如何计算重复的对象属性？
How to count repeated objects attributes?
我有下一个代码，我为每个入口宠物创建对象：
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()
使用该代码，我想计算按日期重复输入宠物的次数，例如，我希望在控制台中：
--------------- 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
我尝试这样做：
if pet.name in list(ls[1]):
counter += 1
但这不起作用，因为我认为我的代码没有进入 if 而我只是进入了控制台：
--------------- 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
所以我来这里寻求帮助。
这可能不是最好的方法，但您可以这样做。应该为每个组重置计数器，正如@Matiiss 所说，ls[1] 是 PetShop class 的实例，因此将其保存在变量中可能会更好，访问其属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()
这是我得到的输出
-------- 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
对 printPets 方法的更改（针对 PEP 8 进行了调整，但仍然应该清楚每个名称指的是什么）（在下面的完整代码示例中有更详细的解释）：
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')
完整的代码示例（它也遵循我建议你也遵循的 PEP 8）；大部分解释在代码注释中：
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()
而您最初的问题是您检查了一个字符串是否在 PetShop 个实例的列表中，因此它总是评估为 False（以及即使正确使用的条件（实际上检查是否例如，宠物名称在宠物名称列表中），它只会给出每个群组的宠物总数）。因此，对于如上所示的计数，我建议使用 collections.Counter，因为它比手动创建算法要容易得多。
有用的资源：
dataclasses library (built-in)collections.Counter docs (built-in)itertools.groupby docs (more for the example)
在 PEP 8 上：
我强烈建议关注 PEP 8 - Style Guide for Python Code。函数和变量名称应在 snake_case 中，class 中的名称应在 CapitalCase 中。如果用作关键字参数 (func(arg='value')) 的一部分，则 = 周围不要有 space，但如果用于 =，则 space 周围有赋值 (variable = 'some value')。在运算符周围有 space（+-/ 等：value = x + y（此处除外 value += x + y））。在函数和 class 声明周围有两个空行。 Class 方法周围有一个空行。
首先使用 entryDate 分组，然后按名称分组
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)
它产生以下结果
-------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}
我可以在上面的代码中看到三个陷阱。
- groupby创建一个生成器，这将只有一次可迭代数据。
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 []
您尝试在群组中使用 list() 是正确的，但您需要先解压缩日期和宠物。这是一个可能的修复方法
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 是一个宠物列表，而不是 pet_names 的列表，所以 if 语句永远不会为真，我们可以使用下面的方法解决这个问题。
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')
- 计数总是等于宠物的数量。您只想检查看不见的宠物以计算重复的宠物，因为每只宠物至少在 pet_list 中出现一次。
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')
还有一个我在 (2) 中包含的小修复，通过取消缩进打印语句来防止打印中间计数。
我有下一个代码，我为每个入口宠物创建对象：
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()
使用该代码，我想计算按日期重复输入宠物的次数，例如，我希望在控制台中：
--------------- 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
我尝试这样做：
if pet.name in list(ls[1]):
counter += 1
但这不起作用，因为我认为我的代码没有进入 if 而我只是进入了控制台：
--------------- 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
所以我来这里寻求帮助。
这可能不是最好的方法，但您可以这样做。应该为每个组重置计数器，正如@Matiiss 所说，ls[1] 是 PetShop class 的实例，因此将其保存在变量中可能会更好，访问其属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()
这是我得到的输出
-------- 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
对 printPets 方法的更改（针对 PEP 8 进行了调整，但仍然应该清楚每个名称指的是什么）（在下面的完整代码示例中有更详细的解释）：
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')
完整的代码示例（它也遵循我建议你也遵循的 PEP 8）；大部分解释在代码注释中：
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()
而您最初的问题是您检查了一个字符串是否在 PetShop 个实例的列表中，因此它总是评估为 False（以及即使正确使用的条件（实际上检查是否例如，宠物名称在宠物名称列表中），它只会给出每个群组的宠物总数）。因此，对于如上所示的计数，我建议使用 collections.Counter，因为它比手动创建算法要容易得多。
有用的资源：
dataclasseslibrary (built-in)collections.Counterdocs (built-in)itertools.groupbydocs (more for the example)
在 PEP 8 上：
我强烈建议关注 PEP 8 - Style Guide for Python Code。函数和变量名称应在 snake_case 中，class 中的名称应在 CapitalCase 中。如果用作关键字参数 (func(arg='value')) 的一部分，则 = 周围不要有 space，但如果用于 =，则 space 周围有赋值 (variable = 'some value')。在运算符周围有 space（+-/ 等：value = x + y（此处除外 value += x + y））。在函数和 class 声明周围有两个空行。 Class 方法周围有一个空行。
首先使用 entryDate 分组，然后按名称分组
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)
它产生以下结果
-------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}
我可以在上面的代码中看到三个陷阱。
- groupby创建一个生成器，这将只有一次可迭代数据。
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 []
您尝试在群组中使用 list() 是正确的，但您需要先解压缩日期和宠物。这是一个可能的修复方法
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 是一个宠物列表，而不是 pet_names 的列表，所以 if 语句永远不会为真，我们可以使用下面的方法解决这个问题。
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')
- 计数总是等于宠物的数量。您只想检查看不见的宠物以计算重复的宠物，因为每只宠物至少在 pet_list 中出现一次。
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')
还有一个我在 (2) 中包含的小修复，通过取消缩进打印语句来防止打印中间计数。