CODEWITHSUNDEEP
pythonmaya
Jared
Jared

Reputation: 555

Iterating through THREE lists at once in Python?

This might be a rather complex question since it's possible a lot of you don't know the software that I'm writing it for: Autodesk Maya 2011. I am trying to speed up a tedious slow process (rigging: giving 3d characters the ability to move) by writing a script that does it automatically.

I'll try my best to explain the situation.

I have a script that takes an object, iterates through the children of that object, stores them in a list, then puts the initial object at the end of the list, reverses the list because it's the wrong way around, then puts the initial object at the front.

Issue: There are three different lists all of the same object TYPE but with different names and they are actually different objects. My goal is to connect them together by generating nodes called 'blendcolors'. But if I have a loop to generate them for each object in list A, then I need loops that also connect them to the objects in other lists and I can not figure this out.

Here is my code, it has been played with so is more incomplete than before as far as the actual loop goes.

    import maya.cmds as cmds

    def crBC(IKJoint, FKJoint, bindJoint, xQuan, switch):

        # gets children joints of the selected joint
        chHipIK = cmds.listRelatives(IKJoint, ad = True, type = 'joint')
        chHipFK = cmds.listRelatives(FKJoint, ad = True, type = 'joint')
        chHipBind = cmds.listRelatives(bindJoint, ad = True, type = 'joint')
        # list is built backwards, this reverses the list
        chHipIK.reverse()
        chHipFK.reverse()
        chHipBind.reverse()
        # appends the initial joint to the list
        chHipIK.append(IKJoint)
        chHipFK.append(FKJoint)
        chHipBind.append(bindJoint)
        # puts the last joint at the start of the list because the initial joint
        # was added to the end
        chHipIK.insert(0, chHipIK.pop())
        chHipFK.insert(0, chHipFK.pop())
        chHipBind.insert(0, chHipBind.pop())


        # pops off the remaining joints in the list the user does not wish to be blended
        chHipBind[xQuan:] = []

        chHipIK[xQuan:] = []

        chHipFK[xQuan:] = []

       # goes through the bind joints, makes a blend colors for each one, connects
       # the switch to the blender

        for a in chHipBind


            rotBC = cmds.shadingNode('blendColors', asUtility = True, n = a + 'rotate_BC')
            tranBC = cmds.shadingNode('blendColors', asUtility = True, n = a + 'tran_BC')
            scaleBC = cmds.shadingNode('blendColors', asUtility = True, n = a + 'scale_BC')

            cmds.connectAttr(switch + '.ikFkSwitch', rotBC + '.blender')
            cmds.connectAttr(switch + '.ikFkSwitch', tranBC + '.blender')
            cmds.connectAttr(switch + '.ikFkSwitch', scaleBC + '.blender')

        # goes through the ik joints, connects to the blend colors

            for b in chHipIK:
                cmds.connectAttr(b + '.rotate', rotBC + '.color1')
                cmds.connectAttr(b + '.translate', tranBC + '.color1')
                cmds.connectAttr(b + '.scale', scaleBC + '.color1')


            # connects FK joints to the blend colors

            for c in chHipFK:
                cmds.connectAttr(c + '.rotate', rotBC + '.color2')
                cmds.connectAttr(c + '.translate', tranBC + '.color2')
                cmds.connectAttr(c + '.scale', scaleBC + '.color2')

        # connects blend colors to bind joints


            cmds.connectAttr(rotBC + '.output', d + '.rotate')
            cmds.connectAttr(tranBC + '.output', d + '.translate')
            cmds.connectAttr(scaleBC + '.output', d + '.scale')                






    # executes function


    crBC('L_hip_IK', 'L_hip_FK', 'L_hip_JNT', 6, 'L_legSwitch_CTRL')

Upvotes: 10

Views: 16044

Answers (5)

Andy Jazz
Andy Jazz

Reputation: 58493

Python's zip_longest function creates an object that allows you to iterate over collections of data (i.e. iterator), that concatenates the elements from each iteration. The iteration continues until the longest iteration is exhausted. By default, a "None" value is used as the empty fill-value of a zip_longest function, but we can use the string "UNKNOWN" here for a nicer formatting and for our own purposes.

Here's the code:

from itertools import zip_longest
 
prims = ["CubeA", "CubeB", "CubeC"]
sizes = [50, 100]
animations = [True, False]
 
for prim, size, animated in zip_longest(prims, sizes, animations, fillvalue="UNKNOWN"):
    print(prim, "has a size", size, "and animated:", animated)

### Results ###

# CubeA has a size 50 and animated: True
# CubeB has a size 100 and animated: False
# CubeC has a size UNKNOWN and animated: UNKNOWN

Upvotes: 0

Rexcirus
Rexcirus

Reputation: 2917

Python3 answer:

for a, b, c in zip(lst1, lst2, lst3):
    ...

Upvotes: 5

Ari Gold
Ari Gold

Reputation: 1548

@florian mayer way with zip or izip works really fine, but you can also use the enumerate to get the list count, 

list_a  = ["x", "y"]
list_b  = ["k", "j"]
list_c  = ["m", "n"]

for count, item in enumerate(list_a):
    print item, list_b[count], list_c[count]

Upvotes: 1

Florian Mayer
Florian Mayer

Reputation: 3081

I do not quite understand the question, are you looking for

import itertools
for a, b, c in itertools.izip(lst1, lst2, lst3):
    ...

?

What izip does is it takes a variable number of arguments and returns an iterator that always yields the respective items of the arguments (a tuple of the first arguments in the first run, a tuple of the second arguments in the second run, and so on and so forth).

Upvotes: 32

S.Lott
S.Lott

Reputation: 391952

Why are there "There are three different lists all of the same object TYPE"? Why can't this be fixed to create one list, where all three objects are matched correctly?

Odds are good that a simple mapping would be better than three parallel lists.

Specifically, you need to fix chHipIK = cmds.listRelatives(IKJoint, ad = True, type = 'joint') to work like this.

chHipIK = [ { 'IK': ik } for ik in mds.listRelatives(IKJoint, ad = True, type = 'joint') ]
for i, fk in enumerate(cmds.listRelatives(FKJoint, ad = True, type = 'joint')):
    chHipIK[i]['FK']= fk
for i, bind in enumerate(cmds.listRelatives(bindJoint, ad = True, type = 'joint')):
    chHipIK[i]['FK']= bind

So that chHipIK is a list of mappings that has all three pieces of information.

Upvotes: 1

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