【发布时间】：2020-11-23 23:44:01
【问题描述】：
抱歉冗长的解释，我有两个嵌套列表，一个包含一系列字段的参考值，称为dl，另一个包含字段名称，称为k_all_tbl。它们看起来像下面这样：
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 列表的长度未知 n，但每个元素 dl[i] 中嵌套了 2 个列表，第一个 (dl[i][0]) 可能替换第二个元素 (dl[i][1])。
k_all_tbl 看起来像这样：
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）也有未知长度m，但每个元素k_all_tbll[j]（每一行）有2个嵌套在其中的列表，有（k_all_tbll[j][0]）和（k_all_tbll[j][1]）（column1和原始表的第 2 列）正在链接。
要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']]]
但是我在下面详细分享的代码和步骤并没有让我在这里，它在没有执行的情况下崩溃，或者给了我错误的排列。任何和所有的帮助将不胜感激。详细解释和代码尝试如下。
尝试的方法：
我首先尝试：对于k_all_tbl[j][0][k] 中与元素dl[i][1] 对应的每个元素，将k_all_tbl[j][0][k] 替换为(dl[i][0])。输出示例为：
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目标：
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用于访问单个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，并为k_all_tbl 中的每个元素创建“产品”，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']]]
我在上面代码的底部尝试了以下两个函数来尝试实现这一点：
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)
【问题讨论】：
-
阅读this article 了解调试代码的技巧。
-
@Code-Apprentice 非常感谢会阅读文档，我可能希望手动执行此操作的方式更少循环、混乱，我可能错过了一个功能或在 itertools 中映射...
-
粗略一瞥，您可能会受益于学习
pandas。这是一个 Python 库，通常用于存储和操作数据表。
标签： python list nested permutation