这可能是一个相当复杂的问题，因为很多人可能不知道我正在编写它的软件：Autodesk Maya 2011。我正在尝试加快一个乏味的缓慢过程（绑定：为 3d 角色提供通过编写自动执行此操作的脚本来移动的能力）。
我会尽力解释情况。
我有一个脚本，它接受一个对象，迭代该对象的子对象，将它们存储在列表中，然后将初始对象放在列表的末尾，反转列表，因为它是错误的方式，然后将初始对象前面有物体。
问题：存在三个不同的列表，所有对象类型相同但名称不同，它们实际上是不同的对象。我的目标是通过生成称为“blendcolors”的节点将它们连接在一起。但是，如果我有一个循环来为列表 A 中的每个对象生成它们，那么我需要循环将它们连接到其他列表中的对象，但我无法弄清楚这一点。
这是我的代码，它已经被使用过，所以就实际循环而言，它比以前更不完整。
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')
我不太明白这个问题，你在找吗
import itertools
for a, b, c in itertools.izip(lst1, lst2, lst3):
...
？
izipPython3答案：
for a, b, c in zip(lst1, lst2, lst3):
...
为什么会出现“同一对象类型存在三个不同的列表”？为什么不能修复此问题以创建一个列表，其中所有三个对象都正确匹配？
一个简单的映射很有可能比三个并行列表更好。
具体来说，您需要修复
chHipIK = cmds.listRelatives(IKJoint, ad = True, type = 'joint')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
因此
chHipIK@florian mayer 使用 zip 或 izip 的方式效果非常好，但您也可以使用枚举来获取列表计数，
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]
Python 的
zip_longest"None"zip_longest"UNKNOWN"这是代码：
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