Python 按循环中的值将列表拆分为嵌套列表（迭代值）

我有这样一个列表：

nestedList = [[0],[0,1,2,3,4,6,7,8,9],[0,1,2,3,4,6,7,8,9],[1,2,3]]

我还有另一个同质（相同长度）的元素列表，我想用它来拆分

nestedList

lengthList = [[1],[5,4],[5,4],[3]]

我试过：

def split(arr, size):
     arrs = []
     while len(arr) > size:
         pice = arr[:size]
         arrs.append(pice)
         arr   = arr[size:]
     arrs.append(arr)
     return arrs

for i,j in zip(nestedList,lengthList):
    for k in j:
        myNewList.append(split(i,k))

但它并不是100%正确

它给出的输出是：

myNewList = [[[0]], [[0, 1, 2, 3, 4], [6, 7, 8, 9]], [[0, 1, 2, 3], [4, 6, 7, 8], [9]], [[0, 1, 2, 3, 4], [6, 7, 8, 9]], [[0, 1, 2, 3], [4, 6, 7, 8], [9]], [[1, 2, 3]]]

而不是

[[[0], [[0, 1, 2, 3, 4], [6, 7, 8, 9]], [[0, 1, 2, 3], [4, 6, 7, 8,9]], [[1, 2, 3]]]

任何帮助都将不胜感激

nestedList = [[0],[0,1,2,3,4,6,7,8,9],[0,1,2,3,4,6,7,8,9],[1,2,3]]
lengthList = [[1],[5,4],[5,4],[3]]

answer = []
for lens,sub in zip(lengthList, nestedList):
    answer.append([])
    top = 0
    for l in lens:
        answer[-1].append(sub[top:top+l])
        top += l

输出：

In [2]: answer
Out[2]: 
[[[0]],
 [[0, 1, 2, 3, 4], [6, 7, 8, 9]],
 [[0, 1, 2, 3, 4], [6, 7, 8, 9]],
 [[1, 2, 3]]]

输出：

In [2]: answer
Out[2]: 
[[[0]],
 [[0, 1, 2, 3, 4], [6, 7, 8, 9]],
 [[0, 1, 2, 3, 4], [6, 7, 8, 9]],
 [[1, 2, 3]]]

---

这是您需要的解决方案

# stackoverflow.com
# Python
# split list into nested lists by values in a loop (iterating values)

nestedList = [[0],[0,1,2,3,4,6,7,8,9],[0,1,2,3,4,6,7,8,9],[1,2,3]]
lengthList = [[1],[5,4],[5,4],[3]]

def toOneList (array):
    oneList = []
    if len(array) > 0:
        for subList in array:
            for x in subList:
                oneList.append(x)
    return oneList

def grouping(array, size):
    newList = []
    i = 0
    for j in size:
        newList.append(array[i:i+j])
        i = i + j
    return newList


nestedList = toOneList(nestedList)
lengthList = toOneList(lengthList)

print(grouping(nestedList, lengthList))

---

这是您需要的解决方案

# stackoverflow.com
# Python
# split list into nested lists by values in a loop (iterating values)

nestedList = [[0],[0,1,2,3,4,6,7,8,9],[0,1,2,3,4,6,7,8,9],[1,2,3]]
lengthList = [[1],[5,4],[5,4],[3]]

def toOneList (array):
    oneList = []
    if len(array) > 0:
        for subList in array:
            for x in subList:
                oneList.append(x)
    return oneList

def grouping(array, size):
    newList = []
    i = 0
    for j in size:
        newList.append(array[i:i+j])
        i = i + j
    return newList


nestedList = toOneList(nestedList)
lengthList = toOneList(lengthList)

print(grouping(nestedList, lengthList))

试试这个代码

# stackoverflow.com
# Python
# split list into nested lists by values in a loop (iterating values)

nestedList = [[0],[0,1,2,3,4,6,7,8,9],[0,1,2,3,4,6,7,8,9],[1,2,3]]
lengthList = [[1],[5,4],[5,4],[3]]

def toOneList (array):
    oneList = []
    if len(array) > 0:
        for subList in array:
            for x in subList:
                oneList.append(x)
    return oneList

def grouping(array, size):
    newList = []
    i = 0
    for j in size:
        newList.append(array[i:i+j])
        i = i + j
    return newList


nestedList = toOneList(nestedList)
lengthList = toOneList(lengthList)

print(grouping(nestedList, lengthList))

试试这个代码

# stackoverflow.com
# Python
# split list into nested lists by values in a loop (iterating values)

nestedList = [[0],[0,1,2,3,4,6,7,8,9],[0,1,2,3,4,6,7,8,9],[1,2,3]]
lengthList = [[1],[5,4],[5,4],[3]]

def toOneList (array):
    oneList = []
    if len(array) > 0:
        for subList in array:
            for x in subList:
                oneList.append(x)
    return oneList

def grouping(array, size):
    newList = []
    i = 0
    for j in size:
        newList.append(array[i:i+j])
        i = i + j
    return newList


nestedList = toOneList(nestedList)
lengthList = toOneList(lengthList)

print(grouping(nestedList, lengthList))

---

输出的确切问题是什么。请描述问题（你正在努力实现的目标）2。如果您希望更正/改进代码，stackexchange上会有CodeReview。您所说的

将列表拆分为同质列表是什么意思？输出的确切问题是什么？1。请描述问题（你正在努力实现的目标）2。如果您想更正/改进代码，stackexchange上会有CodeReview。您所说的
将列表拆分为同质列表是什么意思？谢谢@inspectorG4dget！谢谢@inspectorG4dget！