Python 创建不规则嵌套循环的产品置换_Python_List_Nested_Permutation

Python 创建不规则嵌套循环的产品置换

python list

Python 创建不规则嵌套循环的产品置换,python,list,nested,permutation,Python,List,Nested,Permutation,对于冗长的解释，我深表歉意，我有两个嵌套列表，一个包含一系列字段的引用值，称为dl，另一个包含字段名，称为k_all_tbl。它们看起来像是吼叫： dl = [[['Aegon', 'Aviva', 'HSBC', 'Zurich'], 'S_Field_1_0'], [['Level Life Cover', 'Family Income Benefit', 'Whole of Life Cover'], 'Field_1 policy type

对于冗长的解释，我深表歉意，我有两个嵌套列表，一个包含一系列字段的引用值，称为

dl

，另一个包含字段名，称为

k_all_tbl

。它们看起来像是吼叫：

dl = [[['Aegon', 'Aviva', 'HSBC', 'Zurich'], 'S_Field_1_0'],
            [['Level Life Cover', 'Family Income Benefit', 'Whole of Life Cover'],
            'Field_1 policy type'],
            [['Sum assured', 'Benefit amount'], 'Field_2'],
            [['£ii_Field_52_1', '£ii_Field_53_2 a month'], 'Field_3'],
            [['ii_Field_62_1 years', 'to age ii_Field_63_2', 'Whole of life'], 'Field_4'],
             .....]

dl

列表具有未知长度

，但每个元素

dl[i]

中嵌套有两个列表，第一个（

dl[i][0]

）可能替换第二个元素（

dl[i][1]

）

k\u all\u tbl

如下所示：

k_all_tbl =[[[[['Name']], [['INSERT_3']]],
            [[['Product']], [['S_Field_1_0'], [' '], ['Field_1 policy type']]],
            [[['Field_2']], [['Field_3']]],

           [[[['Name']], [['INSERT_9']]],
            [[['Product']],
             [['INSERT_10'], [' '], ['S_Field_19_18'], [' '], ['INSERT_11']]],
            [[['Sum assured']], [['£'], ['INSERT_12']]],
             .....]

k_all_tbl（从表数据中提取）也具有未知长度

，但每个元素

k_all_tbll[j]

（每行）中嵌套有2个列表，其中（

k_all_tbll[j][0]

）和（

k_all_tbll[j][1]

）（原始表的第1列和第2列）被链接

需求的简短版本是，我需要创建以下输出：

k_all_tbl[0] =[[['Name'], ['INSERT_3']],
                [[['Product']], [['Aegon', ' ', 'Level Life Cover'],
                  ['Aegon', ' ', 'Family Income Benefit'],
                  ['Aegon', ' ', 'Whole of Life Cover'],
                  ['Aviva', ' ', 'Level Life Cover'],
                  ['Aviva', ' ', 'Family Income Benefit'],
                  ['Aviva', ' ', 'Whole of Life Cover'],
                  ['HSBC', ' ', 'Level Life Cover'],
                  ['HSBC', ' ', 'Family Income Benefit'],
                  ['HSBC', ' ', 'Whole of Life Cover'],
                  ['Zurich', ' ', 'Level Life Cover'],
                  ['Zurich', ' ', 'Family Income Benefit'],
                  ['Zurich', ' ', 'Whole of Life Cover']]],

                [[['Sum assured', '£ii_Field_52_1'],
                  ['Sum assured', '£ii_Field_53_2 a month'],
                  ['Benefit amount', '£ii_Field_52_1'],
                  ['Benefit amount', '£ii_Field_53_2 a month']]]

然而，我在下面详细分享的代码和步骤并没有让我明白，它没有执行就崩溃了，或者给了我错误的排列。任何和所有的帮助都将不胜感激。详细说明和代码如下。尝试的方法：

我首先尝试：对于

k_all_tbl[j][0][k]

中对应于元素

dl[I][1]

的每个元素，将

k_all_tbl[j][0][k]

替换为（

dl[I][0]

）。输出的最大值为：

k_all_tbl[0] =[[[[['Name']], [['INSERT_3']]],
            [[['Product']], [[['Aegon', 'Aviva', 'HSBC', 'Zurich']], [' '], [['Level Life Cover', 'Family Income Benefit', 'Whole of Life Cover']]]],
            [[[['Sum assured', 'Benefit amount']]], [[['£ii_Field_52_1', '£ii_Field_53_2 a month']]]]

我通过运行以下代码实现了上述功能：

for ik in range(0,len(k_all_tbl)):
    k_all_tbl[ik] = unlevel(k_all_tbl[ik])
    for ikk in range(0,len(k_all_tbl[ik])):

            for ikkk in range(0,(len (k_all_tbl[ik][ikk]))):

                    for res1 in dropdown_result:
                        if isinstance(k_all_tbl[ik][ikk][ikkk], list)==False: 

                            if  k_all_tbl[ik][ikk][ikkk] == res1[1]:
                                k_all_tbl[ik][ikk][ikkk] = res1[0]


                        if len(k_all_tbl[ik][ikk][ikkk]) >1:
                            for ikkkk in range(0,(len (k_all_tbl[ik][ikk][ikkk]))):
                                if  k_all_tbl[ik][ikk][ikkk][ikkkk] == res1[1]:
                                    k_all_tbl[ik][ikk][ikkk][ikkkk] = res1 [0]

其中函数

unlevel

用于访问单元素嵌套列表中的项目，即

unlevel（[target]，1]]）='[[target]，1]'

，定义为：

def unlevel(obj):
    while isinstance(obj, list) and len(obj) == 1:
        obj = obj[0]
    if isinstance(obj, list):
        return [unlevel(item) for item in obj]
    else:
        return obj

然后，我需要删除不必要的括号，并创建

k_all_tbl

中每个元素的“乘积”，例如

k_all_tbl[0]

：

k_all_tbl[0] =[[['Name'], ['INSERT_3']],
                [[['Product']], [['Aegon', ' ', 'Level Life Cover'],
                  ['Aegon', ' ', 'Family Income Benefit'],
                  ['Aegon', ' ', 'Whole of Life Cover'],
                  ['Aviva', ' ', 'Level Life Cover'],
                  ['Aviva', ' ', 'Family Income Benefit'],
                  ['Aviva', ' ', 'Whole of Life Cover'],
                  ['HSBC', ' ', 'Level Life Cover'],
                  ['HSBC', ' ', 'Family Income Benefit'],
                  ['HSBC', ' ', 'Whole of Life Cover'],
                  ['Zurich', ' ', 'Level Life Cover'],
                  ['Zurich', ' ', 'Family Income Benefit'],
                  ['Zurich', ' ', 'Whole of Life Cover']]],

                [[['Sum assured', '£ii_Field_52_1'],
                  ['Sum assured', '£ii_Field_53_2 a month'],
                  ['Benefit amount', '£ii_Field_52_1'],
                  ['Benefit amount', '£ii_Field_53_2 a month']]]

为了实现这一点，我在上面代码的底部尝试了以下两个功能：

for ik in range(0,len(k_all_tbl)):
        #loop through lists inside nested list -if single element list is nested, remove list before comparison, if more than 1 element keep list, and iterativley compare
        for ikk in range(0,len(k_all_tbl[ik])):

            for ikkk in range(0,(len (k_all_tbl[ik][ikk]))):

                    for res1 in dl:
                        if isinstance(k_all_tbl[ik][ikk][ikkk],list)==True and len(k_all_tbl[ik][ikk][ikkk])==1:
                            k_all_tbl[ik][ikk][ikkk] = unlevel(k_all_tbl[ik][ikk][ikkk])
                            if  k_all_tbl[ik][ikk][ikkk] == res1[1]:
                                k_all_tbl[ik][ikk][ikkk] = res1[0]
                             k_all_tbl[ik][ikk][ikkk] = unlistme(k_all_tbl[ik][ikk][ikkk])
                             k_all_tbl[ik][ikk] = spreadm(k_all_tbl[ik][ikk])


                        if len(k_all_tbl[ik][ikk][ikkk]) >1:
                            for ikkkk in range(0,(len (k_all_tbl[ik][ikk][ikkk]))):
                                if  k_all_tbl[ik][ikk][ikkk][ikkkk] == res1[1]:
                                    k_all_tbl[ik][ikk][ikkk][ikkkk] = res1 [0]
                            k_all_tbl[ik][ikk][ikkk] = unlistme(k_all_tbl[ik][ikk][ikkk])
                            k_all_tbl[ik][ikk] = spreadm(k_all_tbl[ik][ikk])
##function to create permutations
def product(llist):
    result = [[]]
    for lst in llist:
        result = [x + [y] for x in result for y in lst]
    return result


#Function to put single string elements in lists, i.e. "Target" must be changed to ["Target"], otherise the loop will iterate accross "T"-"A"-"R"-"G"-"E"-"T" characters:
def unlistme(ktest):
        ktest2 = []
        for ktt in ktest:
            kttls=[]
            if isinstance(ktt,list)==True:
                ktest2.append(ktt)
            if isinstance(ktt,list)==False:
                kttls.append(ktt)
                ktest2.append(kttls)
        return(ktest2)


#function to open nested list into sublist using product
def spreadm(kx2):
    if isinstance(kx2,list) ==True:
    
        out = product(kx2)
    else:
            out = kx2
    return(out)

为了解决这个问题，我使用了以下两个循环，函数与上面在OP中定义的相同。第一个循环使用参考数据帧dl中的相应字段进行替换，而第二个循环搜索第三级嵌套，如果至少存在一个列表，则尝试创建置换：

for ik in range(0,len(k_all_tbl)):
    k_all_tbl[ik] = unlevel(k_all_tbl[ik])
    for ikk in range(0,len(k_all_tbl[ik])):

            for ikkk in range(0,(len (k_all_tbl[ik][ikk]))):

                    for res1 in dl:
                        if isinstance(k_all_tbl[ik][ikk][ikkk], list)==False: 

                            if  k_all_tbl[ik][ikk][ikkk] == res1[1]:
                                k_all_tbl[ik][ikk][ikkk] = res1[0]


                        if len(k_all_tbl[ik][ikk][ikkk]) >1:
                            for ikkkk in range(0,(len (k_all_tbl[ik][ikk][ikkk]))):
                                if  k_all_tbl[ik][ikk][ikkk][ikkkk] == res1[1]:
                                    k_all_tbl[ik][ikk][ikkk][ikkkk] = res1 [0]



for kj in range(0,len(k_all_tbl)):
    for kj1 in range(0,len(k_all_tbl[kj])):
        for kz in range(0,len(k_all_tbl[kj][kj1])):
            
            if any(isinstance(i, list) for i in k_all_tbl[kj][kj1][kz])==True: #and any(isinstance(i, str) for i in k_all_tbl[kj][kj1])==True:
                k_all_tbl[kj][kj1][kz] = spreadm(unlistme(k_all_tbl[kj][kj1][kz]))

输出：

[[['Name', 'INSERT_3'],
  ['Product',
   [['Aegon', ' ', 'Level Life Cover'],
    ['Aegon', ' ', 'Family Income Benefit'],
    ['Aegon', ' ', 'Whole of Life Cover'],
    ['Aviva', ' ', 'Level Life Cover'],
    ['Aviva', ' ', 'Family Income Benefit'],
    ['Aviva', ' ', 'Whole of Life Cover'],
    ['HSBC', ' ', 'Level Life Cover'],
    ['HSBC', ' ', 'Family Income Benefit'],
    ['HSBC', ' ', 'Whole of Life Cover'],
    ['Zurich', ' ', 'Level Life Cover'],
    ['Zurich', ' ', 'Family Income Benefit'],
    ['Zurich', ' ', 'Whole of Life Cover']]],
  [['Sum assured', 'Benefit amount'],
   ['£ii_Field_52_1', '£ii_Field_53_2 a month']],
  ['Term', ['ii_Field_62_1 years', 'to age ii_Field_63_2', 'Whole of life']],
  ['Premium', ['£', 'INSERT_4', ' a month']],
  ['Payable on', ['Death', 'First death']],
  ['Trust recommended', ['Yes', 'No']]],
 [['Name', 'INSERT_9'],
   .....]

阅读有关调试代码的提示。@code学徒非常感谢您阅读文档，我可能希望有一种不那么复杂、混乱的手动方式，我可能错过了一个函数或在itertools中映射…粗略地看一下，您可能会从学习

pandas

中受益。这是一个Python库，通常用于存储和操作数据表。