列表到嵌套列表和字典

Question

I have this one long list and want to convert it to a nested list and a dictionary.

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

output:

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] } 

Can anyone tell me how to do that in python ? And I can't import anything

Answer 1

I assume the groups are separated by the empty strings. For this you can use 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)

Results:

[['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']}

In the groupby I'm using ''.__ne__ which is the function for "not equal" of an empty string. This way it's only capturing groups of non-empty strings.

EDIT

I just read that you cannot import. Here's a solution just using a loop:

nl = [[]]

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

And for the dict:

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

Here's how to convert L to a nested list:

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)

And here's how to convert L to a dictionary:

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

From my understanding, you are trying to:

1. Split a list by empty string, then
2. Convert the resulting nested list into a dictionary, using first element of each sub-list as the key and the rest as value.

You can certainly accomplish the task without any imports. To split a list, just iterate over it and build the nested list along the way:

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

Then, again, iterate over this nested list to build your desired dictionary:

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

Compose the two functions to get what you want:

>>> 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']}