Develop Reference

Any alternate to slice sling of integers?

I am attempting to make a recursive function that adds the two last numbers until there are none left. For example:
sumDigits(239)
would equate to:
2+3+9=14
It is difficult because the input must be an integer, which cannot be sliced without converting it. I decided to try to turn it into a lists because I thought the pop() method would be useful for this. It appears as though this approach is not working. Any suggestions?
EXECUTION:
>>> sumDigits('234')
9
>>> sumDigits('2343436432424')
8
>>>
CODE:
def sumDigits(n):
l1 = list(str(n))
if len(l1) > 1:
a = int(l1.pop())
b = int(l1.pop())
l1.append(str(a+b))
return sumDigits(int(''.join(l1)))
else:
return n

With functional tools like reduce() the problem is solved by
from functools import reduce
def sumDigits(n):
return reduce((lambda x, y: int(x) + int(y)), list(str(n)))

Instead of passing a string you should pass a list of integers:
def sumDigits(l1):
if len(l1) > 1:
a = l1.pop()
b = l1.pop()
l1.append(a+b)
return sumDigits(l1)
else:
return l1[0]
print sumDigits([2,3, 4])
print sumDigits([2, 3, 4, 3, 4, 3, 6, 4, 3, 2, 4, 2, 4])
The problem with your approach is that:
'23434364324|24|' -> '2343436432|46|' -> '2343436432 | 10',
here now pop will return 0 and 1, instead of 2 and 10 as you would've expected. Hence the wrong output.
Simple solution:
>>> s = '2343436432424'
>>> sum(int(x) for x in s)
44

Since everybody seems intent on solving your homework for you, here's the elegant recursive solution.
def sumDigits(n):
if n < 10:
return n
return n % 10 + sumDigits(n / 10)

EDIT
The simple solution is:
def sumDigits(n):
return sum(int(i) for i in str(n))
Upon your answer of my comment the below solution is not applicable.
def sumDigits(n):
n = [int(i) for i in str(n)]
return sumDigitsRec(n)
def sumDigitsRec(li):
if len(li) > 1:
li[-1] += li.pop()
return sumDigits(''.join(str(i) for i in li))
else:
return li[0]

As strings:
def sumDigits(n):
answer = 0
for num in n:
answer += int(num)
return answer
Without slicing, using only an integer input:
def sumDigits(n):
answer = 0
while n:
answer += n%10
n /= 10
return answer

If I understand your question correctly, you want to stop summing the digits when the sum is a 2 digit number. I think the bug in your program is that you need if len(l1) > 2: not if len(l1) > 1: to make sure you don't recurse when you have just 2 digits.

you can do it recursively in this way:
def sumDigits(n,s=0):
if len(n) == 0:
return s
else:
return sumDigits(n[:-1],s+int(n[-1]))
if you want simpler and pytonic way (not recursive) you can do this
>>> s = 2343436432424
>>> sum(map(int,str(s)))
44

All the solutions provided failed to meet the prereqs which were stated in the problem description. This is the correct answer:
Use **kwargs and exceptions to utilize holding variables in recursion function.
For example:
def recursionFunc(x, **kwargs):
try:
count = kwargs['count']
except:
count = 0
#Code down here to add num from x to count
#remove last index on every iteration from x
#etc
return recursionFunc(x, count = count)

It's working fine, as the following check will show:
>>> 2343436432424 % 9
8
If you didn't mean for it to be called recursively then just don't do so, i.e. stop checking the length and just return the sum.

Find the nth lucky number generated by a sieve in Python

I'm trying to make a program in Python which will generate the nth lucky number according to the lucky number sieve. I'm fairly new to Python so I don't know how to do all that much yet. So far I've figured out how to make a function which determines all lucky numbers below a specified number:
def lucky(number):
l = range(1, number + 1, 2)
i = 1
while i < len(l):
del l[l[i] - 1::l[i]]
i += 1
return l
Is there a way to modify this so that I can instead find the nth lucky number? I thought about increasing the specified number gradually until a list of the appropriate length to find the required lucky number was created, but that seems like a really inefficient way of doing it.
Edit: I came up with this, but is there a better way?
def lucky(number):
f = 2
n = number * f
while True:
l = range(1, n + 1, 2)
i = 1
while i < len(l):
del l[l[i] - 1::l[i]]
i += 1
if len(l) >= number:
return l[number - 1]
f += 1
n = number * f

I came up with this, but is there a better way?
Truth is, there will always be a better way, the remaining question being: is it good enough for your need?
One possible improvement would be to turn all this into a generator function. That way, you would only compute new values as they are consumed. I came up with this version, which I only validated up to about 60 terms:
import itertools
def _idx_after_removal(removed_indices, value):
for removed in removed_indices:
value -= value / removed
return value
def _should_be_excluded(removed_indices, value):
for j in range(len(removed_indices) - 1):
value_idx = _idx_after_removal(removed_indices[:j + 1], value)
if value_idx % removed_indices[j + 1] == 0:
return True
return False
def lucky():
yield 1
removed_indices = [2]
for i in itertools.count(3, 2):
if not _should_be_excluded(removed_indices, i):
yield i
removed_indices.append(i)
removed_indices = list(set(removed_indices))
removed_indices.sort()
If you want to extract for example the 100th term from this generator, you can use itertools nth recipe:
def nth(iterable, n, default=None):
"Returns the nth item or a default value"
return next(itertools.islice(iterable, n, None), default)
print nth(lucky(), 100)
I hope this works, and there's without any doubt more room for code improvement (but as stated previously, there's always room for improvement!).

With numpy arrays, you can make use of boolean indexing, which may help. For example:
>>> a = numpy.arange(10)
>>> print a
[0 1 2 3 4 5 6 7 8 9]
>>> print a[a > 3]
[4 5 6 7 8 9]
>>> mask = np.array([True, False, True, False, True, False, True, False, True, False])
>>> print a[mask]
[0 2 4 6 8]
Here is a lucky number function using numpy arrays:
import numpy as np
class Didnt_Findit(Exception):
pass
def lucky(n):
'''Return the nth lucky number.
n --> int
returns int
'''
# initial seed
lucky_numbers = [1]
# how many numbers do you need to get to n?
candidates = np.arange(1, n*100, 2)
# use numpy array boolean indexing
next_lucky = candidates[candidates > lucky_numbers[-1]][0]
# accumulate lucky numbers till you have n of them
while next_lucky < candidates[-1]:
lucky_numbers.append(next_lucky)
#print lucky_numbers
if len(lucky_numbers) == n:
return lucky_numbers[-1]
mask_start = next_lucky - 1
mask_step = next_lucky
mask = np.array([True] * len(candidates))
mask[mask_start::mask_step] = False
#print mask
candidates = candidates[mask]
next_lucky = candidates[ candidates > lucky_numbers[-1]][0]
raise Didnt_Findit('n = ', n)
>>> print lucky(10)
33
>>> print lucky(50)
261
>>> print lucky(500)
3975
Checked mine and #icecrime's output for 10, 50 and 500 - they matched.
Yours is much faster than mine and scales better with n.

n=input('enter n ')
a= list(xrange(1,n))
x=a[1]
for i in range(1,n):
del a[x-1::x]
x=a[i]
l=len(a)
if i==l-1:
break
print "lucky numbers till %d" % n
print a
lets do this with an example.lets print lucky numbers till 100
put n=100
firstly a=1,2,3,4,5....100
x=a[1]=2
del a[1::2] leaves
a=1,3,5,7....99
now l=50
and now x=3
then del a[2::3] leaving a =1,3,7,9,13,15,.....
and loop continues till i==l-1

looping through loops in python?

I'm trying to solve this problem on the easy section of coderbyte and the prompt is:
Have the function ArrayAdditionI(arr) take the array of numbers stored in arr and return the string true if any combination of numbers in the array can be added up to equal the largest number in the array, otherwise return the string false. For example: if arr contains [4, 6, 23, 10, 1, 3] the output should return true because 4 + 6 + 10 + 3 = 23. The array will not be empty, will not contain all the same elements, and may contain negative numbers.
Here's my solution.
def ArrayAddition(arr):
arr = sorted(arr, reverse=True)
large = arr.pop(0)
storage = 0
placeholder = 0
for r in range(len(arr)):
for n in arr:
if n + storage == large: return True
elif n + storage < large: storage += n
else: continue
storage = 0
if placeholder == 0: placeholder = arr.pop(0)
else: arr.append(placeholder); placeholder = arr.pop(0)
return False
print ArrayAddition([2,95,96,97,98,99,100])
I'm not even sure if this is correct, but it seems to cover all the numbers I plug in. I'm wondering if there is a better way to solve this through algorithm which I know nothing of. I'm thinking a for within a for within a for, etc loop would do the trick, but I don't know how to do that.
What I have in mind is accomplishing this with A+B, A+C, A+D ... A+B+C ... A+B+C+D+E
e.g)
for i in range(len(arr):
print "III: III{}III".format(i)
storage = []
for j in range(len(arr):
print "JJ: II({}),JJ({})".format(i,j)
for k in range(len(arr):
print "K: I{}, J{}, K{}".format(i,j,k)
I've searched all over and found the suggestion of itertool, but I'm wondering if there is a way to write this code up more raw.
Thanks.

A recursive solution:
def GetSum(n, arr):
if len(arr) == 0 and n != 0:
return False
return (n == 0 or
GetSum(n, arr[1:]) or
GetSum(n-arr[0], arr[1:]))
def ArrayAddition(arr):
arrs = sorted(arr)
return GetSum(arrs[-1], arrs[:-1])
print ArrayAddition([2,95,96,97,98,99,100])
The GetSum function returns False when the required sum is non-zero and there are no items in the array. Then it checks for 3 cases:
If the required sum, n, is zero then the goal is achieved.
If we can get the sum with the remaining items after the first item is removed, then the goal is achieved.
If we can get the required sum minus the first element of the list on the rest of the list the goal is achieved.

Your solution doesn't work.
>>> ArrayAddition([10, 11, 20, 21, 30, 31, 60])
False
The simple solution is to use itertools to iterate over all subsets of the input (that don't contain the largest number):
def subsetsum(l):
l = list(l)
target = max(l)
l.remove(l)
for subset_size in xrange(1+len(l)):
for subset in itertools.combinations(l, subset_size):
if sum(subset) == target:
return True
return False
If you want to avoid itertools, you'll need to generate subsets directly. That can be accomplished by counting in binary and using the set bits to determine which elements to pick:
def subsetsum(l):
l = list(l)
target = max(l)
l.remove(l)
for subset_index in xrange(2**len(l)):
subtotal = 0
for i, num in enumerate(l):
# If bit i is set in subset_index
if subset_index & (1 << i):
subtotal += num
if subtotal == target:
return True
return False

Update: I forgot that you want to check all possible combinations. Use this instead:
def ArrayAddition(l):
for length in range(2, len(l)):
for lst in itertools.combinations(l, length):
if sum(lst) in l:
print(lst, sum(lst))
return True
return False
One-liner solution:
>>> any(any(sum(lst) in l for lst in itertools.combinations(l, length)) for length in range(2, len(l)))
Hope this helps!

Generate all the sums of the powerset and test them against the max
def ArrayAddition(L):
return any(sum(k for j,k in enumerate(L) if 1<<j&i)==max(L) for i in range(1<<len(L)))
You could improve this by doing some preprocessing - find the max first and remove it from L

One more way to do it...
Code:
import itertools
def func(l):
m = max(l)
rem = [itertools.combinations([x for x in l if not x == m],i) for i in range(2,len(l)-1)]
print [item for i in rem for item in i if sum(item)==m ]
if __name__=='__main__':
func([1,2,3,4,5])
Output:
[(1, 4), (2, 3)]
Hope this helps.. :)

If I understood the question correctly, simply this should return what you want:
2*max(a)<=sum(a)

counting odd numbers in a list python

This is a part of my homework assignment and im close to the final answer but not quite yet. I need to write a function that counts odd numbers in a list.
Create a recursive function count_odd(l) which takes as its only argument a list of integers. The function will return a count of the number of list elements that are odd, i.e., not evenly divisible by 2.\
>>> print count_odd([])
0
>>> print count_odd([1, 3, 5])
3
>>> print count_odd([2, 4, 6])
0
>>> print count_odd([0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144])
8
Here is what i have so far:
#- recursive function count_odd -#
def count_odd(l):
"""returns a count of the odd integers in l.
PRE: l is a list of integers.
POST: l is unchanged."""
count_odd=0
while count_odd<len(l):
if l[count_odd]%2==0:
count_odd=count_odd
else:
l[count_odd]%2!=0
count_odd=count_odd+1
return count_odd
#- test harness
print count_odd([])
print count_odd([1, 3, 5])
print count_odd([2, 4, 6])
print count_odd([0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144])
Can u help explain what im missing. The first two test harness works fine but i cant get the final two. Thanks!

Since this is homework, consider this pseudo-code that just counts a list:
function count (LIST)
if LIST has more items
// recursive case.
// Add one for the current item we are counting,
// and call count() again to process the *remaining* items.
remaining = everything in LIST except the first item
return 1 + count(remaining)
else
// base case -- what "ends" the recursion
// If an item is removed each time, the list will eventually be empty.
return 0
This is very similar to what the homework is asking for, but it needs to be translate to Python and you must work out the correct recursive case logic.
Happy coding.

def count_odd(L):
return (L[0]%2) + count_odd(L[1:]) if L else 0

Are slices ok? Doesn't feel recursive to me, but I guess the whole thing is kind of against usual idioms (i.e. - recursion of this sort in Python):
def countOdd(l):
if l == list(): return 0 # base case, empty list means we're done
return l[0] % 2 + countOdd(l[1:]) # add 1 (or don't) depending on odd/even of element 0. recurse on the rest
x%2 is 1 for odds, 0 for evens. If you are uncomfortable with it or just don't understand it, use the following in place of the last line above:
thisElement = l[0]
restOfList = l[1:]
if thisElement % 2 == 0: currentElementOdd = 0
else: currentElementOdd = 1
return currentElementOdd + countOdd(restOfList)
PS - this is pretty recursive, see what your teacher says if you turn this in =P
>>> def countOdd(l):
... return fold(lambda x,y: x+(y&1),l,0)
...
>>> def fold(f,l,a):
... if l == list(): return a
... return fold(f,l[1:],f(a,l[0]))

All of the prior answers are subdividing the problem into subproblems of size 1 and size n-1. Several people noted that the recursive stack might easily blow out. This solution should keep the recursive stack size at O(log n):
def count_odd(series):
l = len(series) >> 1
if l < 1:
return series[0] & 1 if series else 0
else:
return count_odd(series[:l]) + count_odd(series[l:])

The goal of recursion is to divide the problem into smaller pieces, and apply the solution to the smaller pieces. In this case, we can check if the first number of the list (l[0]) is odd, then call the function again (this is the "recursion") with the rest of the list (l[1:]), adding our current result to the result of the recursion.

def count_odd(series):
if not series:
return 0
else:
left, right = series[0], series[1:]
return count_odd(right) + (1 if (left & 1) else 0)

Tail recursion
def count_odd(integers):
def iter_(lst, count):
return iter_(rest(lst), count + is_odd(first(lst))) if lst else count
return iter_(integers, 0)
def is_odd(integer):
"""Whether the `integer` is odd."""
return integer % 2 != 0 # or `return integer & 1`
def first(lst):
"""Get the first element from the `lst` list.
Return `None` if there are no elements.
"""
return lst[0] if lst else None
def rest(lst):
"""Return `lst` list without the first element."""
return lst[1:]
There is no tail-call optimization in Python, so the above version is purely educational.
The call could be visualize as:
count_odd([1,2,3]) # returns
iter_([1,2,3], 0) # could be replaced by; depth=1
iter_([2,3], 0 + is_odd(1)) if [1,2,3] else 0 # `bool([1,2,3])` is True in Python
iter_([2,3], 0 + True) # `True == 1` in Python
iter_([2,3], 1) # depth=2
iter_([3], 1 + is_odd(2)) if [2,3] else 1
iter_([3], 1 + False) # `False == 0` in Python
iter_([3], 1) # depth=3
iter_([], 1 + is_odd(3)) if [3] else 1
iter_([], 2) # depth=4
iter_(rest([]), 2 + is_odd(first([])) if [] else 2 # bool([]) is False in Python
2 # the answer
Simple trampolining
To avoid 'max recursion depth exceeded' errors for large arrays all tail calls in recursive functions can be wrapped in lambda: expressions; and special trampoline() function can be used to unwrap such expressions. It effectively converts recursion into iterating over a simple loop:
import functools
def trampoline(function):
"""Resolve delayed calls."""
#functools.wraps(function)
def wrapper(*args):
f = function(*args)
while callable(f):
f = f()
return f
return wrapper
def iter_(lst, count):
#NOTE: added `lambda:` before the tail call
return (lambda:iter_(rest(lst), count+is_odd(first(lst)))) if lst else count
#trampoline
def count_odd(integers):
return iter_(integers, 0)
Example:
count_odd([1,2,3])
iter_([1,2,3], 0) # returns callable
lambda:iter_(rest(lst), count+is_odd(first(lst))) # f = f()
iter_([2,3], 0+is_odd(1)) # returns callable
lambda:iter_(rest(lst), count+is_odd(first(lst))) # f = f()
iter_([3], 1+is_odd(2)) # returns callable
lambda:iter_(rest(lst), count+is_odd(first(lst))) # f = f()
iter_([], 1+is_odd(3))
2 # callable(2) is False

I would write it like this:
def countOddNumbers(numbers):
sum = 0
for num in numbers:
if num%2!=0:
sum += numbers.count(num)
return sum

not sure if i got your question , but as above something similar:
def countOddNumbers(numbers):
count=0
for i in numbers:
if i%2!=0:
count+=1
return count

Generator can give quick result in one line code:
sum((x%2 for x in nums))

My implementation of merging two sorted lists in linear time - what could be improved?

Fromg Google's Python Class:
E. Given two lists sorted in increasing order, create and return a merged
list of all the elements in sorted order. You may modify the passed in lists.
Ideally, the solution should work in "linear" time, making a single
pass of both lists.
Here's my solution:
def linear_merge(list1, list2):
merged_list = []
i = 0
j = 0
while True:
if i == len(list1):
return merged_list + list2[j:]
if j == len(list2):
return merged_list + list1[i:]
if list1[i] <= list2[j]:
merged_list.append(list1[i])
i += 1
else:
merged_list.append(list2[j])
j += 1
First of all, is it okay to use an infinite loop here? Should I break out of the loop using the break keyword when I'm done merging the list, or are the returns okay here?
I've seen similar questions asked here, and all the solutions look quite similar to mine, i.e. very C-like. Is there no more python-like solution? Or is this because of the nature of the algorithm?

This question covers this in more detail than you probably need. ;) The chosen answer matches your requirement. If I needed to do this myself, I would do it in the way that dbr described in his or her answer (add the lists together, sort the new list) as it is very simple.
EDIT:
I'm adding an implementation below. I actually saw this in another answer here which seems to have been deleted. I'm just hoping it wasn't deleted because it had an error which I'm not catching. ;)
def mergeSortedLists(a, b):
l = []
while a and b:
if a[0] < b[0]:
l.append(a.pop(0))
else:
l.append(b.pop(0))
return l + a + b

Here's a generator approach. You've probably noticed that a whole lot of these "generate lists" can be done well as generator functions. They're very useful: they don't require you to generate the whole list before using data from it, to keep the whole list in memory, and you can use them to directly generate many data types, not just lists.
This works if passed any iterator, not just lists.
This approach also passes one of the more useful tests: it behaves well when passed an infinite or near-infinite iterator, eg. linear_merge(xrange(10**9), xrange(10**9)).
The redundancy in the two cases could probably be reduced, which would be useful if you wanted to support merging more than two lists, but for clarity I didn't do that here.
def linear_merge(list1, list2):
"""
>>> a = [1, 3, 5, 7]
>>> b = [2, 4, 6, 8]
>>> [i for i in linear_merge(a, b)]
[1, 2, 3, 4, 5, 6, 7, 8]
>>> [i for i in linear_merge(b, a)]
[1, 2, 3, 4, 5, 6, 7, 8]
>>> a = [1, 2, 2, 3]
>>> b = [2, 2, 4, 4]
>>> [i for i in linear_merge(a, b)]
[1, 2, 2, 2, 2, 3, 4, 4]
"""
list1 = iter(list1)
list2 = iter(list2)
value1 = next(list1)
value2 = next(list2)
# We'll normally exit this loop from a next() call raising StopIteration, which is
# how a generator function exits anyway.
while True:
if value1 <= value2:
# Yield the lower value.
yield value1
try:
# Grab the next value from list1.
value1 = next(list1)
except StopIteration:
# list1 is empty. Yield the last value we received from list2, then
# yield the rest of list2.
yield value2
while True:
yield next(list2)
else:
yield value2
try:
value2 = next(list2)
except StopIteration:
# list2 is empty.
yield value1
while True:
yield next(list1)

Why stop at two lists?
Here's my generator based implementation to merge any number of sorted iterators in linear time.
I'm not sure why something like this isn't in itertools...
def merge(*sortedlists):
# Create a list of tuples containing each iterator and its first value
iterlist = [[i,i.next()] for i in [iter(j) for j in sortedlists]]
# Perform an initial sort of each iterator's first value
iterlist.sort(key=lambda x: x[1])
# Helper function to move the larger first item to its proper position
def reorder(iterlist, i):
if i == len(iterlist) or iterlist[0][1] < iterlist[i][1]:
iterlist.insert(i-1,iterlist.pop(0))
else:
reorder(iterlist,i+1)
while True:
if len(iterlist):
# Reorder the list if the 1st element has grown larger than the 2nd
if len(iterlist) > 1 and iterlist[0][1] > iterlist[1][1]:
reorder(iterlist, 1)
yield iterlist[0][1]
# try to pull the next value from the current iterator
try:
iterlist[0][1] = iterlist[0][0].next()
except StopIteration:
del iterlist[0]
else:
break
Here's an example:
x = [1,10,20,33,99]
y = [3,11,20,99,1001]
z = [3,5,7,70,1002]
[i for i in merge(x,y,z)]

hi i just did this exercise and i was wondering why not use,
def linear_merge(list1, list2):
return sorted(list1 + list2)
pythons sorted function is linear isn't it?

Here's my implementation from a previous question:
def merge(*args):
import copy
def merge_lists(left, right):
result = []
while (len(left) and len(right)):
which_list = (left if left[0] <= right[0] else right)
result.append(which_list.pop(0))
return result + left + right
lists = [arg for arg in args]
while len(lists) > 1:
left, right = copy.copy(lists.pop(0)), copy.copy(lists.pop(0))
result = merge_lists(left, right)
lists.append(result)
return lists.pop(0)

Another generator:
def merge(xs, ys):
xs = iter(xs)
ys = iter(ys)
try:
y = next(ys)
except StopIteration:
for x in xs:
yield x
raise StopIteration
while True:
for x in xs:
if x > y:
yield y
break
yield x
else:
yield y
for y in ys:
yield y
break
xs, ys, y = ys, xs, x

I agree with other answers that extending and sorting is the most straightforward way, but if you must merge, this will be a little faster because it does not make two calls to len every iteration nor does it do a bounds check. The Python pattern, if you could call it that, is to avoid testing for a rare case and catch the exception instead.
def linear_merge(list1, list2):
merged_list = []
i = 0
j = 0
try:
while True:
if list1[i] <= list2[j]:
merged_list.append(list1[i])
i += 1
else:
merged_list.append(list2[j])
j += 1
except IndexError:
if i == len(list1):
merged_list.extend(list2[j:])
if j == len(list2):
merged_list.extend(list1[i:])
return merged_list
edit
Optimized per John Machin's comment. Moved try outside of while True and extended merged_list upon exception.

According to a note here:
# Note: the solution above is kind of cute, but unforunately list.pop(0)
# is not constant time with the standard python list implementation, so
# the above is not strictly linear time.
# An alternate approach uses pop(-1) to remove the endmost elements
# from each list, building a solution list which is backwards.
# Then use reversed() to put the result back in the correct order. That
# solution works in linear time, but is more ugly.
and this link http://www.ics.uci.edu/~pattis/ICS-33/lectures/complexitypython.txt
append is O(1), reverse is O(n) but then it also says that pop is O(n) so which is which? Anyway I have modified the accepted answer to use pop(-1):
def linear_merge(list1, list2):
# +++your code here+++
ret = []
while list1 and list2:
if list1[-1] > list2[-1]:
ret.append(list1.pop(-1))
else:
ret.append(list2.pop(-1))
ret.reverse()
return list1 + list2 + ret

This solution runs in linear time and without editing l1 and l2:
def merge(l1, l2):
m, m2 = len(l1), len(l2)
newList = []
l, r = 0, 0
while l < m and r < m2:
if l1[l] < l2[r]:
newList.append(l1[l])
l += 1
else:
newList.append(l2[r])
r += 1
return newList + l1[l:] + l2[r:]