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Insert specific Node in desired Position Linked List on Python

i have sort of problem imagining or solving this, i have this node and list
class node:
def __init__(self, info):
self.info = info
self.next = None
class list:
def __init__(self):
self.__first = None
self.__last = None
and i have to create a function, called indexL(self, info, position), that put the exact node in the desired position, here is what i have for now
(i have created already a method called self.length() to give the size of the list)
def indexL(self, info, position):
longit = self.length()
n = node(info)
p = self.__first
if position == 0:
self.__first = n
else:
if position > 0 and position < longit:
if position == longit-1:
self.__last = n
else:
and now i'm stuck there, by the way i can't add any method to the node or use other method of the list, if someone want to help or any recommendation, i would be really grateful
EDIT: in my last comment is more clear if you are confused :P

You don't really need a length() method... all you need to do is iterate over the list until you reach position and update the info:
def indexL(self, info, position):
if not self.__first:
raise Error("Empty List")
item = self.__first
i = 0
while item:
if i == position:
break
item = item.next
i += 1
else:
raise Error("Position is outside of the size of the list")
item.info = info # Replace info

Making a linked list without built in functions

I have a project to make a linked list in python.
My program needs to add to the list, remove from it and get elements. Sounds easy right? Wrong! We aren't allowed to use normal lists or built-in functions (other than the basic print, str...)
I have one problem with my code, I have to initialise a blank list then add the elements, 1 by 1. Everything else works fine.
My questions:
Is this how a normal python list works?
Is it possible to add items to a linked list with a loop? (without another list)
Here's the code:
class Node: # the node class
def __init__(self, cargo = None, next = None): # __init__ stands for initialize
self.cargo = cargo # e.g. steve
self.next = next # represents the next node or None if its the last
def __str__(self): # __str__ is called when the node is printed or converted to a string
return str(self.cargo) # return a string
class List: # the main list class
def __init__(self): # default is to initialize an empty list
self.first_node = None
self.last_node = None
self.length = 0
def get(self, position, length): # function for efficiency
if position == "end":
position = length - 1 # last
if position > length - 1: # can't go beyond last
raise ValueError("List index out of range")
prv_node = self.first_node
node = self.first_node # start at the first
num = 0
while num < position: # go up to position
prv_node = node # remember the previous node
node = node.next # next node!
num = num + 1
return prv_node, node, position
def add_node(self, cargo, position = "end"): # adds a node
prv_node, node, position = self.get(position, self.length + 1) # +1 because the length is being increased
print("adding node at "+str(position)+": "+str(cargo))
if position == 0: # first
self.first_node = Node(cargo, next = self.first_node) # the first node is the new node
if self.length == 0: # first node to be added
self.last_node = self.first_node # there is only one node
elif position == self.length: # last
self.last_node.next = Node(cargo, next = None) # last_node.next was None, it is now a new node
self.last_node = self.last_node.next # last node is now the new last node
else: # normal
prv_node.next = Node(cargo, next = node) # stick it in between
self.length = self.length + 1 # length is now + 1
def get_node(self, position): # gets a node
...
def remove_node(self, position): # removes a node
...
def __str__(self): # when the list is printed
node = self.first_node # start from the first
string = ""
while node != self.last_node: # go to the end
string = string + str(node) + ", " # print each node
node = node.next
string = string + str(self.last_node) # last node hasn't been added yet
return string
# initialize
mylist = List()
mylist.add_node("steve")
mylist.add_node("james")
mylist.add_node("tom")
mylist.add_node("david")
mylist.add_node("hoe-yin")
mylist.add_node("daniel")
print(mylist)
[EDIT] second question re-phrased

Here's how Python lists are implemented in CPython: http://www.laurentluce.com/posts/python-list-implementation/
If you have your values in some other iterable, then yes:
for item in list_of_items:
mylist.add_node(item)

Python: Create a Binary search Tree using a list

The objective of my code is to get each seperate word from a txt file and put it into a list and then making a binary search tree using that list to count the frequency of each word and printing each word in alphabetical order along with its frequency. Each word in the can only contain letters, numbers, -, or ' The part that I am unable to do with my beginner programming knowledge is to make the Binary Search Tree using the list I have (I am only able to insert the whole list in one Node instead of putting each word in a Node to make the tree). The code I have so far is this:
def read_words(filename):
openfile = open(filename, "r")
templist = []
letterslist = []
for lines in openfile:
for i in lines:
ii = i.lower()
letterslist.append(ii)
for p in letterslist:
if p not in ['a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z',"'","-",' '] and p.isdigit() == False:
letterslist.remove(p)
wordslist = list("".join(letterslist).split())
return wordslist
class BinaryTree:
class _Node:
def __init__(self, value, left=None, right=None):
self._left = left
self._right = right
self._value = value
self._count = 1
def __init__(self):
self.root = None
def isEmpty(self):
return self.root == None
def insert(self, value) :
if self.isEmpty() :
self.root = self._Node(value)
return
parent = None
pointer = self.root
while (pointer != None) :
if value == pointer._value:
pointer._count += 1
return
elif value < pointer._value:
parent = pointer
pointer = pointer._left
else :
parent = pointer
pointer = pointer._right
if (value <= parent._value) :
parent._left = self._Node(value)
else :
parent._right = self._Node(value)
def printTree(self):
pointer = self.root
if pointer._left is not None:
pointer._left.printTree()
print(str(pointer._value) + " " + str(pointer._count))
if pointer._right is not None:
pointer._right.printTree()
def createTree(self,words):
if len(words) > 0:
for word in words:
BinaryTree().insert(word)
return BinaryTree()
else:
return None
def search(self,tree, word):
node = tree
depth = 0
count = 0
while True:
print(node.value)
depth += 1
if node.value == word:
count = node.count
break
elif word < node.value:
node = node.left
elif word > node.value:
node = node.right
return depth, count
def main():
words = read_words('sample.txt')
b = BinaryTree()
b.insert(words)
b.createTree(words)
b.printTree()

Since you're a beginner I'd advice to implement the tree methods with recursion instead of iteration since this will result to simpler implementation. While recursion might seem a bit difficult concept at first often it is the easiest approach.
Here's a draft implementation of a binary tree which uses recursion for insertion, searching and printing the tree, it should support the functionality you need.
class Node(object):
def __init__(self, value):
self.value = value
self.left = None
self.right = None
self.count = 1
def __str__(self):
return 'value: {0}, count: {1}'.format(self.value, self.count)
def insert(root, value):
if not root:
return Node(value)
elif root.value == value:
root.count += 1
elif value < root.value:
root.left = insert(root.left, value)
else:
root.right = insert(root.right, value)
return root
def create(seq):
root = None
for word in seq:
root = insert(root, word)
return root
def search(root, word, depth=1):
if not root:
return 0, 0
elif root.value == word:
return depth, root.count
elif word < root.value:
return search(root.left, word, depth + 1)
else:
return search(root.right, word, depth + 1)
def print_tree(root):
if root:
print_tree(root.left)
print root
print_tree(root.right)
src = ['foo', 'bar', 'foobar', 'bar', 'barfoo']
tree = create(src)
print_tree(tree)
for word in src:
print 'search {0}, result: {1}'.format(word, search(tree, word))
# Output
# value: bar, count: 2
# value: barfoo, count: 1
# value: foo, count: 1
# value: foobar, count: 1
# search foo, result: (1, 1)
# search bar, result: (2, 2)
# search foobar, result: (2, 1)
# search bar, result: (2, 2)
# search barfoo, result: (3, 1)

To answer your direct question, the reason why you are placing all of the words into a single node is because of the following statement inside of main():
b.insert(words)
The insert function creates a Node and sets the value of the node to the item you pass in. Instead, you need to create a node for each item in the list which is what your createTree() function does. The preceeding b.insert is not necessary.
Removing that line makes your tree become correctly formed, but reveals a fundamental problem with the design of your data structure, namely the printTree() method. This method seems designed to traverse the tree and recursively call itself on any child. In your initial version this function worked, because there the tree was mal-formed with only a single node of the whole list (and the print function simply printed that value since right and left were empty).
However with a correctly formed tree the printTree() function now tries to invoke itself on the left and right descendants. The descendants however are of type _Node, not of type BinaryTree, and there is no methodprintTree() declared for _Node objects.
You can salvage your code and solve this new error in one of two ways. First you can implement your BinaryTree.printTree() function as _Node.printTree(). You can't do a straight copy and paste, but the logic of the function won't have to change much. Or you could leave the method where it is at, but wrap each _left or _right node inside of a new BinaryTree so that they would have the necessary printTree() method. Doing this would leave the method where it is at, but you will still have to implement some kind of helper traversal method inside of _Node.
Finally, you could change all of your _Node objects to be _BinaryTree objects instead.
The semantic difference between a node and a tree is one of scope. A node should only be aware of itself, its direct children (left and right), and possibly its parent. A tree on the other hand can be aware of any of its descendents, no matter how far removed. This is accomplished by treating any child node as its own tree. Even a leaf, without any children at all can be thought of as a tree with a depth of 0. This behavior is what lets a tree work recursively. Your code is mixing the two together.

how to get min or max value from a linked list?

I am trying to define a function that can get min value from a linked list of ints.
Given Function(not allowed to be modified):
class LN:
def __init__(self,value,next=None):
self.value = value
self.next = next
def list_to_ll(l):
if l == []:
return None
front = rear = LN(l[0])
for v in l[1:]:
rear.next = LN(v)
rear = rear.next
return front
Function list_to_ll convert a normal list to linked list:
A recursive function I am trying to define:
def get_min(ll):
if ll == None:
return None
else:
if ll.value < ll.next.value:
return ll.value
return get_min(ll.next)
For example:
get_min(list_to_ll([7, 3, 5, 2, 0]))--> 0
But my function gives me:
RuntimeError: maximum recursion depth exceeded in comparison
Please help. Actual codes would be really appreciated.

Implement __iter__ for your data structure so you can iterate over it. Then you can use the regular min() and max() functions (as well as for loops, the any() and all() functions, map() and list comprehensions... etc.).
def __iter__(self):
ptr = self
while ptr is not None:
yield ptr.value
ptr = ptr.next

Your get_min function contains the following mistakes:
the base case should be if ll.next == None and not if ll == None. Indeed the minimum of the empty list is not well-defined. But if ll.next is None it means that your list only contains one item. In that case the minimum of the list is the item itself, i.e. ll.value
when the list has more than one element, the minimum of the list can be obtained by comparing the first element of the list (ll.value) to the minimum of the remaining list starting at ll.next (the tail of the list).
finally it is a better practice to use the is operator to test if a Python variable is None.
A working code could be the following:
def get_min(ll):
if ll.next is None:
return ll.value
else:
tail_min = get_min(ll.next)
if ll.value < tail_min:
return ll.value
else:
return tail_min
If you can use the min function to compare two numbers, a more concise version is:
def get_min(ll):
if ll.next is None:
return ll.value
else:
return min(ll.value, get_min(ll.next))
Finally, you could raise an exception when the list is empty to warn the user of the function that he is using it in a non-applicable case:
def get_min(ll):
if ll is None:
raise ValueError("Cannot compute the minimum of the empty list.")
elif ll.next is None:
return ll.value
else:
return min(ll.value, get_min(ll.next))

I have done this in python.
Here fist we created a circular link list and then we
print minNode() value and maxNode() value.
#Represents the node of list.
class Node:
def __init__(self,data):
self.data = data;
self.next = None;
class CreateList:
#Declaring head and tail pointer as null.
def __init__(self):
self.head = Node(None);
self.tail = Node(None);
self.head.next = self.tail;
self.tail.next = self.head;
#This function will add the new node at the end of the list.
def add(self,data):
newNode = Node(data);
#Checks if the list is empty.
if self.head.data is None:
#If list is empty, both head and tail would point to new node.
self.head = newNode;
self.tail = newNode;
newNode.next = self.head;
else:
#tail will point to new node.
self.tail.next = newNode;
#New node will become new tail.
self.tail = newNode;
#Since, it is circular linked list tail will point to head.
self.tail.next = self.head;
#Finds out the minimum value node in the list
def minNode(self):
current = self.head;
#Initializing min to initial node data
minimum = self.head.data;
if(self.head == None):
print("List is empty");
else:
while(True):
#If current node's data is smaller than min
#Then replace value of min with current node's data
if(minimum > current.data):
minimum = current.data;
current= current.next;
if(current == self.head):
break;
print("Minimum value node in the list: "+ str(minimum));
#Finds out the maximum value node in the list
def maxNode(self):
current = self.head;
#Initializing max to initial node data
maximum = self.head.data;
if(self.head == None):
print("List is empty");
else:
while(True):
#If current node's data is greater than max
#Then replace value of max with current node's data
if(maximum < current.data):
maximum = current.data;
current= current.next;
if(current == self.head):
break;
print("Maximum value node in the list: "+ str(maximum));
class CircularLinkedList:
cl = CreateList();
#Adds data to the list
cl.add(5);
cl.add(20);
cl.add(10);
cl.add(1);
#Prints the minimum value node in the list
cl.minNode();
#Prints the maximum value node in the list
cl.maxNode();
Source:coderforevers.com

Creating a tree/deeply nested dict from an indented text file in python

Basically, I want to iterate through a file and put the contents of each line into a deeply nested dict, the structure of which is defined by the amount of whitespace at the start of each line.
Essentially the aim is to take something like this:
a
b
c
d
e
And turn it into something like this:
{"a":{"b":"c","d":"e"}}
Or this:
apple
colours
red
yellow
green
type
granny smith
price
0.10
into this:
{"apple":{"colours":["red","yellow","green"],"type":"granny smith","price":0.10}
So that I can send it to Python's JSON module and make some JSON.
At the moment I'm trying to make a dict and a list in steps like such:
{"a":""} ["a"]
{"a":"b"} ["a"]
{"a":{"b":"c"}} ["a","b"]
{"a":{"b":{"c":"d"}}}} ["a","b","c"]
{"a":{"b":{"c":"d"},"e":""}} ["a","e"]
{"a":{"b":{"c":"d"},"e":"f"}} ["a","e"]
{"a":{"b":{"c":"d"},"e":{"f":"g"}}} ["a","e","f"]
etc.
The list acts like 'breadcrumbs' showing where I last put in a dict.
To do this I need a way to iterate through the list and generate something like dict["a"]["e"]["f"] to get at that last dict. I've had a look at the AutoVivification class that someone has made which looks very useful however I'm really unsure of:
Whether I'm using the right data structure for this (I'm planning to send it to the JSON library to create a JSON object)
How to use AutoVivification in this instance
Whether there's a better way in general to approach this problem.
I came up with the following function but it doesn't work:
def get_nested(dict,array,i):
if i != None:
i += 1
if array[i] in dict:
return get_nested(dict[array[i]],array)
else:
return dict
else:
i = 0
return get_nested(dict[array[i]],array)
Would appreciate help!
(The rest of my extremely incomplete code is here:)
#Import relevant libraries
import codecs
import sys
#Functions
def stripped(str):
if tab_spaced:
return str.lstrip('\t').rstrip('\n\r')
else:
return str.lstrip().rstrip('\n\r')
def current_ws():
if whitespacing == 0 or not tab_spaced:
return len(line) - len(line.lstrip())
if tab_spaced:
return len(line) - len(line.lstrip('\t\n\r'))
def get_nested(adict,anarray,i):
if i != None:
i += 1
if anarray[i] in adict:
return get_nested(adict[anarray[i]],anarray)
else:
return adict
else:
i = 0
return get_nested(adict[anarray[i]],anarray)
#initialise variables
jsondict = {}
unclosed_tags = []
debug = []
vividfilename = 'simple.vivid'
# vividfilename = sys.argv[1]
if len(sys.argv)>2:
jsfilename = sys.argv[2]
else:
jsfilename = vividfilename.split('.')[0] + '.json'
whitespacing = 0
whitespace_array = [0,0]
tab_spaced = False
#open the file
with codecs.open(vividfilename,'rU', "utf-8-sig") as vividfile:
for line in vividfile:
#work out how many whitespaces at start
whitespace_array.append(current_ws())
#For first line with whitespace, work out the whitespacing (eg tab vs 4-space)
if whitespacing == 0 and whitespace_array[-1] > 0:
whitespacing = whitespace_array[-1]
if line[0] == '\t':
tab_spaced = True
#strip out whitespace at start and end
stripped_line = stripped(line)
if whitespace_array[-1] == 0:
jsondict[stripped_line] = ""
unclosed_tags.append(stripped_line)
if whitespace_array[-2] < whitespace_array[-1]:
oldnested = get_nested(jsondict,whitespace_array,None)
print oldnested
# jsondict.pop(unclosed_tags[-1])
# jsondict[unclosed_tags[-1]]={stripped_line:""}
# unclosed_tags.append(stripped_line)
print jsondict
print unclosed_tags
print jsondict
print unclosed_tags

Here is an object oriented approach based on a composite structure of nested Node objects.
Input:
indented_text = \
"""
apple
colours
red
yellow
green
type
granny smith
price
0.10
"""
a Node class
class Node:
def __init__(self, indented_line):
self.children = []
self.level = len(indented_line) - len(indented_line.lstrip())
self.text = indented_line.strip()
def add_children(self, nodes):
childlevel = nodes[0].level
while nodes:
node = nodes.pop(0)
if node.level == childlevel: # add node as a child
self.children.append(node)
elif node.level > childlevel: # add nodes as grandchildren of the last child
nodes.insert(0,node)
self.children[-1].add_children(nodes)
elif node.level <= self.level: # this node is a sibling, no more children
nodes.insert(0,node)
return
def as_dict(self):
if len(self.children) > 1:
return {self.text: [node.as_dict() for node in self.children]}
elif len(self.children) == 1:
return {self.text: self.children[0].as_dict()}
else:
return self.text
To parse the text, first create a root node.
Then, remove empty lines from the text, and create a Node instance for every line, pass this to the add_children method of the root node.
root = Node('root')
root.add_children([Node(line) for line in indented_text.splitlines() if line.strip()])
d = root.as_dict()['root']
print(d)
result:
{'apple': [
{'colours': ['red', 'yellow', 'green']},
{'type': 'granny smith'},
{'price': '0.10'}]
}
I think that it should be possible to do it in one step, where you simply call the constructor of Node once, with the indented text as an argument.

Here is a recursive solution. First, transform the input in the following way.
Input:
person:
address:
street1: 123 Bar St
street2:
city: Madison
state: WI
zip: 55555
web:
email: boo#baz.com
First-step output:
[{'name':'person','value':'','level':0},
{'name':'address','value':'','level':1},
{'name':'street1','value':'123 Bar St','level':2},
{'name':'street2','value':'','level':2},
{'name':'city','value':'Madison','level':2},
{'name':'state','value':'WI','level':2},
{'name':'zip','value':55555,'level':2},
{'name':'web','value':'','level':1},
{'name':'email','value':'boo#baz.com','level':2}]
This is easy to accomplish with split(':') and by counting the number of leading tabs:
def tab_level(astr):
"""Count number of leading tabs in a string
"""
return len(astr)- len(astr.lstrip('\t'))
Then feed the first-step output into the following function:
def ttree_to_json(ttree,level=0):
result = {}
for i in range(0,len(ttree)):
cn = ttree[i]
try:
nn = ttree[i+1]
except:
nn = {'level':-1}
# Edge cases
if cn['level']>level:
continue
if cn['level']<level:
return result
# Recursion
if nn['level']==level:
dict_insert_or_append(result,cn['name'],cn['value'])
elif nn['level']>level:
rr = ttree_to_json(ttree[i+1:], level=nn['level'])
dict_insert_or_append(result,cn['name'],rr)
else:
dict_insert_or_append(result,cn['name'],cn['value'])
return result
return result
where:
def dict_insert_or_append(adict,key,val):
"""Insert a value in dict at key if one does not exist
Otherwise, convert value to list and append
"""
if key in adict:
if type(adict[key]) != list:
adict[key] = [adict[key]]
adict[key].append(val)
else:
adict[key] = val

The following code will take a block-indented file and convert into an XML tree; this:
foo
bar
baz
ban
bal
...becomes:
<cmd>foo</cmd>
<cmd>bar</cmd>
<block>
<name>baz</name>
<cmd>ban</cmd>
<cmd>bal</cmd>
</block>
The basic technique is:
Set indent to 0
For each line, get the indent
If > current, step down and save current block/ident on a stack
If == current, append to current block
If < current, pop from the stack until you get to the matching indent
So:
from lxml import builder
C = builder.ElementMaker()
def indent(line):
strip = line.lstrip()
return len(line) - len(strip), strip
def parse_blockcfg(data):
top = current_block = C.config()
stack = []
current_indent = 0
lines = data.split('\n')
while lines:
line = lines.pop(0)
i, line = indent(line)
if i==current_indent:
pass
elif i > current_indent:
# we've gone down a level, convert the <cmd> to a block
# and then save the current ident and block to the stack
prev.tag = 'block'
prev.append(C.name(prev.text))
prev.text = None
stack.insert(0, (current_indent, current_block))
current_indent = i
current_block = prev
elif i < current_indent:
# we've gone up one or more levels, pop the stack
# until we find out which level and return to it
found = False
while stack:
parent_indent, parent_block = stack.pop(0)
if parent_indent==i:
found = True
break
if not found:
raise Exception('indent not found in parent stack')
current_indent = i
current_block = parent_block
prev = C.cmd(line)
current_block.append(prev)
return top

First of all, don't use array and dict as variable names because they're reserved words in Python and reusing them may end up in all sorts of chaos.
OK so if I get you correctly, you have a tree given in a text file, with parenthood indicated by indentations, and you want to recover the actual tree structure. Right?
Does the following look like a valid outline? Because I have trouble putting your current code into context.
result = {}
last_indentation = 0
for l in f.xreadlines():
(c, i) = parse(l) # create parse to return character and indentation
if i==last_indentation:
# sibling to last
elif i>last_indentation:
# child to last
else:
# end of children, back to a higher level
OK then your list are the current parents, that's in fact right - but I'd keep them pointed to the dictionary you've created, not the literal letter
just starting some stuff here
result = {}
parents = {}
last_indentation = 1 # start with 1 so 0 is the root of tree
parents[0] = result
for l in f.xreadlines():
(c, i) = parse(l) # create parse to return character and indentation
if i==last_indentation:
new_el = {}
parents[i-1][c] = new_el
parents[i] = new_el
elif i>last_indentation:
# child to last
else:
# end of children, back to a higher level