Python3- __init__()从1到5个位置参数，但给出了6个
Python3- __init__()从1到5个位置参数，但给出了6个
提问于 2020-10-14 04:55:35
我第一次尝试在python中实现linked_lists，但我的附加函数似乎不能正常工作。以下是目前为止的代码：
class node:
def __init__(self, Name = None, Contact_number = None, Type = None, Urgency = None):
self.Name = Name
self.Contact_number = Contact_number
self.Type = Type
self.Urgency = Urgency
self.next = None
class linked_list:
def __init__(self):
self.head = node()
def append(self, Name, Contact_number, Type, Urgency):
new_node = node(self,Name,Contact_number,Type,Urgency)
cur = self.head
while cur.next != None:
cur = cur.next
cur.next = new_node
def length(self):
cur = self.head()
total = 0
while cur.next != None:
total+=1
cur=cur.next
return total
def display(self):
Appointments = []
cur_node = self.head
while cur_node.next != None:
cur_node = cur_node.next
aptmnt = (cur_node.Name, cur_node.Contact_number, cur_node.Type, cur_node.Urgency)
Appointments.append(aptmnt)
print(Appointments)
general_surgeon = linked_list()
general_surgeon.display()
general_surgeon.append("Akash", "827xxxxxx1", "Min Surgery", "no")下面是我尝试使用附加函数时遇到的错误：
TypeError Traceback (most recent call last)
<ipython-input-27-0ffb4f93a9c0> in <module>
39 general_surgeon.display()
40
---> 41 general_surgeon.append("Akash", "827xxxxxx1", "Min Surgery", "no")
<ipython-input-27-0ffb4f93a9c0> in append(self, Name, Contact_number, Type, Urgency)
12
13 def append(self, Name, Contact_number, Type, Urgency):
---> 14 new_node = node(self,Name,Contact_number,Type,Urgency)
15 cur = self.head
16 while cur.next != None:
TypeError: __init__() takes from 1 to 5 positional arguments but 6 were given我一直在直接从youtube视频中复制链接列表的基本结构和功能，除了我的代码有更多的数据变量之外，我注意到我的代码和他的没有区别。
请帮帮忙？
回答 2
Stack Overflow用户
回答已采纳
发布于 2020-10-14 04:57:40
def append(self, Name, Contact_number, Type, Urgency):
new_node = node(self,Name,Contact_number,Type,Urgency)因为要调用node类，所以不需要传递self参数。
Stack Overflow用户
发布于 2020-10-14 05:01:29
下面是错误，它已经显示为一个TypeError
TypeError: __init__() takes from 1 to 5 positional arguments but 6 were given所以你在line # 14中给出了6个参数，而不是5个
无需将self作为参数传递到node class at line # 14中
试着用这个：
def append(self, Name, Contact_number, Type, Urgency):
new_node = node(Name, Contact_number, Type, Urgency)
cur = self.head
while cur.next != None:
cur = cur.next
cur.next = new_node页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/64346826
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