how would one go about using recursion in order to take a list of random values and make it a linked list? Where each value is a node. As of right now, i've tried implementing the following...
def pyListToMyList(pylst):
lists = mkMyList()
lists.head = pyListToMyListRec(pylst)
return lists
def pyListToMyList(pylst):
if pylst:
return mkEmptyNode()
else:
return mkNode(pylst[0], pyLstToMyListRec(pylst[1:]))
The problem is the the else statement which returns an error saying that the index is out of range.
def pyListToMyList(pylst):
if not pylst:
return mkEmptyNode()
else:
return mkNode(pylst[0], pyLstToMyListRec(pylst[1:]))
EDIT: Though this is O(n^2) because of all the list copying.
I would do
def pyListToMyList(pylst, i=0):
if i > len(pylst):
return mkEmptyNode()
else:
return mkNode(pylst[i], pyLstToMyListRec(pylst, i+1))
or even more efficient and less likely to overflow stack (though this does not use recursion):
def pyListToMyList(pylst):
lst = mkEmptyNode()
for x in reversed(pylist):
lst = mkNode(x, lst)
return lst
Related
Now I have the source code above:
class Stats(object):
def __init__(self):
self._pending = []
self._done = []
#property
def pending(self):
return self._pending
The way those lists are filled is not important for my question.
The situation is that I'm getting a sublist of these lists this way:
stats = Stats()
// code to fill the lists
stats.pending[2:10]
The problem here is that I expect to get as many elements as I retrieved.
In the example above I expect a sublist that contains 8 elements (10-2).
Of course, actually I'll get less than 8 elements if the list is shorter.
So, what I need is:
When the list has enough items, it returns the corresponding sublist.
When the list is shorter, it returns a sublist with the expected length, filled with the last elements of the original lists and a default value (for example None) for the extra items.
This way, if I did:
pending_tasks = stats.pending[44:46]
And the pending list only contains 30 elements, it should returns a list of two default elements, for example: [None, None]; instead of an empty list ([]) which is the default behaviour of the lists.
I guess I already know how to do it inside a normal method/function, but I want to do it in the most clean way, trying to follow the #property approach, if possible.
Thanks a lot!
This is not easy to do because the slicing operation is what you want to modify, and that happens after the original list has been returned by the property. It's not impossible though, you'll just need to wrap the regular list with another object that will take care of padding the slices for you. How easy or difficult that will be may depend on how much of the list interface you need your wrapper to implement. If you only need indexing and slicing, it's really easy:
class PadSlice(object):
def __init__(self, lst, default_value=None):
self.lst = lst
self.default_value
def __getitem__(self, index):
item = getitem(self.lst, index)
if isinstance(index, slice):
expected_length = (index.stop - index.start) // (index.step or 1)
if len(item) != expected_length:
item.extend([default_value] * (expected_length - len(item)))
return item
This code probably won't work right for negative step slices, or for slices that don't specify one of the end points (it does have logic to detect an omitted step, since that's common). If this was important to you, you could probably fix up those corner cases.
This is not easy. How would the object (list) you return know how it will be sliced later? You could subclass list, however, and override __getitem__ and __getslice__ (Python2 only):
class L(list):
def __getitem__(self, key):
if isinstance(key, slice):
return [list(self)[i] if 0 <= i < len(self) else None for i in xrange(key.start, key.stop, key.step or 1)]
return list(self)[key]
def __getslice__(self, i, j):
return self.__getitem__(slice(i, j))
This will pad all slices with None, fully compatible with negative indexing and steps != 1. And in your property, return an L version of the actual list:
#property
def pending(self):
return L(self._pending)
You can construct a new class, which is a subclass of list. Then you can overload the __getitem__ magic method to overload [] operator to the appropriate behavior. Consider this subclass of list called MyList:
class MyList(list):
def __getitem__(self, index):
"""Modify index [] operator"""
result = super(MyList, self).__getitem__(index)
if isinstance(index, slice):
# Get sublist length.
if index.step: # Check for zero to avoid divide by zero error
sublist_len = (index.stop - index.start) // index.step
else:
sublist_len = (index.stop - index.start)
# If sublist length is greater (or list is shorter), then extend
# the list to length requested with default value of None
if sublist_len > len(self) or index.start > len(self):
result.extend([None for _ in range(sublist_len - len(result))])
return result
Then you can just change the pending method to return a MyList type instead of list.
class Stats(object):
#property
def pending(self):
return MyList(self._pending)
Hopefully this helps.
I'm attempting to generate all n choose k combinations of a list (not checking for uniqueness) recursively by following the strategy of either include or not include an element for each recursive call. I can definitely print out the combinations but I for the life of me cannot figure out how to return the correct list in Python. Here are some attempts below:
class getCombinationsClass:
def __init__(self,array,k):
#initialize empty array
self.new_array = []
for i in xrange(k):
self.new_array.append(0)
self.final = []
self.combinationUtil(array,0,self.new_array,0,k)
def combinationUtil(self,array,array_index,current_combo, current_combo_index,k):
if current_combo_index == k:
self.final.append(current_combo)
return
if array_index >= len(array):
return
current_combo[current_combo_index] = array[array_index]
#if current item included
self.combinationUtil(array,array_index+1,current_combo,current_combo_index+1,k)
#if current item not included
self.combinationUtil(array,array_index+1,current_combo,current_combo_index,k)
In the above example I tried to append the result to an external list which didn't seem to work. I also tried implementing this by recursively constructing a list which is finally returned:
def getCombinations(array,k):
#initialize empty array
new_array = []
for i in xrange(k):
new_array.append(0)
return getCombinationsUtil(array,0,new_array,0,k)
def getCombinationsUtil(array,array_index,current_combo, current_combo_index,k):
if current_combo_index == k:
return [current_combo]
if array_index >= len(array):
return []
current_combo[current_combo_index] = array[array_index]
#if current item included & not included
return getCombinationsUtil(array,array_index+1,current_combo,current_combo_index+1,k) + getCombinationsUtil(array,array_index+1,current_combo,current_combo_index,k)
When I tested this out for the list [1,2,3] and k = 2, for both implementations, I kept getting back the result [[3,3],[3,3],[3,3]]. However, if I actually print out the 'current_combo' variable within the inner (current_combo_index == k) if statement, the correct combinations print out. What gives? I am misunderstanding something to do with variable scope or Python lists?
The second method goes wrong because the line
return [current_combo]
returns a reference to current_combo. At the end of the program, all the combinations returned are references to the same current_combo.
You can fix this by making a copy of the current_combo by changing the line to:
return [current_combo[:]]
The first method fails for the same reason, you need to change:
self.final.append(current_combo)
to
self.final.append(current_combo[:])
Check this out: itertools.combinations. You can take a look at the implementation as well.
So for an assignment I have to create a bunch of different functions in one Python file. One of the functions calls for inputting a list (sorted_list) and a string from that list (item). What the function does is reads the list and removes any duplicates of the specified string from the list.
def remove_duplicates(sorted_list, item):
list_real = []
for x in range(len(sorted_list)-1):
if(sorted_list[i] == item and sorted_list[i+1] == item):
list_real = list_real + [item]
i+1
else:
if(sorted_list[i] != item):
list_real = list_real + [sorted_list[i]]
i+=1
return list_real
So
remove_duplicates(['a','a','a','b','b','c'] 'a') would return ['a','b','b','c']
This probably isn't the most efficient way to do something like this, but that isn't my problem.
The next function I have to define is similar to the one above except it only takes sorted_list and it has to remove duplicates for each item instead of a specified one. The only thing I know is that you have to use a for loop that makes the remove_duplicates run for each item in a given list, but I have no idea how to actually implement a function inside of another function. Can anyone help me out?
This works nicely:
from itertools import ifilterfalse
def remove_duplicates(sorted_list, item):
idx = sorted_list.index(item)
list_real = sorted_list[:idx+1]
if len(list_real) != len(sorted_list):
for item in ifilterfalse (lambda x: x is item, sorted_list[idx:]):
list_real.append(item)
return list_real
I am supposed to write the sorting function. Define
a function sort list that takes in a list and returns a list sorted in ascending order.
(4 marks)
Note: your code should make use of the insert list function defined earlier.
With my previous defined variables:
def search(x , seq):
for i in seq:
if x > max(seq):
return len(seq)
elif x < min(seq):
return 0
return search(x, seq[:-1])
def insert_tup(x, tup):
searchtuple = search(x, tup)
ontheleft = tup[:searchtuple]
ontheright = tup[ searchtuple:]
altogether = ontheleft + (x,) + ontheright
return altogether
def insert_list(x, lst):
A = search(x, lst)
lst.insert(A, x)
return lst
I did a trial
def sort_list(lst):
result = []
for i in lst:
result += [i]
return insert_list(result[0], lst)
But i keep getting Runtime error where
File "", line 7, in search
return search(x, seq[:-1])
The output is desired to be
sort_list([5, 1, 9])
[1, 5, 9]
You are recursively calling the search() method and never breaking out of this loop. I expect this is raising a RuntimeError: maximum recursion depth exceeded exception.
Python has a recursion limit to protect against infinite recursion. FYI you can modify the recursion limit via sys.setrecursionlimit - but this is not something you should be doing.
Instead you can fix this by re-writing your algorithm iteratively.
I also suggest you read the Python sorting wiki, as this demonstrates the pythonic way to sort things with the built in sorted() function and the list method sort(). You don't need to re-invent the wheel.
You need a base case for the recursion. Something like:
if len(seq)==1:
return seq
Otherwise it calls itself forever. And the result is the error that you see.
I am trying to run a sorting function recursively in python. I have an empty list that starts everything but everytime I try to print the list I get an empty list. here is my code. Any help would be greatly appreciated
def parse(list):
newParse = []
if len(list) == 0:
return newParse
else:
x = min(list)
list.remove(x)
newParse.append(x)
return sort(list)
The value of newParse is not preserved between invocations of the function; you're setting it equal to [] (well, you're creating a new variable with the value []).
Since the only time you return is
newParse = []
if len(list) == 0:
return newParse`
you will always be returning [] because that is the value of newParse at that time.
Because you are doing this recursively, you are calling the function anew, without keeping the function's own state. Take a moment to consider the implications of this on your code.
Instead of initialising newParse = [], add an optional parameter newParse defaulting to a bogus value, and set newParse = [] if you receive that bogus value for newParse. Otherwise, you'll actually be getting the same list every time (i.e. the contents of the list object are being mutated). And newParse through in your tail call.
You also seem to have the problem that your definition and and the supposedly-recursive call refer to different functions.
def sort(list, newParse = None):
if newParse is None:
newParse = []
if len(list) == 0:
return newParse
else:
x = min(list)
list.remove(x)
newParse.append(x)
return sort(list, newParse)
Here is what I think you are trying to do:
def recursive_sort(a_list):
def helper_function(list_to_be_sorted, list_already_sorted):
new = []
if len(list_to_be_sorted) == 0:
return list_already_sorted
else:
x = min(list_to_be_sorted)
list_to_be_sorted.remove(x)
new.append(x)
return helper_function(list_to_be_sorted, list_already_sorted + new)
return helper_function(a_list, [])
You shouldn't name variables list, as that is a builtin.
Also, if you are trying to implement a recursive sort function, you might want to look at quicksort, which is a very common (and fast) recursive sorting algorithm. What you have tried to implement is a recursive version of selection sort, which is much slower.
Also, if you actually need a sorting function, rather than just wanting to implement a recursive one, you should use the list method sort, or the function on an iterable sorted, both of which will be a lot faster than anything you could make in Python.