Question

我有一个长长的列表，想将其转换为嵌套列表和字典。

L= ["a","abc","de","efg","", "b","ijk","lm","op","qr","", "c","123","45","6789"]

输出：

nested list:
[["a","abc","de","efg"], ["b","ijk","lm","op","qr"], ["c","123","45","6789"]] 

dictionary: 
{"a":["abc","de","efg"],
"b":["ijk","lm","op","qr"], "c":["123","45","6789] }

有人可以告诉我如何在 python 中做到这一点吗？而且我什么也不能导入

Answer 1

我假设这些组由空字符串分隔。为此，您可以使用itertools.groupby：

from itertools import groupby

data = ["a","abc","de","efg","", "b","ijk","lm","op","qr","", "c","123","45","6789"] 

nl = [list(g) for k, g in groupby(data, ''.__ne__) if k]
d = {next(g): list(g) for k, g in groupby(data, ''.__ne__) if k}

print(nl)
print(d)

结果：

[['a', 'abc', 'de', 'efg'], ['b', 'ijk', 'lm', 'op', 'qr'], ['c', '123', '45', '6789']]
{'a': ['abc', 'de', 'efg'], 'b': ['ijk', 'lm', 'op', 'qr'], 'c': ['123', '45', '6789']}

在分组方式中，我使用的是''.__ne__，该功能用于“不等于”一个空字符串。这样，它仅捕获非空字符串组。

编辑

我刚刚读到您无法导入。这是一个使用循环的解决方案：

nl = [[]]

for s in data:
    if s:
        nl[-1].append(s)
    else:
        nl.append([])

对于字典：

itr = iter(data)
key = next(itr)
d = {key: []}

while True:
    try: val = next(itr)
    except StopIteration: break
    if val:
        d[key].append(val)
    else:
        key = next(itr)
        d[key] = []

Answer 2

以下是将L转换为嵌套列表的方法：

L= ["a","abc","de","efg","","b","ijk","lm","op","qr","","c","123","45","6789"] 

nested_list_L = []

temp = []
for item in L:
    if item != "":
        temp.append(item)
    else:
        nested_list_L.append(temp)
        temp = []
        
nested_list_L.append(temp)

以下是将L转换为字典的方法：

L= ["a","abc","de","efg","","b","ijk","lm","op","qr","","c","123","45","6789"] 

dict_L = {}

temp = []
key = ""
for item in L:
    if len(item) == 1:
        key = item
    elif len(item) > 1:
        temp.append(item)
    else:
        dict_L[key] = temp
        temp = []
        key = ""
        
dict_L[key] = temp

Answer 3

据我了解，您正在尝试：

用空字符串分割列表，然后
使用每个子列表的第一个元素作为键，其余的作为值，将生成的嵌套列表转换为字典。

您当然可以完成任务，而无需任何导入。要拆分列表，只需对其进行迭代并按照以下方式构建嵌套列表：

def split(data, on):
    nested = []
    curr = []
    for x in data:
        if x == on:
            nested.append(curr)
            curr = []
        else:
            curr.append(x)
    if curr != [] or data[-1:] == [on]:
        nested.append(curr)
    return nested

然后再次遍历此嵌套列表以构建所需的字典：

def build_dict(key_valss):
    d = {}
    for key_vals in key_valss:
        if key_vals != []:
            key = key_vals[0]
            vals = key_vals[1:]
            d[key] = vals
    return d

组合两个函数以获取所需的内容：

>>> build_dict( split(data = ["a","abc","de","efg","", "b","ijk","lm","op","qr","", "c","123","45","6789"] , on = '') )
{'a': ['abc', 'de', 'efg'], 'b': ['ijk', 'lm', 'op', 'qr'], 'c': ['123', '45', '6789']}

列表到嵌套列表和字典

3 个答案: