I have a linked stack class and I'm having the problem of printing the elements in the stack, with each element in a new line. The str function in the linked stack class is printing every element in a new line which is what i wanted but it even prints an extra new line at the end.
class Node:
def __init__(self,item,the_next = None):
self.item = item
self.next = the_next
def __str__(self):
return str(self.item)
class LinkedStack:
def __init__(self):
self.top = None
self.count = 0
def __len__(self):
return self.count
def is_empty(self):
return self.count == 0
def isFull(self):
return False
def reset(self):
self.top = None
self.count = 0
def __str__(self): #im having the issue here whereby it prints a newline even after printing the last element
current = self.top
ans = ""
while not (current is None):
ans += str(current)
ans += '\n'
current = current.next
return ans
if __name__ == "__main__":
L = LinkedStack()
L.push(1)
L.push(2)
L.push(3)
print(L)
The output i get is:
3
2
1
#an extra newline printed here which is not wanted
I'm looking for a way to improvise the str function in the Linked Stack class in order to get rid of the redundant new line at the end. Any help is much appreciated.
Why not simply trim the return value of __str__ like so return ans[:-1] ? Since you always append a new line after adding an element to the string.
On a side note, it could be nicer to write your function like so:
def __str__(self):
strs = []
cur = self.top
while cur is not None:
strs = [str(cur)] + strs
cur = cur.next
return '\n'.join(strs)
You could move the addition of the newline until after you get the next item, and check it is not None before you then append it.
A further optimisation then is to move the break condition to that point:
while True:
ans += str(current)
current = current.next
if current is None:
break
ans += '\n'
Related
EDITED.
I am learning about Linked Lists. For each process applied by a Method, it is printed out to the console. So, adding, removing, searching (i.e, displaying the result of a search), are all streamed to stdout, but I cannot seem to do this for the insertion Method even though the insert Method is executed.
Some Methods have a return statement, while others rely on the __repr__() for conversion to string, to then be streamed to the console. The insertion Method (not mine, but a course worked example) takes two arguments and does not have a return statement. The most consistent error message I get when attempting to print is TypeError: %d format: a real number is required, not NoneType, or TypeError: not enough arguments for format string, where I have replaced %d with %s.
What I do not understand is, why I am unable to display test data for the insert Method, while I can do so for all others.
The code,
#!/usr/bin/env python3
class Node:
data = None
next_node = None
def __init__(self, data):
self.data = data
def __repr__(self):
return "<Node data: {}>".format(self.data)
# Linked List
class LinkedList:
def __init__(self):
self.head = None
def is_empty(self):
return self.head == None # corrected
def size(self):
current = self.head
count = 0
while current:
count += 1
current = current.next_node
return count
# Adding a node
def add(self, data):
new_node = Node(data)
new_node.next_node = self.head
self.head = new_node
# Searching the List
def search(self, key):
current = self.head
while current:
if current.data == key:
return current
else:
current = current.next_node
return None
# Inserting into the List
def insert(self, data, index):
if index == 0:
self.add(data)
if index > 0:
new_data = Node(data)
position = index
current = self.head
while position > 1:
current = current.next_node
position -= 1
past_node = current
future_node = current.next_node
past_node.next_node = new_data
new_data = current.next_node
# Removing a node from the List
def remove(self, key):
current = self.head
previous = None
found = False
while current and not found:
if current.data == key and current == self.head:
found = True
self.head = current.next_node
elif current.data == key:
found = True
previous.next_node = current.next_node
return current
def __repr__(self):
nodes = []
current = self.head
while current:
if current is self.head:
nodes.append("[Head: {}]".format(current.data))
elif current.next_node is None:
nodes.append("[Tail {}]".format(current.data))
else:
nodes.append("[{}]".format(current.data))
current = current.next_node
return '-> '.join(nodes)
Test output;
l = LinkedList()
l.add(1)
l.add(2)
l.add(3)
l.add(5)
l.add(6)
l.add(7)
length = l.size()
print("Size of list: {}".format(length)) # Size of list: 6
print(l) # [Head: 7]-> [6]-> [5]-> [3]-> [2]-> [Tail: 1]
seek = l.search(7)
print("Found: {}".format(seek)) # Found: <Node data: 7>
between = l.insert(4, 3)
if between is not None:
print(f"Inserted {between} at index {between}")
else:
print("A problem with code") # A problem with code
gone = l.remove(1)
print("Removed: {}".format(gone)) # Removed: <Node data: 1>
# Note the insertion of '4' at index 3
print(l) # [Head: 7]-> [6]-> [5]-> [4]-> [3]-> [Tail: 2]
THIS CODE WORKS!
Other variants of the print format have been tried f"{}", .format() and even an attempt at conversion to string str() was made, with no luck. Could someone explain exactly what the problem is (though, I believe it to be a NoneType issue) and how to resolve it?
I hope my question is clearer. Thank you.
There are several issues with the code you presented, including the following:
The Node class should not define data and next_node as class attributes. They should be instance attributes. Luckily, the constructor creates an instance attribute data (hiding the class attribute), but for next_node this is not done, which makes your linked list unusable.
In line with the previous comment, you should have self.next_node = None in your constructor.
[You corrected this in an edit to your question: The method name is_empty suggests that it will return a boolean indicating whether the list is empty or not. But instead it makes the list empty. That seems wrong.]
[You corrected this in an edit to your question: insert can call a method add which is not defined.]
In insert, when index is 0, the code will still continue after the first if and reference a variable new_data that has not been defined (since the second if condition was not true). You should avoid that any of the other code is executed when index is 0. You can do this with a return.
In insert, in the while loop there is no verification whether current is None. If that happens, current = current.next_node will raise an error.
new_data = current.next_node is useless and leaves the next_node attribute of next_node uninitialised.
Not an issue, but in remove, the found name is not very useful. Just break out of the loop when the node has been found and removed. Also, avoid having the current.data == key condition executed twice for the same node.
In remove, in the loop, you never change current nor previous, and so the loop hangs.
[You corrected this in an edit to your question: In the main code, the list is empty at the moment that l.insert is called, so it is strange to pass 3 as value for the index parameter, as that index is out of range. As mentioned in a previous bullet, this will trigger an error. If you want to add a node at index 3, you'll first have to add nodes at indexes 0, 1, and 2.]
The insert method does not return anything, so capturing its return value is not going to give you anything else than None. If you really want to get some feedback from it, then do like you did for the remove method: have it return the relevant node. In that case you should also let add have a return value.
Here is some working code with the above issues addressed and more:
class Node:
def __init__(self, data):
self.data = data
self.next_node = None # next_node neads to be an instance attribute, not a class attribute
def __repr__(self):
return "<Node data: {}>".format(self.data)
class LinkedList:
def __init__(self):
self.head = None
def is_empty(self):
return self.head == None # Don't MAKE it empty!
def size(self):
current = self.head
count = 0
while current:
count += 1
current = current.next_node
return count
def add(self, data):
new_node = Node(data)
new_node.next_node = self.head
self.head = new_node
return new_node # Return the new node
def search(self, key):
current = self.head
while current:
if current.data == key:
return current
else:
current = current.next_node
return None
def insert(self, data, index):
if index == 0:
# Don't continue after this call to self.add
return self.add(data) # Return the new node
current = self.head
while index > 1 and current: # Protect against out of range index
current = current.next_node
index -= 1
if current: # Protect against out of range index
new_data = Node(data)
# Make sure the new node gets a next_node assignment
new_data.next_node = current.next_node
current.next_node = new_data
return new_data # Return the new node
def remove(self, key):
current = self.head
previous = None
while current:
if current.data == key: # Check this only once per node
if current == self.head:
self.head = current.next_node
else:
previous.next_node = current.next_node
break # No need for variable - just exit
previous = current # Update previous
current = current.next_node # Move to next node
return current
def __repr__(self):
nodes = []
current = self.head
while current:
if current is self.head:
nodes.append("[Head: {}]".format(current.data))
elif current.next_node is None:
nodes.append("[Tail {}]".format(current.data))
else:
nodes.append("[{}]".format(current.data))
current = current.next_node
return '-> '.join(nodes)
l = LinkedList()
l.add(1)
l.add(2)
l.add(3)
l.add(5)
l.add(6)
l.add(7)
length = l.size()
print("Size of list: {}".format(length)) # Size of list: 6
print(l) # [Head: 7]-> [6]-> [5]-> [3]-> [2]-> [Tail: 1]
seek = l.search(7)
print("Found: {}".format(seek)) # Found: <Node data: 7>
node = l.insert(4, 3)
print("Inserted {}".format(node)) # Inserted: <Node data: 4>
gone = l.remove(1)
print("Removed: {}".format(gone)) # Removed: <Node data: 1>
# Note the insertion of '4' at index 3
print(l) # [Head: 7]-> [6]-> [5]-> [4]-> [3]-> [Tail: 2]
insert does not "have a problem" outputting data - just like standard Python lists, it is an in-place operation. You are modifying the list on which it is applied.
insert() does not need to return anything, as all the information you need is provided by you when calling it - you need to pass a list, you need to pass data to insert and you need to pass an index at which the element is to be placed - there is no new information to be gained from returning anything.
Related question:
Why don't list operations return the resulting list?
I think you're confusing 2 things here. The value after the return statement is what the function call is replaced with when the function is called. So for example:
def square(x):
return x*x
square(4)
here the square(4) would be replaced with 4*4. And if you don't explicitly use a return statement than a None is returned after the last command in the function/method.
Whereas repr() is a way to specifiy the string representation of that object. So for example:
class A:
pass
a = A()
print(a)
might create a cryptic output of <main.A at 0x7fbc841c9490>. So if you want it to be more descriptive you could add a repr() method:
class Point:
def __init__(self, x,y):
self.x = x
self.y = y
def __repr__(self):
return f"Coordinates of the point are x: {self.x}, y: {self.y}"
p = Point(2,4)
And instead of the cryptic default message you'd get:
Coordinates of the point are x: 2, y: 4
So the representation is how the obj is converted to a string whereas the return value is what the function call is replaced with.
print is TypeError: %d format: a real number is required, not
NoneType, or TypeError: not enough arguments for format string, where
I have replaced %d with %s.
So this creates errors because %d and %s expect numbers and strings when the return type of a method without return is None.
I am having some trouble with linked lists in Python. For a problem I am supposed to do trivia like question and answer with linked lists, each node having a question and answer. I am supposed to have the program go through the list over and over until each question is answered correctly twice. The following code is the method I made to do this.
class trivia(object):
def __init__(self, question, answer):
self.question = question
self.answer = answer
self.next = None
def getQuestion(self):
return self.question
def getAnswer(self):
return self.answer
def getNext(self):
return self.next
def setNext(self, next):
self.next = next
class triviaDeck(object):
def __init__(self):
self.head = None
def size(self):
current = self.head
count = 0
while current != None: # while not at the end of the list
count += 1
current = current.getNext()
return count
def showTrivia(self, pos=None):
if pos == None:
pos = self.size() - 1
if pos < self.size() and pos >= 0:
current = self.head
previous = None
index = 0
count = 0
while count != 2:
print(current.getQuestion())
answer = input()
if answer == current.getAnswer():
print("Correct")
count = count + 1
if current.getNext() == 2:
current = self.head
else:
current = current.getNext()
def add(self, question, answer):
temp = trivia(question, answer)
temp.setNext(self.head)
self.head = temp
if __name__ == '__main__':
deck = triviaDeck()
deck.add("A", "A")
deck.add("B", "B")
deck.add("C", "C")
deck.showTrivia()
Currently this code just goes through the list once and then exits.
For a circular linked list:
The first node inserted should point to itself
For additional nodes, adjust the pointers to insert the new node after the head
Try this code:
def add(self, question, answer):
temp = trivia(question, answer)
if not self.head: # first node
self.head = temp
temp.setNext(self.head) # single node, loop to itself
else: # add node
temp.setNext(self.head.getNext()) # shift loop
self.head.setNext(temp) # insert just after head
Also - For the count = 2 logic:
The node (trivia) object should have a counter property which counts correct responses to that question
If the response counter = 2, remove that node from the linked list
When the last node is removed, game over
I have looked at many very similar questions and cannot figure it out so:
I have a string like this:
{121{12}12{211}2}
I want to read the string into a tree like this:
I am confused as how to tell python to add a whole list as a child node?
I would also like to know how to change the current node to the parent of the old current node?
Here is my code so far:
class Node:
def __init__(self,val):
self.value = val
self.children = []
#init Node class so we can pass in values as nodes and set children to empty list
def add_child(self, obj):
self.children.append(obj)
s=[]
for i in filedata:
if i == leftbrace:
n = Node(i)
#create new child of current node
s = []
#reset list s to blank
if i == rightbrace:
n.add_child(s)
#add list s to current node
#make parent of current node the new current node
else:
s.append(i)
#add i to list s
for c in n.children:
print (c.data)
To make something like this work, it is easiest if you use recursion. Here is one way that this can be done.
Code:
class Node:
def __init__(self, stream):
val = []
children = []
while True:
try:
# get the next character from the stream
ch = next(stream)
# if this is an open brace, then recurse to a child
if ch == '{':
children.append(Node(stream))
# if this is a close brace, we are done on this level
elif ch == '}':
break
# otherwise add this character to our value
else:
val.append(ch)
# stream is empty, we are done
except StopIteration:
break
self.value = ''.join(val)
self.children = children
#classmethod
def from_string(cls, string):
stream = iter(string)
tree_top = Node(stream)
# assert that the string started with '{' and was one top node
assert len(tree_top.children) == 1 and tree_top.value == ''
return tree_top.children[0]
def __str__(self):
return self.value
def __repr__(self):
return "Node('%s', <%d children>)" % (
self.value, len(self.children))
def tree_string(self, level=0):
yield '-' + " " * level + str(self)
for child in self.children:
for child_string in child.tree_string(level+1):
yield child_string
tree = '{121{12}12{211}2}'
for line in Node.from_string(tree).tree_string():
print(line)
Results:
-121122
- 12
- 211
I got this school assignment, here is my code:
class Doubly_linked_node():
def __init__(self, val):
self.value = val
self.next = None
self.prev = None
def __repr__(self):
return str(self.value)
class Deque():
def __init__(self):
self.header = Doubly_linked_node(None)
self.tailer = self.header
self.length = 0
def __repr__(self):
string = str(self.header.value)
index = self.header
while not (index.next is None):
string+=" " + str(index.next.value)
index = index.next
return string
def head_insert(self, item):
new = Doubly_linked_node(item)
new.next=self.header
self.header.prev=new
self.header=new
self.length+=1
if self.tailer.value==None:
self.tailer = self.header
def tail_insert(self, item):
new = Doubly_linked_node(item)
new.prev=self.tailer
self.tailer.next=new
self.tailer=new
self.length+=1
if self.header.value==None:
self.header = self.tailer
it builds a stack, allowing you to add and remove items from the head or tail (I didn't include all the code only the important stuff).
When I initiate an object, if I return self.next it prints None, but if I return self.prev, it prints nothing, just skips, I don't understand why since they are both defined exactly the same as you see, and if I insert only head several times for example for i in range(1,5): D.head_insert(i) and then I print D it prints 5 4 3 2 1 None but if I do tail insert for example for i in range(1,5): D.tail_insert(i) and print D it prints 1 2 3 4 5"as it should without the None. Why is that?
I have included an image:
Keep in mind that you create a Deque which is not empty. You're initializing it with a Node with value None
You're interchanging the value and the Node object. When you're checking if self.tailer.value==None: it's probably not what you're meaning
Following to point 2 is a special handling for the empty Deque, where header and tailer is None
Here is what I have in mind, if I would implement the Deque. I'm slightly changed the return value of __repr__.
class Deque():
def __init__(self):
self.header = None
self.tailer = None
self.length = 0
def __repr__(self):
if self.header is None:
return 'Deque<>'
string = str(self.header.value)
index = self.header.next
while index!=None:
string+=" " + str(index.value)
index = index.next
return 'Deque<'+string+'>'
def head_insert(self, item):
new = Doubly_linked_node(item)
new.next=self.header
if self.length==0:
self.tailer=new
else:
self.header.prev=new
self.header=new
self.length+=1
def tail_insert(self, item):
new = Doubly_linked_node(item)
new.prev=self.tailer
if self.length==0:
self.header=new
else:
self.tailer.next=new
self.tailer=new
self.length+=1
Following Günthers advice, I have modified the __repr__ to this:
def __repr__(self):
string = str(self.header.value)
index = self.header
while not (str(index.next) == "None"):
string += (" " + str(index.next.value))
index = index.next
return string
that did solve the problem, but it is the ugliest solution I have ever seen.
does anyone know a better way?
Following to the question of a better __repr__ method here my proposal. Extend the Deque class with an __iter__ method. So you can iterate over the Deque which is nice to have, e.g.:
for item in D:
print item
Based on that the __repr__ method is easy. Here is the whole change:
def __repr__(self):
return 'Deque<'+' '.join([str(item.value) for item in self])+'>'
def __iter__(self):
index=self.header
while index is not None:
yield index.value
index=index.next
I have a class
class List:
def __init__(self,head,tail):
self.head = head
self.tail = tail
def cons(self,item):
return List(item,self)
def isEmpty(self):
return self.head == None
def display(self):
s = "["
first = True
list = self
while not list.isEmpty():
if not first:
s=s+","
first = False
s=s+str(list.head)
list = list.tail
s=s+"]"
return s`
which creates a List object. I have a function (not sure if it works yet);
def sorted(list):
sort = False
i = 0
while i < range(len(list))+1:
if list[i] < list[i+1]:
sort = True
return sort
else:
return sort
and I want to run this function on the List object without adding another method to the class. I know if this was in the class it would just be List.sorted() but how can I run this function on the object without it being the objects method?
sorted(List) doesn't seem to work either. Help please.
Please, please, PLEASE: don't use sorted as a function name. There is a standard (built-in) function sorted() already, that returns a sorted version of an iterable.
Given that your code just checks to see if the List is in order, perhaps you could call it in_order or ordered or is_ascending instead?
That said, let's try to get your code working:
class Node:
def __init__(self, data=None, next=None):
self.data = data
self.next = next
def __str__(self):
return str(self.data)
class List:
def __init__(self, head:Node=None, tail:Node=None):
self.head = tail if head is None else head
self.tail = head if tail is None else tail
def cons(self, node:Node):
node.next = self.head
return List(node, self.tail)
def isEmpty(self):
return self.head == None
def display(self):
return str(self)
def __str__(self):
result = "["
first = True
n = self.head
while n is not None:
if not first:
result += ", "
result += str(n)
if n == self.tail:
break
n = n.next
first = False
result += "]"
return result
def ascendingp(self) -> bool:
n = self.head
last = None
while last != self.tail:
if last is not None and not last.data < n.data:
return False
last = n
n = n.next
return True
tests = (
(1,2,3),
(1,3,2),
(),
)
for t in tests:
last = None
first = None
for i in reversed(t):
n = Node(data=i)
first = n if first is None else first
n.next = last
last = n
l = List(head=last, tail=first)
print(str(l), "Ascending?", l.ascendingp())