Deque | A double-ended queue type in pure Swift
kandi X-RAY | Deque Summary
kandi X-RAY | Deque Summary
A double-ended queue type in pure Swift
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Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of Deque
Deque Key Features
Deque Examples and Code Snippets
import PackageDescription
let package = Package(
name: "MyPackage",
dependencies: [
.Package(url: "https://github.com/attaswift/SipHash.git", from: "3.1.1")
]
)
def pop(self) -> Any:
"""
Removes the last element of the deque and returns it.
Time complexity: O(1)
@returns topop.val: the value of the node to pop.
>>> our_deque = Deque([1, 2, 3, 15182])
def popleft(self) -> Any:
"""
Removes the first element of the deque and returns it.
Time complexity: O(1)
@returns topop.val: the value of the node to pop.
>>> our_deque = Deque([15182, 1, 2, 3])
def generate_dequeue_op(self, tpu_device=0):
"""Generate TPU dequeue ops.
Args:
tpu_device: The TPU device ordinal where the infeed instruction should be
placed.
Returns:
A list of Outputs corresponding to a partitio
Community Discussions
Trending Discussions on Deque
QUESTION
I know that it is possible to modify a list inside a function by using assignment as followslis[:] = new_list
, however the question is, is it possible to modify a deque inside a function as it is also an iterable?
Of course without using return
inside the function.
It not possible to use 'deq[:] = new_deq ' as it gives the following error: TypeError: sequence index must be integer, not 'slice'
ANSWER
Answered 2021-Jun-10 at 21:47deque
does not support a slice as an index, so to achieve the effect of lis[:] = new_list
with a deque
, you can clear it first before extending it:
QUESTION
The following program seems to be an error associated with an explicit constructor. However, I'm unable to find that out.
Using Visual Stduio 2017, the following error comes up on build:
...ANSWER
Answered 2021-Jun-11 at 07:15packaged_task
needs to be able to call SumUp::operator()
so that needs to be public not private:
QUESTION
I have a problem to write code / function that will check that number which provides user belongs to std::stack
or not. I tried making it in different ways but I haven't found any working solutions.
I tried:
...ANSWER
Answered 2021-Jun-07 at 14:15There're no find
in std::stack
, but you can use std::deque
, which provides the same functionality. Use push_back()
for push()
and pop_back()
for pop()
. std::vector
also provides such functionalities.
Example code:
QUESTION
I have a 2D container whose first dimension is deque, and second dimensional is vector.
How to translate it to the new container whose first and second dimensional is the same vector ?
...ANSWER
Answered 2021-Jun-01 at 06:46QUESTION
The code below gives the error error C2039: 'value_type': is not a member of 'Child_Container'
on line 7. This happens in MSVC and Clang, but not with GCC. Thereby when using std::deque
, but not std::set
, std::vector
. Does anyone know why? Thank you!
ANSWER
Answered 2021-Jun-01 at 05:32The variable here is simply whether std::deque
requires its element type to be complete when it is instantiated. (Of course it must be complete when certain member functions are instantiated, but that’s separate.) If it does, you end up needing your value_type
before it’s declared, which produces the error observed. C++17 requires that std::vector
support incomplete types, but says nothing about std::deque
, which is why this varies per standard library.
QUESTION
I have a concern in understanding the Cartpole code as an example for Deep Q Learning. The DQL Agent part of the code as follow:
...ANSWER
Answered 2021-May-31 at 22:21self.model.predict(state)
will return a tensor of shape of (1, 2) containing the estimated Q values for each action (in cartpole the action space is {0,1}).
As you know the Q value is a measure of the expected reward.
By setting self.model.predict(state)[0][action] = target
(where target is the expected sum of rewards) it is creating a target Q value on which to train the model. By then calling model.fit(state, train_target)
it is using the target Q value to train said model to approximate better Q values for each state.
I don't understand why you are saying that the loss becomes 0: the target is set to the discounted sum of rewards plus the current reward
QUESTION
I made a simple replay buffer that when I sample from it gives me the error TypeError: 'type' object is not iterable
ANSWER
Answered 2021-May-30 at 15:52You're adding a type to your list, not an instance of the type. What you're doing is essentially the same as this:
QUESTION
I am working on the LeetCode problem 104. Maximum Depth of Binary Tree:
Given the
root
of a binary tree, return its maximum depth.A binary tree's maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
My attempt is not working: I first add the root
to a queue (if root
is not None
), and then process it, by adding its children to the queue.
While doing this, I keep a counter, and each time I add a child node, I increment the counter by 1. When both left and right child exist, I will only increment the counter by 1.
...ANSWER
Answered 2021-May-22 at 11:35Is it because I have structured the function to not take a list as an input?
No. This may be confusing, but on LeetCode the raw list representation of the input is translated to an instance of TreeNode
before your function is called. So you should never have to deal with this list structure. It is merely the common input format that LeetCode uses across the different programming languages. But the conversion to the target language's data structure is done for you before your implementation is called.
Your code produces an error on the first call of queue.append
because of this line:
QUESTION
D'Escopo-Pape algorithm is very similar in implementation to the Dijkstra's algorithm and works with negative weight edges but it doesn't work with negative cycles. It is apparently faster then Dijkstra's algorithm and Bellman-Ford algorithm in most cases. But there is apparently special cases where this algorithm takes exponential time, can someone provide some example or point me to some material that analyses this algorithm more thoroughly.
This is the implementation:
ANSWER
Answered 2021-May-21 at 19:19This won't be a full proof, you can read "A Note on Finding Shortest Path Trees" (Aaron Kershenbaum, 1981, https://doi.org/10.1002/net.3230110410) if you want that. An other source you may find interesting is "Properties of Labeling Methods for Determining Shortest Path Trees".
The intuitive reason why this algorithm can go bad is that a node in set M0 is pulled out from it again to be re-examined later if an edge that points to it is found. That already sounds quadratic because there could be |V|-1
edges pointing to it, so every node could potentially be "resurrected" that many times, but it's even worse: that effect is self-amplifying because every time a node is "resurrected" in that way, the edges outgoing from that node can cause more resurrections, and so on. In a full proof, some care must be taken with the edge weights, to ensure that enough of those "resurrections" can actually happen, because they are conditional, so [Kershenbaum 1981] presents a way to build an actual example on which Pape's algorithm requires an exponential number of steps.
By the way, in the same paper the author says:
I have used this algorithm to find routes in very large, very sparse real networks (thousands of nodes and average nodal degree between 2 and 3) with a variety of length functions (generally distance-related) and have found it to outperform all others.
(but since no one uses this algorithm, there are not many available benchmarks, except this one in which Pape's algorithm does not compare so favourably)
In contrast, triggering the exponential behaviour requires a combination of a high degree, a particular order of edges in the adjacency list, and unusual edge weights. Some mitigations are mentioned, for example sorting the adjacency lists by weight before running the running the algorithm.
Why is it rarely usedI could not find any real source on this, and don't expect it to exist, after all you would not have to defend not-using some unusual algorithm that has strange properties to boot. There is the exponential worst case (though on closer inspection it seems unlikely to be triggered by accident, and anyway the Simplex algorithm has an exponential worse case, and it is used a lot), it is relatively unknown, and the only available actual benchmark casts doubt on the claim that the algorithm is "usually efficient" (though I will note they used a degree of 4, which still seems low, but it is definitely higher than the "between 2 and 3" used in the claim that the algorithm is efficient). Also, it should not be expected that Pape's algorithm will perform well on dense graphs in general, even if the exponential behaviour is not triggered.
QUESTION
So, what is the difference between these two statement:
...ANSWER
Answered 2021-May-19 at 07:31Community Discussions, Code Snippets contain sources that include Stack Exchange Network
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