I am a bit new to python and I am trying to get a list containing all root parent existing in a scene of type joint.
for example, my scene outliner is something like that:
group1>>group2>>joint1>>joint2>>joint3
group3>>joint4>>joint5
joint16>>joint17>>joint18
I want a script that travels through the outliner and returns a list, in my example:
[joint1, joint4, joint16]
Any tips would be really appreciated. thank you so much.
Im not sure if it is any of use has Haggi Krey solution works fine but
You can use also the flag : -long from cmds.ls
# list all the joints from the scene
mjoints = cmds.ls(type='joint', l=True)
# list of the top joints from chain
output = []
# list to optimise the loop counter
exclusion = []
# lets iterate joints
for jnt in mjoints:
# convert all hierarchy into a list
pars = jnt.split('|')[1:]
# lets see if our hierarchy is in the exclusion list
# we put [1:] because maya root is represented by ''
if not set(pars) & set(exclusion):
# we parse the hierarchy until we reach the top joint
# then we add it to the output
# we add everything else to the exclusion list to avoid
for p in pars:
if cmds.nodeType(p) == 'joint':
output.append(p)
exclusion+=pars
break
print(output)
I just put this because there is not one way to go. I hope the construction of this code could help your python skills. It is exactly the same, just the way to find the parent nodes is different !
I've used DrWeeny's idea before where you traverse the hierarchy by the object's long name. The difference in this answer is that the script won't crash if there's objects with duplicate names in the scene. What I mean by that is let's say you have a situation where you have 2 hierachies:
group1>>joint1>>joint2>>group2>>joint3
and
group3>>joint1>>joint2>>group2>>joint3
Maya easily allows this, like when duplicating a top node, so we need to prevent the script from crashing in this case. When there's multiple objects with duplicate names Maya will crash if you try to access the object's short name (it doesn't know what one you're referring to!), so instead we must always use its long name:
import maya.cmds as cmds
jnts = cmds.ls(type="joint", l=True) # Collect all joints in the scene by their long names.
output = set() # Use a set to avoid adding the same joint.
for jnt in jnts:
pars = jnt.split("|") # Split long name so we can traverse its hierarchy.
root_jnt = None
while pars:
obj = "|".join(pars)
del pars[-1] # Remove last word to "traverse" up hierarchy on next loop.
# If this is a joint, mark it as the new root joint.
if obj and cmds.nodeType(obj) == "joint":
root_jnt = obj
# If a root joint was found, append it to our final list.
if root_jnt is not None:
output.add(root_jnt)
print(list(output))
Using this script on the hierarchies above would return
[u'|group1|joint1', u'|group3|joint1']
I'd suggest to list all joints and for every joint you can check if it's parent is not a joint. In your definition, these joints should be your root joints.
I use this method to get a joint hierarchy. I have given up on trying to figure out a sexier way to do this.
myItems = cmds.ls(selection = True, type='joint')
theParentJnt = cmds.listRelatives(myItems, parent = True)
jntRel = cmds.listRelatives(myItems, allDescendents = True)
allJnt = jntRel + myItems
#Green Cell
Your method worked once, and never worked again. Restarted maya 2020 more then 5 times and only displays to the top node joint, never again does it return the all the joints in one list.
I haven't gotten a response on my other post since I updated it and I really need some help, so I'm reposting. The code below is the important bit, but a link to the full code is here: https://github.com/amstrudy/nao-ncsu/blob/master/oop_a_star.py
I've been following along the Wikipedia pseudocode for an A * implementation in Python. My code reaches the goal, but I can't seem to figure out how to reconstruct the actual path. It doesn't work because the piece of code where cameFrom is updated is only updated once at the end when the goal is reached. I don't understand that really but it's how the wikipedia article had the pseudocode.
def a_star ():
# create node object for home
home_node = Node(home, goal, home)
# set of nodes already evaluated
closedSet = []
# set of currently discovered nodes that are not evaluated yet
# initially, only the start node is known
openSet = [home_node]
while len(openSet) != 0:
minIndex = 0;
for i in list(range(len(openSet))):
if openSet[i] < openSet[minIndex]:
minIndex = i
cur_node = openSet[minIndex] # expand on node with smallest fScore
cur_node.generateNeighbors() # make new nodes to check
if cur_node.location == home_node.goal:
print("Made it to the goal!")
return reconstruct(cur_node)
openSet.pop(minIndex)
closedSet.append(cur_node)
for neighbor in cur_node.neighbors:
if neighbor in closedSet:
continue
if neighbor not in openSet:
openSet.append(neighbor)
if neighbor.gScore >= cur_node.gScore:
print(neighbor.gScore)
print(cur_node.gScore)
continue # this is not eh better path
# this path is the best for now, so record it
neighbor.cameFrom = cur_node
print(cur_node)
printMap(cur_node.location)
def reconstruct(target):
path = []
while target:
print("here")
path.append(target.location)
target = target.cameFrom
return path
For those commenting that this is a duplicate: yeah it is. I'm not getting a response on the old post, so I edited it and reposted. I will delete the old post.
I am trying to develop a python program to check if there is a loop in the graph when a new line is added.
I am storing different lines in a list ordered by shortest length first, and the lines are a class:
class Line():
def __init__(self,node1,node2,length):
self.node1 = node1
self.node2 = node2
self.length = int(length)
self.drawn = False
And the nodes are stored in a list:
nodes = ["A","B","C","D","E","F"]
When my program has run it stores the route as a list:
route = [class(Line),class(Line)...]
What i want it to do is to check that when a new line is added that it does not form a cycle. I plan to use a method inside of a bigger class:
something like:
def check_loop(new_line,graphs):
add new line to graph
if there is a loop in graphs:
return False
else:
return True
(sorry this is one of my first posts so the format is rubbish)
To identify if a cycle is created in a tree is pretty simple, all you need to do is pick a node and then breadth first search the entire tree from that node. If any node has been visited before by the time you reach it, then you know that there is a cycle because there was an alternative path that reached that node beside the one that you have taken.
Yes, this is related to homework and no I'm not looking for you to code it for me just a point in the right direction.
I'm trying to implement a linked list to represent the list of the current possible words in the list after the user has guessed a letter.
Currently I am doing the following to read in the lenght of the word the player wants to guess upon and then appending all possible words to the linked list.
However, I fail to see how I could continually update the linked list to compute the biggest word family and print the current word progress
(Ex. the dashed word with all of the current letters the user has guessed in there).
Mainly having problems because I fail to fully understand how linked lists work and how to implement them in my programs.
How I'm making the initial linked list of all possible words.
def make_ll(self):
linked_list = ll.LinkedList()
print(linked_list)
for word in self.init_list:
if len(word) == int(self.word_length):
linked_list.append(str(word))
self.linked_list = linked_list
def init_family(self, guess):
self.dictionary = {}
cursor = self.linked_list.head
while cursor != None:
word = cursor.data
cursor = cursor.next
word = self.make_families(word, guess)
def make_families(self, word, guess):
count = 0
new_word = ""
for c in word:
if guess == c:
count += 1
else:
new_word = word.replace(c, "-")
if count == 0:
self.linked_list.delete(word)
key = new_word
if key not in self.dictionary:
self.dictionary[key] = []
self.dictionary[key].append(word)
max = max(self.dictionary.values(), key = len)
new_linked_list = ll.linked_list
for word in max:
new_linked_list.append(word)
self.linked_list = new_linked_list
return word
Skeletor,
This can be done using a Linked List, but it is not the optimal solution.
You can think of a Linked List as a bunch of train cars. Each car does two things:
1) It holds something (in programming, this is the data - in your case it is a word).
2) It connects to other cars (in programming this is usually something called a pointer).
Both of these things are a part of the train car. In Linked Lists, this is called the node.
Node1 Node2 Node3
----- ----- -----
|Data1| -> |Data2| -> |Data3|
----- ----- -----
There are a few operations you can perform on train cars:
1) You can add a new car anywhere if you adjust the connectors appropriately.
2) You can walk through cars, depending on the rules for traffic, and look at the data inside.
3) Being in a car let's you easily find the next: you just have to exit and board the next car.
Linked lists have these same rules:
1) They allow you to add a node anywhere you like, but you have to adjust the references to the node (the pointers.)
2) When you're looking at the node you can easily access the data.
3) You can access the next node easily by looking at that node's reference. If you have a doubly linked list, you can look at both the previous and next node.
Most confusion understanding linked lists is the linked part of the definition. Like I mentioned, this is usually a pointer. To understand how that works you'll have to understand pointers, however, you can still understand linked lists by walking through one yourself.
First, let's define our train car(or our node):
class TrainCar:
def __init__(self, data):
# this is the contents of our train car
self.data = data
# This is the reference to the next node, or the train car's connector.
# Accessing it gets you the next TrainCar, if one exists
self.next_car = None
Once you get this, it already starts to fall into place. Basically you have a bunch of nodes with each one referencing the next node! All that's left to do is to implement our operations:
# making a new linked list
class Train:
def __init__(self):
self.head_car = None
def add_car(self, data):
# make new car/node
new_car = TrainCar()
# add our data to the new car as desired
new_car.data = data
# This is where the magic happens! We're updating the reference inside
# the node we just made to point to the last car we were on.
# We are just adding a new car to the end of the train.
new_car.next_car = self.head_car
# Now we're setting the head car to be the one we just made!
self.head_car = new_car
Now, other operations - such as traversing the linked list and deleting nodes - are left to you.
Okay, sorry if my problem seems a bit rough. I'll try to explain it in a figurative way, I hope this is satisfactory.
10 children. 5 boxes. Each child chooses three boxes. Each box is opened:
- If it contains something, all children selected this box gets 1 point
- Otherwise, nobody gets a point.
My question is about what I put in bold. Because in my code, there are lots of kids and lots of boxes.
Currently, I proceed as follows:
children = {"child_1" : 0, ... , "child_10": 0}
gp1 = ["child_3", "child_7", "child_10"] #children who selected the box 1
...
gp5 = ["child_2", "child_5", "child_8", "child_10"]
boxes = [(0,gp1), (0,gp2), (1,gp3), (1,gp4), (0,gp5)]
for box in boxes:
if box[0] == 1: #something inside
for child in box[1]:
children[child] += 1
I worry mainly about the for loop that assigns each child an extra point. Because in my final code, I have many many children, I fear that doing so would slow the program too.
Is there a more efficient way for all children of the same group may have their point faster?
Represent children as indices into arrays, not as strings:
childrenScores = [0] * 10
gp1 = [2,6,9] # children who selected box 1
...
gp5 = [1,4,7,9]
boxes = [(0,gp1), (0,gp2), (1,gp3), (1,gp4), (0,gp5)]
Then, you can store childrenScores as a NumPy array and use advanced indexing:
childrenScores = np.zeros(10, dtype=int)
...
for box in boxes:
if box[0]:
childrenScores[box[1]] += 1 # NumPy advanced indexing
This still involves a loop somewhere, but the loop is deep inside NumPy instead, which should provide a meaningful speedup.
The only speed up that I can think of is to use numpy arrays and stream the sum operation.
children[child] += np.ones(len(children[child]))
You should benchmark the operation and see if that is too slow for your business case.
What I would do
In the gpX lists don't save the "name of the child" (e.g. "child_10") but save a reference to the child's number of points.
How to do that
Using the fact that lists are objects in python, you can:
Change the children dict to look like: children = {"child_0": [0], "child_1": [0], ...} and so on.
When you assign to group, don't assign the key but assign the value (e.g. gp1.append(children["child_0"])).
The loop should then look like: for child in box[1]: child[0]+=1. This WILL update the children dict.
EDIT:
Why this is faster:
Because you leave out the part where you search for children[child], which might be costly.
This technique works because by storing the totals in a mutable type, and appending those values to the group lists, both the dict value and each box's list value will point to the same list entries, and changing one will change the other.
Two general points:
(1) Based on what you've told us, there's no reason to focus your energy on minor performance optimizations. Your time would be better spent thinking about ways to make your data structures less awkward and more communicative. A bunch of interrelated dicts, lists, and tuples quickly becomes difficult to maintain. For an alternative, see the example below.
(2) As the game designer, you understand that events follow a certain sequence: first the kids select their boxes, and later they discover whether they get points for them. But you don't have to implement it that way. A kid can choose a box and get points (or not) immediately. If there's a need to preserve the child's ignorance about such outcomes, the parts of your algorithm that depend on such ignorance can enforce that veil of secrecy as needed. The upshot: there is no need for a box to loop through its children, awarding points to each one; instead, award the points immediately to kids as boxes are selected.
import random
class Box(object):
def __init__(self, name):
self.name = name
self.prize = random.randint(0,1)
class Child(object):
def __init__(self, name):
self.name = name
self.boxes = []
self.score = 0
self._score = 0
def choose(self, n, boxes):
bs = random.sample(boxes, n)
for b in bs:
self.boxes.append(b)
self._score += b.prize
def reveal_score(self):
self.score = self._score
boxes = [Box(i) for i in range(5)]
kids = [Child(i) for i in range(10)]
for k in kids:
k.choose(3, boxes)
# Later in the game ...
for k in kids:
k.reveal_score()
print (k.name, k.score), '=>', [(b.name, b.prize) for b in k.boxes]
One way or another, you're going to be looping over the children, and your answer appears to avoid looping over children who don't get any points.
It might be slightly faster to use filter or itertools.ifilter to pick the boxes that have something in them:
import itertools
...
for box in itertools.ifilter(lambda x: x[0], boxes):
for child in box[1]
children[child] += 1
If you don't need to immediately print the number of points for every child, you could calculate it on demand, thus saving time. This could help if you only need to query a child every now and then for how many points it has. You can cache each result as you obtain is so you don't go about calculating it again the next time you need it.
Firstly, you'll need to know which groups a child belongs to. We'll store this information as map we'll call childToGroupsMap, which will map each child to an array holding the boxes it belongs to, like so:
childToGroupsMap = {}
for child in children:
childToGroupsMap[child[0]] = []
for box in boxes:
for child in box[1]:
if (box[1] not in childToGroupsMap[child]):
childToGroupsMap[child].append(box[1])
This constructs the reverse map from children to boxes.
It'll also help to have a map from each box to a boolean representing whether it's been opened:
boxToOpenedMap = {}
for box in boxes:
boxToOpenedMap[box[1]] = box[0]
Now, when someone queries how many points a child has, you can go through each of the boxes it belongs to (using childToGroupsMap, of course), and simply count how many of those boxes have been mapped to 1 in the map boxes:
def countBoxesForChild(child):
points = 0
for box in childToGroupsMap[child]
if boxToOpenedMap[box] == 1:
points += 1
return points
To make this better, you can cache the resulting number of points. Make a map like so:
childToPointsCalculated = {}
for child in children:
childToPointsCalculated[child[0]] = -1
Where the -1 denotes that we don't yet know how many points this child has.
Finally, you can modify your countBoxesForChild function to exploit the cache:
def countBoxesForChild(child):
if childToPointsCalculated[child] != -1
return childToPointsCalculated[child]
points = 0
for box in childToGroupsMap[child]
if boxToOpenedMap[box] == 1:
points += 1
childToPointsCalculated[child] = points
return points