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Efficient methods of accessing the outputs of generators

I'm working through some exercises in my python book and was asked to implement an infinite product to estimate the sine function depending on some amount of iterations k. The code I wrote is the following:
def approx_sin(x,k):
def u(k):
if k==0:
return x
else:
return (1-((x**2)/((k**2)*(pi**2))))
u0=u(0)
yield u0
for n in range(1,k+1):
u0=u(n)*u0
yield u0
Which works as intended. Now I'm asked to sketch this function on some interval for different values of k, meaning we have to extract some values from the generator. To do this, I create the following function:
def sin_sketch(x,k):
return list(itertools.islice(approx_sin(x,k),k,k+1))[0]
And I can now plot sin_sketch(x,n) for some linspace x and a given value for n. My question boils down to, there has to be a better/more time efficient way to extract the value produced by itertools.islice(approx_sin(x,k),k-1,k) rather then converting it to a list, then taking its only element. Any tips?

Creating a list of a single element should not be much of a problem, performance-wise, but you could e.g. also use next to get the first (and only) element in that islice. In this case, the upper bound does not matter at all.
def sin_sketch(x, k):
return next(itertools.islice(approx_sin(x, k), k-1, k))
Or, instead of islice, use next with enumerate and check the index. I did not time which of those is faster, but this might be a bit clearer.
def sin_sketch(x, k):
return next(v for i, v in enumerate(approx_sin(x, k)) if i == k-1)

It seems like a case of Get the nth item of a generator in Python[*]. But I don't particularly like approaches described there, so here's mine.
From what I understand, you want to get k'th value generated by approx_sin(x, k). So let's do just that:
def sin_sketch(x, k):
gen = approx_sin(x, k)
for _ in range(k):
v = next(gen)
return v
This might look a bit bulkier, but I think it's way more explicit and transparent, without requiring itertools.
[*]: Looking through answers there, I found the one that I like even more than my own: https://stackoverflow.com/a/72817329/9288580

I would simplify your appox function a little to become
from math import pi
from itertools import count
def approx_sin(x):
product = 1
for n in count(1):
product = product * (1-((x**2) / ((n**2) * (pi**2))))
yield x * product
that might make the second part of the question more straightforward.

Does one for loop mean a time complexity of n in this case?

So, I've run into this problem in the daily coding problem challenge, and I've devised two solutions. However, I am unsure if one is better than the other in terms of time complexity (Big O).
# Given a list of numbers and a number k,
# return whether any two numbers from the list add up to k.
#
# For example, given [10, 15, 3, 7] and k of 17, return true since 10 + 7 is 17.
#
# Bonus: Can you do this in one pass?
# The above part seemed to denote this can be done in O(n).
def can_get_value(lst=[11, 15, 3, 7], k=17):
for x in lst:
for y in lst:
if x+y == k:
return True
return False
def optimized_can_get_value(lst=[10, 15, 3, 7], k=17):
temp = lst
for x in lst:
if k-x in temp:
return True
else:
return False
def main():
print(can_get_value())
print(optimized_can_get_value())
if __name__ == "__main__":
main()
I think the second is better than the first since it has one for loop, but I'm not sure if it is O(n), since I'm still running through two lists. Another solution I had in mind that was apparently a O(n) solution was using the python equivalent of "Java HashSets". Would appreciate confirmation, and explanation of why/why not it is O(n).

The first solution can_get_value() is textbook O(n^2). You know this.
The second solution is as well. This is because elm in list has O(n) complexity, and you're executing it n times. O(n) * O(n) = O(n^2).
The O(n) solution here is to convert from a list into a set (or, well, any type of hash table - dict would work too). The following code runs through the list exactly twice, which is O(n):
def can_get_value(lst, k):
st = set(lst) # make a hashtable (set) where each key is the same as its value
for x in st: # this executes n times --> O(n)
if k-x in st: # unlike for lists, `in` is O(1) for hashtables
return True
return False
This is thus O(n) * O(1) = O(n) in most cases.

In order to analyze the asymptotic runtime of your code, you need to know the runtime of each of the functions which you call as well. We generally think of arithmetic expressions like addition as being constant time (O(1)), so your first function has two for loops over n elements and the loop body only takes constant time, coming out to O(n * n * 1) = O(n^2).
The second function has only one for loop, but checking membership for a list is an O(n) function in the length of the list, so you still have O(n * n) = O(n^2). The latter option may still be faster (Python probably has optimized code for checking list membership), but it won't be asymptotically faster (the runtime still increases quadratically in n).
EDIT - as #Mark_Meyer pointed out, your second function is actually O(1) because there's a bug in it; sorry, I skimmed it and didn't notice. This answer assumes a corrected version of the second function like
def optimized_can_get_value(lst, k=17):
for x in lst:
if k - x in lst:
return True
return False
(Note - don't have a default value for you function which is mutable. See this SO question for the troubles that can bring. I also removed the temporary list because there's no need for that; it was just pointing to the same list object anyway.)
EDIT 2: for fun, here are a couple of O(n) solutions to this (both use that checking containment for a set is O(1)).
A one-liner which still stops as soon as a solution is found:
def get_value_one_liner(lst, k):
return any(k - x in set(lst) for x in lst)
EDIT 3: I think this is actually O(n^2) because we call set(lst) for each x. Using Python 3.8's assignment expressions could, I think, give us a one-liner that is still efficient. Does anybody have a good Python <3.8 one-liner?
And a version which tries not to do extra work by building up a set as it goes (not sure if this is actually faster in practice than creating the whole set at the start; it probably depends on the actual input data):
def get_value_early_stop(lst, k):
values = set()
for x in lst:
if x in values:
return True
values.add(k - x)
return False

What is the time complexity in my single number code

The question is "Given an array of integers, every element appears twice except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?"
My code is below:
def singleNumber(nums):
for i in range(len(nums)):
if nums.count(nums[i]) == 1:
return nums[i]
Why my code is not O(N)? did it calculate by for loop, which takes n rounds?
Thanks.

We can do this using an xor command:
a = [0,0,1,1,2,2,3,3,4,5,5,6,6,7,7,8,8,9,9]
ans = 0
for i in a:
ans = i^ans
ans
4
This works, as the xor is effectively doing (1^1)^(2^2)^4^(5^5), and we cancel all the doubles out.

You code runs in O(N^2) because the count method runs in O(N) and it is executed inside of a for loop. Which in total gives you O(N^2).
If you want to make it run in O(N).
You can do the following:
Loop through the array and set its values as keys to a dict and the number of times it appears in the list as it key's value.
Then iterate through the key-value pairs of the dict again and look for the key whose value is 1. This is will give O(2*N) which is O(N) time complexity.

What is the average case performance of this permutation-generating algorithm?

I'm trying to determine the average case performance of this permutation-generating algorithm. It uses the recursive approach, in which the first element is swapped with each other element, producing a new set of permutations - these sets then go through the same routine, but with the first element fixed.
Here's the code in Python:
# Returns a list of all permutations of the given list.
def permutate(set):
# A list of all found permutations, to be returned
permutations = []
# Takes a set which has all elements below index i fixed and finds all permutations
def recurse(set, i):
# If all elements are fixed, store the current permutation
if i + 1 == len(set):
permutations.append(set)
else:
# Swap the "first" element with each other element to generate new permutations
for element in xrange(i, len(set)):
set[element], set[i] = set[i], set[element]
recurse(set, i + 1)
set[element], set[i] = set[i], set[element]
# Use the recursive algorithm to find all permutations, starting with no fixed elements
recurse(set, 0)
return permutations
print permutate([1, 2, 3])
I don't have much experience with analyzed recursive function performances, so I don't know how to solve this. If I had to make a guess, I would say that the runtime is Θ(n!), because a set with n elements has n! permutations (so there must be that much effort put in by the algorithm, right?)
Any help would be appreciated.

First of all, the complexity is O(n!) for the reason mentioned in the comment to the question.
But there are two other things.
Do not use set as a variable name, because you shadow a built-in data type
Your algorithm is not correct because of python implementation details
at the bottom of recursion, you append a resulting permutation to the permutations variable. But list in python is not passed by a value, so you actually append a reference to the input list. Because after recurse finishes its work, the input set is in the same order that it was at the beginning, so the permutation variable will store n! references to the same list. To fix that, you can use deepcopy method of copy module, the resulting code is (notice that you can stop resursion when i == len(s)):
import copy
# Returns a list of all permutations of the given list.
def permutate(s):
# A list of all found permutations, to be returned
permutations = []
# Takes a set which has all elements below index i fixed and finds all permutations
def recurse(s, i):
# If all elements are fixed, store the current permutation
if i == len(s):
# append a deepcopy of s
permutations.append(copy.deepcopy(s))
else:
# Swap the "first" element with each other element to generate new permutations
for element in xrange(i, len(s)):
s[element], s[i] = s[i], s[element]
recurse(s, i + 1)
s[element], s[i] = s[i], s[element]
# Use the recursive algorithm to find all permutations, starting with no fixed elements
recurse(s, 0)
return permutations
print permutate([1, 2, 3])

Python lists, dictionary optimization

I have been attending a couple of hackathons. I am beginning to understand that writing code is not enough. The code has to be optimized. That brings me to my question. Here are two questions that I faced.
def pairsum(numbers, k)
"""Write a function that returns two values in numbers whose sum is K"""
for i, j in numbers:
if i != j:
if i+j == k
return i, j
I wrote this function. And I was kind of stuck with optimization.
Next problem.
string = "ksjdkajsdkajksjdalsdjaksda"
def dedup(string):
""" write a function to remove duplicates in the variable string"""
output = []
for i in string:
if i not in output:
output.append(i)
These are two very simple programs that I wrote. But I am stuck with optimization after this. More on this, when we optimize code, how does the complexity reduce? Any pointers will help. Thanks in advance.

Knowing the most efficient Python idioms and also designing code that can reduce iterations and bail out early with an answer is a major part of optimization. Here are a few examples:
List list comprehensions and generators are usually fastest:
With a straightforward nested approach, a generator is faster than a for loop:
def pairsum(numbers, k):
"""Returns two unique values in numbers whose sum is k"""
return next((i, j) for i in numbers for j in numbers if i+j == k and i != j)
This is probably faster on average since it only goes though one iteration at most and does not check if a possible result is in numbers unless k-i != i:
def pairsum(numbers, k):
"""Returns two unique values in numbers whose sum is k"""
return next((k-i, i) for i in numbers if k-i != i and k-i in numbers)
Ouput:
>>> pairsum([1,2,3,4,5,6], 8)
(6, 2)
Note: I assumed numbers was a flat list since the doc string did not mention tuples and it makes the problem more difficult which is what I would expect in a competition.
For the second problem, if you are to create your own function as opposed to just using ''.join(set(s)) you were close:
def dedup(s):
"""Returns a string with duplicate characters removed from string s"""
output = ''
for c in s:
if c not in output:
output += c
return output
Tip: Do not use string as a name
You can also do:
def dedup(s):
for c in s:
s = c + s.replace(c, '')
return s
or a much faster recursive version:
def dedup(s, out=''):
s0, s = s[0], s.replace(s[0], '')
return dedup(s, n + s0) if s else out + s0
but not as fast as set for strings without lots of duplicates:
def dedup(s):
return ''.join(set(s))
Note: set() will not preserve the order of the remaining characters while the other approaches will preserve the order based on first occurrence.

Your first program is a little vague. I assume numbers is a list of tuples or something? Like [(1,2), (3,4), (5,6)]? If so, your program is pretty good, from a complexity standpoint - it's O(n). Perhaps you want a little more Pythonic solution? The neatest way to clean this up would be to join your conditions:
if i != j and i + j == k:
But this simply increases readability. I think it may also add an additional boolean operation, so it might not be an optimization.
I am not sure if you intended for your program to return the first pair of numbers which sum to k, but if you wanted all pairs which meet this requirement, you could write a comprehension:
def pairsum(numbers, k):
return list(((i, j) for i, j in numbers if i != j and i + j == k))
In that example, I used a generator comprehension instead of a list comprehension so as to conserve resources - generators are functions which act like iterators, meaning that they can save memory by only giving you data when you need it. This is called lazy iteration.
You can also use a filter, which is a function which returns only the elements from a set for which a predicate returns True. (That is, the elements which meet a certain requirement.)
import itertools
def pairsum(numbers, k):
return list(itertools.ifilter(lambda t: t[0] != t[1] and t[0] + t[1] == k, ((i, j) for i, j in numbers)))
But this is less readable in my opinion.
Your second program can be optimized using a set. If you recall from any discrete mathematics you may have learned in grade school or university, a set is a collection of unique elements - in other words, a set has no duplicate elements.
def dedup(mystring):
return set(mystring)
The algorithm to find the unique elements of a collection is generally going to be O(n^2) in time if it is O(1) in space - if you allow yourself to allocate more memory, you can use a Binary Search Tree to reduce the time complexity to O(n log n), which is likely how Python sets are implemented.
Your solution took O(n^2) time but also O(n) space, because you created a new list which could, if the input was already a string with only unique elements, take up the same amount of space - and, for every character in the string, you iterated over the output. That's essentially O(n^2) (although I think it's actually O(n*m), but whatever). I hope you see why this is. Read the Binary Search Tree article to see how it improves your code. I don't want to re-implement one again... freshman year was so grueling!

The key to optimization is basically to figure out a way to make the code do less work, in terms of the total number of primitive steps that needs to be performed. Code that employs control structures like nested loops quickly contributes to the number of primitive steps needed. Optimization is therefore often about replacing loops iterating over the a full list with something more clever.
I had to change the unoptimized pairsum() method sligtly to make it usable:
def pairsum(numbers, k):
"""
Write a function that returns two values in numbers whose sum is K
"""
for i in numbers:
for j in numbers:
if i != j:
if i+j == k:
return i,j
Here we see two loops, one nested inside the other. When describing the time complexity of a method like this, we often say that it is O(n²). Since when the length of the numbers array passed in grows proportional to n, then the number of primitive steps grows proportional to n². Specifically, the i+j == k conditional is evaluated exactly len(number)**2 times.
The clever thing we can do here is to presort the array at the cost of O(n log(n)) which allows us to hone in on the right answer by evaluating each element of the sorted array at most one time.
def fast_pairsum(numbers, k):
sortedints = sorted(numbers)
low = 0
high = len(numbers) - 1
i = sortedints[0]
j = sortedints[-1]
while low < high:
diff = i + j - k
if diff > 0:
# Too high, let's lower
high -= 1
j = sortedints[high]
elif diff < 0:
# Too low, let's increase.
low += 1
i = sortedints[low]
else:
# Just right
return i, j
raise Exception('No solution')
These kinds of optimization only begin to really matter when the size of the problem becomes large. On my machine the break-even point between pairsum() and fast_pairsum() is with a numbers array containing 13 integers. For smaller arrays pairsum() is faster, and for larger arrays fast_pairsum() is faster. As the size grows fast_pairsum() becomes drastically faster than the unoptimized pairsum().
The clever thing to do for dedup() is to avoid having to linearly scan through the output list to find out if you've already seen a character. This can be done by storing information about which characters you've seen in a set, which has O(log(n)) look-up cost, rather than the O(n) look-up cost of a regular list.
With the outer loop, the total cost becomes O(n log(n)) rather than O(n²).
def fast_dedup(string):
# if we didn't care about the order of the characters in the
# returned string we could simply do
# return set(string)
seen = set()
output = []
seen_add = seen.add
output_append = output.append
for i in string:
if i not in seen:
seen_add(i)
output_append(i)
return output
On my machine the break-even point between dedup() and fast_dedup() is with a string of length 30.
The fast_dedup() method also shows another simple optimization trick: Moving as much of the code out of the loop bodies as possible. Since looking up the add() and append() members in the seen and output objects takes time, it is cheaper to do it once outside the loop bodies and store references to the members in variables that is used repeatedly inside the loop bodies.

To properly optimize Python, one needs to find a good algorithm for the problem and a Python idiom close to that algorithm. Your pairsum example is a good case. First, your implementation appears wrong — numbers is most likely a sequence of numbers, not a sequence of pairs of numbers. Thus a naive implementation would look like this:
def pairsum(numbers, k)
"""Write a function that returns two values in numbers whose sum is K"""
for i in numbers:
for j in numbers:
if i != j and i + j != k:
return i, j
This will perform n^2 iterations, n being the length of numbers. For small ns this is not a problem, but once n gets into hundreds, the nested loops will become visibly slow, and once n gets into thousands, they will become unusable.
An optimization would be to recognize the difference between the inner and the outer loops: the outer loop traverses over numbers exactly once, and is unavoidable. The inner loop, however, is only used to verify that the other number (which has to be k - i) is actually present. This is a mere lookup, which can be made extremely fast by using a dict, or even better, a set:
def pairsum(numbers, k)
"""Write a function that returns two values in numbers whose sum is K"""
numset = set(numbers)
for i in numbers:
if k - i in numset:
return i, k - i
This is not only faster by a constant because we're using a built-in operation (set lookup) instead of a Python-coded loop. It actually does less work because set has a smarter algorithm of doing the lookup, it performs it in constant time.
Optimizing dedup in the analogous fashion is left as an excercise for the reader.

Your string one, order preserving is most easily and should be fairly efficient written as:
from collections import OrderedDict
new_string = ''.join(OrderedDict.fromkeys(old_string))