ostrich | A stats collector & reporter for Scala servers | Analytics library
kandi X-RAY | ostrich Summary
kandi X-RAY | ostrich Summary
Ostrich is a library for scala servers that makes it easy to:. The idea is that it should be simple and straightforward, allowing you to plug it in and get started quickly.
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Community Discussions
Trending Discussions on ostrich
QUESTION
The xml format is a good way to store any hierarchical data. As an example we are using the classification of animals
...ANSWER
Answered 2021-May-28 at 10:32What we can see in the html table, is that the first row holds a cell for every hierarchy level which is represented as a column. This means it has to be generated a row with all elements from the highest till the deepest hierarchy level. The element on the deepest hierarchy level is the one without further descendants. To get these we can use this xpath expression
QUESTION
I'm essentially making a counter and it counts the number of times a name appears in a list. I'm trying to use a function so I can easily do it for all the names. It works fine when I don't make the code a function but as soon as I do it no longer returns the value of y.
...ANSWER
Answered 2021-Apr-18 at 02:39The assignment inside a function does not modify the global variable. To modify a global variable from inside a function, use the global keyword as shown below.
QUESTION
This is an incomplete tree diagram I made to classify this list of animals:
horse, cow, sheep, pig, dog, cat, lion, tiger, whale, dolphin, seal, penguin, ostrich, sparrow, spider, ant, bee, wasp, termite, octopus, squid
I havent finished putting in dog, tiger, lion, etc. just because the application required me to buy some subscription for more shapes and i wasn't gonna do that, but that doesn't matter cos I can visualise the rest. My question is; In python code, how can I make a program that asks the user yes/no questions continuously until it can make out what animal it is out of the list. I can obviously do this with lots of IF statements, or with OOP and using attributes, however both solutions require me to ask EVERY single question, which would amount to quite a lot of lines of code, and it would be quite ugly. How do I make it, for example, so that if the user says that their animal is aquatic, it no longer asks any of the questions that don't apply to the animal. For example:
If I pick wasp, and I answer yes to the question "Is your animal a land animal?", then no to "Is your animal a mammal?", then yes to it being a carnivore and being able to fly, how do i make it so the program will only branch to those questions? Basically, how do i code a tree diagram that follows the user's inputs? (I dont need any GUI)
...ANSWER
Answered 2020-Oct-26 at 20:59You could define the tree with a classical Node
class:
QUESTION
Consider the following two data frames:
...ANSWER
Answered 2020-Oct-04 at 01:49Here might be some simplifications in terms of expression (not check the performance yet)
- Using
asplit
+outer
QUESTION
As a follow-up question to Efficient ways of creating matrix based on vector matches, how would I go about if I would like to produce a boolean matrix with matches between each column in Y
and X
:
ANSWER
Answered 2020-Sep-28 at 16:31I'm not entirely sure what you are after, but perhaps something like this?
QUESTION
SCENARIO
I have a program in Java that randomly puts out text as follows:
1 egg1 chicken and 2 ducks
2 cows1 pig1 egg and 2 cats
1 hour1 dog1 ostrich 2 cows and 3 chickens
QUESTION:
How do I add a comma and a space within a string immediately before the occurrence of all numbers between the first and last ones? That is, how can I format these strings so that these cases look as follows:
1 egg, 1 chicken and 2 ducks
2 cows, 1 pig, 1 egg and 2 cats
1 tractor, 1 dog, 1 ostrich, 2 cows and 3 chickens
As you will see, in each case I:
WHAT I HAVE SO FAR:
I already have it successfully handles cases of:
1 object:
1 cat
2 objects:
2 cats
Variants of Multiple Objects:
3 cats and 1 ostrich
and
1 truck and 6 gnomes
What I need to do is:
- Identify is a string has 3 or more numbers.
- If so, skip the first number in the string.
- Find the next occurring number that is not the last one.
- Add " ," to it.
I can successfully go through the string and count the numbers it has in it:
...ANSWER
Answered 2020-Jul-04 at 14:41You could do it like:
QUESTION
I'm trying to extract the text from this html structure:
...ANSWER
Answered 2020-Jun-06 at 13:52We can pull the details separately and merge them after :
QUESTION
I wonder why my first code doesn't work and the second does. I thought they're basically the same.... Looks like var inside if statement doesn't seem to be processed as expected. Could anyone please clarify why it doesn't work?
...ANSWER
Answered 2020-May-31 at 02:20The first code is trying to get the position [i] from the array, but the variable i does not exist there, it is created in the for loop that is below.
QUESTION
I have a Cucumber-Selenium
based test written using Spring Boot
. The problem is that if I have just one step definition file GoolgeCalcStepDefinition.java
then the program works and test passes without any issue but as soon as I added BingSearchStepDefinition.java
along with feature file then I get following error.
I googled around on how to configure Spring Boot
with Cucumber
but most of the examples/articles available online shows only one step definition file.
mvn verify
...ANSWER
Answered 2020-May-12 at 21:37Both of your runner classes extend DemoApplicationTests.java
which is itself a @SpringBootTest
.
Cucumber cannot determine which SpringBootContext to load when it fires up.. remove the class extends from your stepdefs and have TestRunner
extend DemoApplicationTests
instead
QUESTION
I have some items in a nested dictionary and I need to match if any word in a sentence matches the values in the dictionary. If it does, it returns all the keys from the nested list.
What I have tried so far:
...ANSWER
Answered 2020-Apr-07 at 12:38Could make a recursive function that keeps track of the path, and prints the path when an animal is found in line
.
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