s2 | Node.js JavaScript / TypeScript bindings for Google S2
kandi X-RAY | s2 Summary
kandi X-RAY | s2 Summary
S2 is a library from Google for easily storing, indexing, and retrieving geographic locations. Geographic regions can be indexed by S2 cell ids of various levels in a data store and then later retrieved by these ids for extremely quick geolocation lookups.
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Lines of Code : 78License : Non-SPDX (Apache License 2.0)
Copydef split(value, num_or_size_splits, axis=0, num=None, name="split"):
"""Splits a tensor `value` into a list of sub tensors.
See also `tf.unstack`.
If `num_or_size_splits` is an `int`, then it splits `value` along the
dimension `axis` into def repeat_elements(x, rep, axis):
"""Repeats the elements of a tensor along an axis, like `np.repeat`.
If `x` has shape `(s1, s2, s3)` and `axis` is `1`, the output
will have shape `(s1, s2 * rep, s3)`.
Args:
x: Tensor or variable.
def row_partitions(self):
"""A tuple of `RowPartition`s defining the shape of this `StructuredTensor`.
When `self.rank <= 1`, this tuple will be empty.
When `self.rank > 1`, these `RowPartitions` define the shape of the
`Struc Community Discussions
Trending Discussions on s2
QUESTION
I recently came across this problem:
You are given two strings, s1 and s2, comprised entirely of lowercase letters 'a' through 'r', and need to process a series of queries. Each query provides a subset of lowercase English letters from 'a' through 'r'. For each query, determine whether s1 and s2, when restricted only to the letters in the query, are equal. s1 and s2 can contain up to 10^5 characters, and there are up to 10^5 queries.
For instance, if s1 is "aabcd" and s2 is "caabd", and you are asked to process a query with the subset "ac", then s1 becomes "aac" while s2 becomes "caa". These don't match, so the query would return false.
I was able to solve this in O(N^2) time by doing the following: For each query, I checked if s1 and s2 would be equal by iterating through both strings, one character at a time, skipping the characters that do not lie within the subset of allowed characters, and checking to see if the "allowed" characters from both s1 and s2 match. If at some point, the characters don't match, then the strings are not equal. Otherwise, the s1 and s2 are equal when restricted only to letters in the query. Each query takes O(N) time to process, and there are N queries, for a total of O(N^2) time.
However, I was told that there was a way to solve this faster in O(N). Does anyone know how this might be done?
...ANSWER
Answered 2022-Mar-28 at 11:30The first obvious speedup is to ensure your set membership test is O(1). To do that, there's a couple of options:
- Represent every letter as a single bit -- now every character is an 18-bit value with only one bit set. The set of allowed characters is now a mask with these bits ORed together and you can test membership of a character with a bitwise-AND;
- Alternatively, you can have an 18-value array and index it by character (
c - 'a'would give a value between 0 and 17). The test for membership is then basically the cost of an array lookup (and you can save operations by not doing the subtraction -- instead just make the array larger and index directly by character.
The next potential speedup is to recognize that any character which does not appear exactly the same number of times in both strings will instantly be a failed match. You can count all character frequencies in both strings with a histogram which can be done in O(N) time. In this way, you can prune the search space if such a character were to appear in the query, and you can test for this in constant time.
Of course, that won't help for a real stress-test which will guarantee that all possible letters have a frequency matched in both strings. So, what do you do then?
Well, you extend the above premise by recognizing that for any position of character x in string 1 and some position of that character in string 2 that would be a valid match (i.e the same number of character x appears in both strings up to their respective positions), then the total count of any other character up to those positions must also be equal. For any character where that is not true, it cannot possibly be compatible with character x.
Let's start by thinking about this in terms of a technique known as memoization where you can leverage precomputed or partially-computed information and get a whole lot out of it. So consider two strings like this:
QUESTION
I would need help in order to draw a complex plot with geom_segments along a circular figure.
I have a dataframe such as
...ANSWER
Answered 2022-Mar-17 at 12:12This gets you pretty close. You may want to tweak the angle values in geom_text() to get it closer.
QUESTION
Clang compiles this fine, but GCC and MSVC complain that operator= cannot be defaulted:
ANSWER
Answered 2022-Mar-14 at 10:40Given the lack of any indications to the contrary, I'm going to answer my own question and say that, as far as I've been able to find relevant clauses in the standard, I think the code is legal and thus GCC and MSVC are complaining erroneously.
As someone pointed out above, there appears to be a bug report tracking this issue.
QUESTION
I'm tackling a exercise which is supposed to exactly benchmark the time complexity of such code.
The data I'm handling is made up of pairs of strings like this hbFvMF,PZLmRb, each string is present two times in the dataset, once on position 1 and once on position 2 . so the first string would point to zvEcqe,hbFvMF for example and the list goes on....
I've been able to produce code which doesn't have much problem sorting these datasets up to 50k pairs, where it takes about 4-5 minutes. 10k gets sorted in a matter of seconds.
The problem is that my code is supposed to handle datasets of up to 5 million pairs. So I'm trying to see what more I can do. I will post my two best attempts, initial one with vectors, which I thought I could upgrade by replacing vector with unsorted_map because of the better time complexity when searching, but to my surprise, there was almost no difference between the two containers when I tested it. I'm not sure if my approach to the problem or the containers I'm choosing are causing the steep sorting times...
Attempt with vectors:
...ANSWER
Answered 2022-Feb-22 at 07:13You can use a trie data structure, here's a paper that explains an algorithm to do that: https://people.eng.unimelb.edu.au/jzobel/fulltext/acsc03sz.pdf
But you have to implement the trie from scratch because as far as I know there is no default trie implementation in c++.
QUESTION
I have a string with data that looks like this:
...ANSWER
Answered 2022-Feb-28 at 15:19You can use
QUESTION
As far as I know cpython implementation keeps the same object for some same values in order to save memory. For example when I create 2 strings with the value hello, cpython does not create 2 different PyObject:
ANSWER
Answered 2022-Jan-25 at 18:49Mutable objects always create a new object, otherwise the data would be shared. There's not much to explain here, as if you append an item to an empty list, you don't want all of the empty lists to have that item.
Immutable objects behave in a completely different manner:
Strings get interned. If they are smaller than 20 alphanumeric characters, and are static (consts in the code, function names, etc), they get cached and are accessed from a special mapping reserved for these. It is to save memory but more importantly used to have a faster comparison. Python uses a lot of dictionary access operations under the hood which require string comparison. Being able to compare 2 strings like attribute or function names by comparing their memory address instead of the actual value, is a significant runtime improvement.
Booleans simply return the same object. Considering there are only 2 available, it makes no sense creating them again and again.
Small integers (from -5 to 256) by default, are also cached. These are used quite often, just about everywhere. Every time an integer is in that range, CPython simply returns the same object.
Floats however are not cached. Unlike integers, where the numbers 0-10 are extremely common, 1.0 isn't guaranteed to be more used than 2.0 or 0.1. That's why float() simply returns a new float. We could have optimized the empty float(), and we can check for speed benefits but it might not have made such a difference.
The confusion starts to arise when float(0.0) is float(0.0). Python has numerous optimizations built in:
First of all, consts are saved in each function's code object.
0.0 is 0.0simply refers to the same object. It is a compile-time optimization.Second of all,
float(0.0)takes the0.0object, and since it's a float (which is immutable), it simply returns it. No need to create a new object if it's already a float.Lastly,
1.0 + 1.0 is 2.0will also work. The reason is that1.0 + 1.0is calculated on compile time and then references the same2.0object:
QUESTION
I want to create a polynomial ring which has float Coefficients like this. I can create with integers but, Floats does not work.
...ANSWER
Answered 2022-Jan-18 at 23:30While I do not have previous experience with this particular (from appearances, rather sophisticated) package Oscar.jl, parsing this error message tells me that the function you are trying to call is being given a BigFloat as input, but simply does not have a method for that type.
At first this was a bit surprising given that there are no BigFloats in your input, but after a bit of investigation, it appears that the culprit is the following
QUESTION
My question is about this kata on Codewars. The function takes two sorted lists with distinct elements as arguments. These lists might or might not have common items. The task is find the maximum path sum. While finding the sum, if there any common items you can choose to change your path to the other list.
The given example is like this:
...ANSWER
Answered 2021-Dec-05 at 10:58Once you know the items shared between the two lists, you can iterate over each list separately to sum up the items in between the shared items, thus constructing a list of partial sums. These lists will have the same length for both input lists, because the number of shared items is the same.
The maximum path sum can then be found by taking the maximum between the two lists for each stretch between shared values:
QUESTION
I would like to remove all characters from a first string s1 exactly the number of times they appear in another string s2, i.e. if s1 = "AAABBBCCCCCCD" and s2 = "ABBCCC" then the result should be s = "AABCCCD". (The order of the characters in the resulting string is actually irrelevant but it's a plus if it can be preserved.)
The following rather crude code can do this:
...ANSWER
Answered 2021-Nov-28 at 22:28You can use counter objects. Subtract one against the other and join the remaining elements together.
QUESTION
I am writing a lot of parser code where string_view excels, and have gotten fond of the type. I recently read ArthurO'Dwyer's article std::string_view is a borrow type, where he concludes that string_view (and other 'borrow types') are fine to use as long as they "... appear only as function parameters and for-loop control variables." (with a couple of exceptions).
However, I have lately started to use string_view as return value for functions that convert enum to string (which I use a lot), like this Compiler Explorer:
...ANSWER
Answered 2021-Oct-23 at 08:58is returning the string_view this way unsafe (or UB) in any way, or can I keep on doing this with good conscience?
Yes. The way you use it is perfectly ok. The string_view returned by your toString function forms a view on data that will remain intact until the program terminates.
Alternatively, is there a better (faster/safer) way of solving this general problem of enum-to-string?
You could make a constexpr function with a switch-statement inside it, like so:
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