tulip | customizable application | Frontend Framework library
kandi X-RAY | tulip Summary
kandi X-RAY | tulip Summary
Tulip is a web interface built on top of Simple Map D3. It allows you to make a custom map using any GeoJSON or TopoJSON data source. If you data source has numerical properties, you can create a "choropleth" style map. (coming soon) Tulip is fully customizable so that it can be deployed for an organization or individual with specific configuration options already set or removed so that the design of maps can be controlled.
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Community Discussions
Trending Discussions on tulip
QUESTION
The following link shows how to add multiple EntityRuler with spaCy. The code to do that is below:
...ANSWER
Answered 2021-Jun-15 at 09:55Imagine that your dataframe is
QUESTION
import os
import random
import time
import math
def stringmanipulator(xy, y=40):
xy= xy.lower()
x = []
x = list(xy)
length = len(x)
y = int(math.floor(length * (y/100)))
while(y):
r =int(random.random()*(length-1))
if(x[r] != '_' and x[r] != ' '):
x[r] = '_'
y = y-1
return x
def printcomplement():
x = int(random.random()*11)
if(x == 0):
print("well done!!")
elif(x == 1):
print("keep going!!")
elif(x == 2):
print("YOU can save him!!")
elif(x == 3):
print("You are the hero no one wanted but everyone deserves.")
elif(x == 4):
print("Genius kid.")
elif(x == 5):
print("You are Smart, not kidding.")
elif(x == 6):
print("You are one who will destroy my carrer using your intellect.")
elif(x == 7):
print("The most kind hearted person I have ever seen till now. Yes I am talking about you")
elif(x == 8):
print("You nailed it.")
elif(x == 9):
print("AND I thought the game was hard.")
elif(x == 10):
print("I will find more difficult words to challenge you with.")
elif(x == 11):
print("How about you put another life on risk after this round.")
def printdis():
x = int(random.random()*11)
if(x == 0):
print("Fool")
elif(x == 1):
print("You will end up killing the fool and then I will hang you next.")
elif(x == 2):
print("What a piece of shit you are.")
elif(x == 3):
print("Hey disgrace to humanity.")
elif(x == 4):
print("Don't cry after the man is dead. You killed him, I gave you a chance to save him.")
elif(x == 5):
print("Dumbass!!")
elif(x == 6):
print("You know what it was my mistake to let such an idiot play.")
elif(x == 7):
print("This is your last game. I don't want fools playing this game.")
elif(x == 8):
print("I see you are already crying.")
elif(x == 9):
print("Even the guy who's life is line is laughing at your stupidity.")
elif(x == 10):
print("My 120 years old grandma has a sharper brain than yours.")
elif(x == 11):
print("Get lost, YOU useless, moronic, unworthy pile of garbage.")
def hangman(i = 0):
if(i == 0):
print("___________")
print("| |")
print("| |")
print("| ")
print("| ")
print("| ")
print("| ")
print("| ")
print("|")
elif(i == 1):
print("___________")
print("| |")
print("| |")
print("| ( ) ")
print("| ")
print("| ")
print("| ")
print("| ")
print("|")
elif(i == 2):
print("___________")
print("| |")
print("| |")
print("| ( ) ")
print("| | ")
print("| | ")
print("| ")
print("| ")
print("|")
elif(i == 3):
print("___________")
print("| |")
print("| |")
print("| ( ) ")
print("| \\ | / ")
print("| | ")
print("| ")
print("| ")
print("|")
elif(i == 4):
print("___________")
print("| |")
print("| |")
print("| \\ ( ) /")
print("| \\ | / ")
print("| ")
print("| ")
print("| ")
print("|")
elif(i == 5):
print("___________")
print("| |")
print("| |")
print("| \\ ( ) /")
print("| \\ | / ")
print("| | ")
print("| / \\")
print("| ")
print("|")
elif(i == 6):
print("___________")
print("| |")
print("| |")
print("| \\ ( ) /")
print("| \\ | / ")
print("| | ")
print("| / \\")
print("| / \\")
print("|")
print("\n\nGAME OVER. You have succesfully killed a person. Better luck next time")
def game(xy, y):
x=[]
i = 0
letter = ''
x = stringmanipulator(xy, y)
xy = xy.lower()
# os.system('cls')
for index in range(len(x)):
if(x[index] == '_'):
while(letter != x[index]):
_= os.system('cls')
hangman(i)
for char in range(len(x)):
print(x[char], end=' ')
print("\n")
letter = input("Enter the letter in the first blank: ")
print(letter+str(i))
if(letter == xy[index]):
print("complement")
x[index] = letter
else:
printdis()
i+=1
dictionary ={}
dictionary["films"] = ["A Space OdysseY", "The GodFather", "Citizen Kane", "Raiders of the lost Ark", "Seven Samurai", "There will be Blood", "Casablanca", "Vertigo", "Notorious", "City Lights"]
dictionary["cities"] = ["Tokyo", "Mecca", "Beijing", "London", "Kolkata", "Washington DC", "Mumbai", "Mexico City", "Delhi", "Shanghai"]
dictionary["fruits"] = ["Damson Plum", "Pomelo", "Blood Orange", "Kumquat", "Blackcurrant", "Acerola", "Avocado", "Pomegrenate", "Apple", "Mango"]
dictionary["country"] = ["Djibouti", "Azerbaijan Azerbaijan,", "Venzuela", "Armenia", "Khazakhstan", "Bangladesh", "Saudi Arabia", "United Kingdom", "United States of America", "India"]
dictionary["flowers"] = ["Monkey Face Orchid", "Naked Man Orchid", "Dancing Girls", "Chamber Maids", "Hibiscus", "Marigold", "Tulip", "Lilies", "Daisy", "Hydrangea"]
print("WELCOME TO THE GAME HANGMAN.\n TAKE THE GAME SERIOUSLY SINCE THE LIFE OF A MAN IS DEPENDING ON YOUR KNOWLEDGE. \n\nI DON'T KNOW HOW MANY CHANCE YOU WILL GET, NOT MANY THAT I CAN CONFIRM.\n SO TRY TO SAVE YOUR FELLOW HUMAN OR LET IT BE MY FOOD. HAHAHAHAHAHAHAHAHAH!!!!!!!")
# x = input("Press 1 for films, 2 for cities, 3 for fruits, 4 for country and 5 for flowers (The most beautiful are usually the hardest): ")
# x = int(x)
x = int(input("Enter a number between 1 and 5: "))
if((x < 1) or(x > 5)):
print("What a moron you are. You couldn't even choose one of the option properly game over good bye, tata, cya")
x = random.randint(1,5)
time.sleep(10)
print("Just kidding you still get to play the game but now I will decide what kind of object you have to guess.")
y = int(input("Enter 40 for easy, 60 for medium and 80 for hard: "))
i = 0
xy = ""
r = random.randint(0,9)
if(x == 1):
xy = dictionary["films"][r]
print("FILMS:")
elif(x == 2):
xy = dictionary["cities"][r]
print("CITIES:")
elif(x == 3):
xy = dictionary["fruits"][r]
print("FRUITS:")
elif(x == 4):
xy = dictionary["country"][r]
print("COUNTRY:")
elif(x == 5):
xy = dictionary["flowers"][r]
print("FLOWERS:")
# hangman(0)
game(xy, y)
...ANSWER
Answered 2021-May-31 at 14:43Running your code os.system('cls') is clearing the screen before the input is read in the loop. This makes it seem that there is no output is being displayed when it's really being overwritten.
A quick test can be done to confirm that this is the problem. To do this we add another input read in the game function. like so:
QUESTION
So I have a small list that I am trying to manipulate and organize.
...ANSWER
Answered 2021-May-28 at 22:03Try out :
QUESTION
I have a list of 3 dataframes which contain various names and numbers :
...ANSWER
Answered 2021-May-13 at 11:53Is this what you're looking for?:
QUESTION
I am trying to test a Python library called Tulip dynamically. To do it I need to call the proper ti.
and pass the arguments to call the method.
The problem is, each method has fixed number of parameters and I don't know how to pass it properly.
Let's say I want to test ti.sma
method that requires two arguments real and period
ANSWER
Answered 2021-May-06 at 20:56You actually can use **kwargs
:
File test.py
QUESTION
I can't tell if this is a bug or not.
...ANSWER
Answered 2021-Apr-19 at 19:52$(".mon").attr("style", "border")
sets the border to nothing - you are not testing if it is set.
Explanation:
You click and
QUESTION
@DCTLib, do you recall this discussion below? You suggested a recursive equation, which was the right approach.
Cudd_PrintMinterm, accessing the individual minterms in the sum of products
Now, I am considering multistate reliability, where we can have either not fail or fail to n-1 different states, with n >= 2. Tulip-dd implements MDDs as described in:
https://github.com/tulip-control/dd/blob/master/doc.md#multi-valued-decision-diagrams-mdd
https://github.com/tulip-control/dd/issues/71
https://github.com/tulip-control/dd/issues/66
In the diagrams in the drawings below, we have defined an MDD declared by:
aut.declare_variable(x=(0,3)) u = aut.add_expr(‘x=i’)
Each value/state of the multi-value variable (MSV) x, x=0, x=1, x=2, or x=3 leads to a specific BDD as shown in the diagrams at the bottom, taking a four-state variable x as example here. The notation is that state 0 represents the normal state and x can fail to different states 1, 2, and 3. The failure probabilities are assigned in table below. In the BDDs below, we (and tulip as well) use the binary coding with two bits x_1 and x_0 to represent each state/value of the MSV. The least significant bit (LSB), i.e., x_0, is always the ancestor. Each of the BDD diagrams below is a representation of a specific value, or state.
To quantify the BDD of a specific state, i.e., the top node, we must know probabilities of binary variables x_0 and x_1 taking different branches (then or else) in the BDD. These branch probabilities are not given directly but need to be calculated according to the BDD structure.
The key here is that the child node probabilities and the branch probabilities of the parent node must be known prior to the calculation of the parent node probability. In the previous BDD quantification, we knew the probabilities of branches from node x_1 to leaf nodes when calculating node x_1 probability. We did not need to know how node x_1 was connected to node x_0. Now, for this four-state variable x, we need to know how node x_1 is connected to node x_0, the binary variable representing the least significant bit, to determine the probabilities of branches from node x_1 to leaf nodes. The question is how to implement this?
...ANSWER
Answered 2021-Apr-05 at 12:58The key here is that the child node probabilities and the branch probabilities of the parent node must be known prior to the calculation of the parent node probability.
Yes, exactly. In this case, a fully recursive bottom-up computation, like normally done with BDDs, will not work for the reason that you wrote.
However, the approach will start to work again when you treat the variables that together form a state to be a block. So in your recursive function for the probability calculation, whenever you encounter a variable for a block, you treat the node and the successor nodes for the same state component as a block and only recurse when you encounter a node not belonging to the block.
Note that this approach requires that the variables for the state appear continuously in the variable ordering. For the CUDD library, you can constrain the automatic variable reordering to guarantee this.
The following code is a modification of yours implementing this idea:
QUESTION
I wrote these lines to import zip file with log files in it:
...ANSWER
Answered 2021-Apr-05 at 05:09In the for loop you write the variable lines
but because it loops you over write it every time it loops which is why you are only getting the last values, a better way is to create a list and add to it every time you loop which will not over write it and store it nicely.
QUESTION
I currently have a seemingly easy task, but as the data set is quite large, it is not feasible to do the task manually by hand.
The data.table
stores one column of name
and numerous other columns storing data corresponding to that name object, say var1
, var2
and var3
. The name
variable is type character and the others are type numeric.
Now to the question. Names can start the same (same prefix) but might end differntly (diverging suffix). Imagine for example plant species of different classes or company names with differnt legal entitity form.
My generic question is, whether there is a simple method to match all values of the name
variable that have the same prefix and assign them this "same" prefix while discarding the "diverging" suffix. See below for "abstract" example data.
EDIT Names are not separated by an underscore but by a space and can have several words (not solely constrained to two words)
EDIT 2 The longest shared prefix should be retained. Added some examples to illustrate that point. Proposed solutions would result in the last to entries being "American" instead "American Tulip".
...ANSWER
Answered 2021-Apr-03 at 16:01I think this will also help you if you are also interested in alternative solutions:
QUESTION
I use this schema of addresses and use it in several schemas. All works fine, only when I update a address
...ANSWER
Answered 2021-Mar-07 at 15:35I am not sure is there any straight way to do this, but you can create pre midddlewares for your all parent schemas customerSchema
and providerSchema
and put condition like is address updated or not,
- create a function to handle operation on address schema if its in update
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Install tulip
Go into the template directory: cd tulip
Install NodeJS packages: npm install
Install Bower components: bower install
Run: grunt
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