Iterating through THREE lists at once in Python?

Question

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')

Answer 1

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).

Answer 2

Python3 answer:

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

Answer 3

@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]