Consider the following code snippet.
from typing import Iterable
def geometric_progression(
start: float, multiplier: float, num_elements: int
) -> Iterable[float]:
assert num_elements >= 0
if num_elements > 0:
yield start
yield from geometric_progression(
start * multiplier, multiplier, num_elements - 1
)
This function returns the first num_elements
of the geometric progression starting with start
and multipliying by multiplier
each time. It's easy to see that the last element will be passed through one yield-statement and num_elements-1
yield-from-statements. Does this function have O(num_elements)
time complexity, or does it have O(num_elements**2)
time complexity due to a "ladder" of nested yield-from-statements of depths 0, 1, 2, ..., num_elements-2
, num_elements-1
?
EDIT: I've come up with a simpler code snippet to demonstrate what I am asking.
from typing import Iterable, Any
def identity_with_nested_yield_from(depth: int, iterable: Iterable[Any]) -> Iterable[Any]:
assert depth >= 1
if depth == 1:
yield from iterable
else:
yield from identity_with_nested_yield_from(depth-1, iterable)
Is this function O(depth + length of iterable)
, or is it O(depth * length of iterable)
?
Worst time complexity for inserting into list is O(n-i) , where n is the length of the list and i is the index at which you are inserting. So in case if you are inserting at the last index, that makes it O(1) .
The yield keyword in Python is similar to a return statement used for returning values in Python which returns a generator object to the one who calls the function which contains yield, instead of simply returning a value.
Yield is generally used to convert a regular Python function into a generator. Return is generally used for the end of the execution and “returns” the result to the caller statement. It replace the return of a function to suspend its execution without destroying local variables.
The average time complexity of the in operator for lists is O(n) .
For example: We have an algorithm that has Ω (n²) running time complexity, then it is also true that the algorithm has an Ω (n) or Ω (log n) or Ω (1) time complexity. It describes the limiting behavior of a function, when the argument tends towards a particular value or infinity.
When you say that Time complexity of algorithm is O (1) that does mean that all the steps that are present in your algorithm will complete in constant amount of time and each line of the code will be solved in a single step . These steps could be like equals to , addition , subtraction .. etc
When the time complexity increases linearly with the input size then the algorithm is supposed to have a Linear time complexity. Consider that we have an algorithm, and we are calculating the time it takes to sort items.
So, to get desired results from the algorithm in optimum amount of time, we take time complexity into consideration. How to measure time complexity? Do we measure absolute time? No, we consider number of steps in algorithm and input size.
I could've sworn there was an optimization in place to shortcut these kinds of yield from
chains, but testing shows no such optimization, and I couldn't find anything in the places I thought the optimization was implemented either.
The generators on each level of a yield from
chain must be suspended and resumed individually to pass yield
and send
values up and down the chain. Your function has O(num_elements**2)
time complexity. Also, it hits a stack overflow once the call stack reaches a depth of 1000.
yield from
is formally equivalent to a loop of response = yield child.send(response)
, plus error propagation and handling. When consumed in iteration, the response
is always None
and no errors are propagated/handled. This is equivalent to a for
loop.
# `yield from child` without error handling/response
for x in child:
yield x
Thus, each yield from
has the time/space complexity of iterating its argument. Stacking yield from
of a size n
child a total of m
times thus has a time complexity of O(nm)
.
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