    ('12/22/2018', 'Teddy', 'Carl Johnson'),
    ('07/10/2020', 'Kenny', 'Richard Campbell'),
    ('04/13/2021', 'Max', 'Bryan Miller'),
    ('07/10/2020', 'Buddy', 'Kathy Brown'),
    ('07/10/2020', 'Kenny', 'John Brown')
]

for e_date, p_name, o_name in pet_shops:
    pets.add_pet(e_date, p_name, o_name)

pets.print_all_pets()

你最初的問題是你檢查了一個字串是否在一個PetShop實體串列中，所以它總是會評估為False（即使正確使用條件（例如，實際上檢查寵物名稱是否在寵物名稱串列中），它只會給出每個分組的寵物總數）。因此，對于如上所示的計數，我建議使用，collections.Counter因為它比手動制作演算法要容易得多。

有用的來源：

dataclasses 圖書館（內置）
collections.Counter 檔案（內置）
itertools.groupby 檔案（更多示例）

關于 PEP 8：
我強烈建議遵循PEP 8 - Python 代碼風格指南。函式名和變數名應該在snake_case，類名應該在CapitalCase. 沒有足夠的空間周圍=，如果它被用作關鍵字引數的一部分，（ func(arg='value')），但周圍有空間，=如果是用于分配的值（variable = 'some value'）。在運算子周圍留出空間（ -/等value = x y：（此處除外value = x y））。在函式和類宣告周圍有兩個空行。類方法周圍有一個空行。

uj5u.com熱心網友回復：

首先使用 entryDate 然后按名稱分組

for i, j in groupby(pet.petsList, key=lambda p: p.entryDate):
    print(f"-------{i}-----------")
    for item1, item2  in groupby(list(j), key=lambda k: k.name):
        dic = {}
       for k1 in list(item2):
       if k1.name in dic:
            dic[k1.name] = dic[k1.name] 1
       else:
           dic[k1.name] = 1
       print(f"Name: {k1.name}")
       print(f"Owner name: {k1.ownerName}")
       print(dic)

它產生以下結果

-------04/13/2021-----------
Name: Pinky
Owner name: David Smith
Name: Pinky
Owner name: Daniel Trincot
{'Pinky': 2}
Name: Max
Owner name: Bryan Miller
{'Max': 1}
-------07/10/2020-----------
Name: Charlie
Owner name: Joe Davis
{'Charlie': 1}
Name: Kenny
Owner name: Susan Jones
Name: Kenny
Owner name: Richard Campbell
{'Kenny': 2}
Name: Buddy
Owner name: Kathy Brown
{'Buddy': 1}
Name: Kenny
Owner name: John Brown
{'Kenny': 1}
-------12/22/2018-----------
Name: Teddy
Owner name: Carl Johnson
{'Teddy': 1}

uj5u.com熱心網友回復：

我可以在上面的代碼中看到三個問題。

groupby 創建了一個生成器，它只會有一次可迭代的資料。

    def printPets(self):
        self.petsList.sort(key=lambda p: p.entryDate)
        counter = 0
        for group in groupby(self.petsList, key=lambda p: p.entryDate):
            ls = list(group)
            print(f'first {list(ls[1])}')
            print(f'second {list(ls[1])}')
#first [<__main__.Petshop object at 0x7fcd1001e430>, <__main__.Petshop object at 0x7fcd504428b0>, <__main__.Petshop object at 0x7fcd50446d00>]
#second []

您嘗試在組上使用 list() 是正確的，但您需要先解壓日期和寵物。這是一個可能的修復

    def printPets(self):
        self.petsList.sort(key=lambda p: p.entryDate)
        counter = 0
        for date, pets in groupby(self.petsList, key=lambda p: p.entryDate):
            pet_list = list(pets)
            print(f'first {pet_list}')
            print(f'second {pet_list}')
#first [<__main__.Petshop object at 0x7fcd1001e430>, <__main__.Petshop object at 0x7fcd504428b0>, <__main__.Petshop object at 0x7fcd50446d00>]
#second [<__main__.Petshop object at 0x7fcd1001e430>, <__main__.Petshop object at 0x7fcd504428b0>, <__main__.Petshop object at 0x7fcd50446d00>]

petlist 是一個寵物串列，而不是一個 pet_names 串列，所以 if 陳述句永遠不會為真，我們可以使用下面的方法解決這個問題。

    def printPets(self):
        self.petsList.sort(key=lambda p: p.entryDate)
        counter = 0
        for date, pets in groupby(self.petsList, key=lambda p: p.entryDate):
            pet_list = list(pets)
            pet_names = [p.name for p in pet_list]
            print("---------------", ls[0], '------------------')
            for pet in pet_list:
                print("Name:", pet.name)
                print("Owner name:", pet.ownerName)
                if pet.name in pet_names:
                    counter  = 1
            print('There are ',counter,'pets with the same name')

計數總是等于寵物的數量。您只想檢查未見過的寵物以計算重復的寵物，因為每只寵物在 pet_list 中至少出現一次。

    def printPets(self):
        self.petsList.sort(key=lambda p: p.entryDate)
        counter = 0
        for agg, vals in groupby(self.petsList, key=lambda p: p.entryDate):
            _vals = list(vals)
            name_list =[x.name for x in _vals]
            print("---------------", agg, '------------------')
            for i, pet in enumerate(_vals):
                print("Name:", pet.name)
                print("Owner name:", pet.ownerName)
                if pet.name in name_list[:i]:
                    counter  = 1
            print('There are ',counter,'pets with the same name')

我在 (2) 中還包含了一個小修復，通過取消縮進列印陳述句來防止列印中間計數。

uj5u.com熱心網友回復：

也許這不是最好的方法，但在這里。應該為每個組重置計數器，正如@Matiiss 所說，ls[1] 是 PetShop 類的實體，因此將其保存在變數中可能會更好，訪問其屬性可以使其更加清晰。

from itertools import groupby

class Petshop():
    def __init__(self, entryDate, name, ownerName):
        self.entryDate = entryDate
        self.name = name
        self.ownerName = ownerName
        
class Pets():
    def __init__(self):
        self.petsList = []
        
    def addPets(self, entryDate, name, ownerName ):
        entry_pet = Petshop(entryDate, name, ownerName)
        self.petsList.append(entry_pet)
        
    def printPets(self):
        self.petsList.sort(key=lambda p: p.entryDate)
        
        for group in groupby(self.petsList, key=lambda p: p.entryDate):
            ls = list(group)
            data = list(ls[1])
            names= []
            print("--------", ls[0], "--------")
            index = 0
            for pet in list(data):   
                if pet.name in names :
                    repeatedName = pet.name
                index = 1
                
                print("Name:", pet.name)
                names.append(pet.name)
                print("Owner:", pet.ownerName)
               
                if index == len(data):
                   print("There are", names.count(repeatedName), "pets with the same name")
            names = []
            
pet = Pets()

pet.addPets('04/13/2021','Pinky', 'David Smith')
pet.addPets('07/10/2020', 'Charlie', 'Joe Davis')
pet.addPets('04/13/2021','Pinky', 'Daniel Trincot')
pet.addPets('07/10/2020', 'Kenny', 'Susan Jones')
pet.addPets('12/22/2018', 'Teddy', 'Carl Johnson')
pet.addPets('07/10/2020', 'Kenny', 'Richard Campbell')
pet.addPets('04/13/2021','Max', 'Bryan Miller')
pet.addPets('07/10/2020', 'Buddy', 'Kathy Brown')
pet.addPets('07/10/2020', 'Kenny', 'John Brown')
pet.printPets()

這是我得到的輸出

-------- 04/13/2021 --------
Name: Pinky
Owner: David Smith
Name: Pinky
Owner: Daniel Trincot
Name: Max
Owner: Bryan Miller
There are 2 pets with the same name
-------- 07/10/2020 --------
Name: Charlie
Owner: Joe Davis
Name: Kenny
Owner: Susan Jones
Name: Kenny
Owner: Richard Campbell
Name: Buddy
Owner: Kathy Brown
Name: Kenny
Owner: John Brown
There are 3 pets with the same name
-------- 12/22/2018 --------
Name: Teddy
Owner: Carl Johnson
There are 0 pets with the same name

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