In Python can you add elements from a list to the values of a dictionary?

I am trying to add the elements of a list of lists to the values of a dictionary.

I have created a list with elements from a file that looks like this:

list_of_lists = [[966], [1513, 2410], [2964, 1520, 604]....]

I am trying to add this list to a dictionary that I have made to look like this:

{'Youngstown': ['OH', 4110, 8065, 115436], 'Yankton': ['SD', 4288, 9739, 
12011], 'Yakima': ['WA', 4660, 12051, 49826]....]

I have tried the following code:

 x = 1
 for x in d2.values():
     d2.append(list_of_list)
 print(d2)

I am not even sure that this is something that is possible, but I am trying to get the dictionary to be:

{'Youngstown': ['OH', 4110, 8065, 115436], 'Yankton': ['SD', 4288, 9739, 
12011, [966]], 'Yakima': ['WA', 4660, 12051, 49826, [1513, 2410]]....]

How can I accomplish this?

Answers:

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Method 1

I know there are more ways to do that, But I think this is more readable and understandable code.

list_of_lists = [[966], [1513, 2410], [2964, 1520, 604]] 

dict_ = {'Youngstown': ['OH', 4110, 8065, 115436], 'Yankton': ['SD', 4288, 9739, 
12011], 'Yakima': ['WA', 4660, 12051, 49826]}

i = 0
# list(dict_.items())[1:] is the list of all keys and values except first one.
for key,value in list(dict_.items())[1:]:
    dict_[key] = value+[list_of_lists[i]]
    i+=1
print(dict_)

Method 2

You do x = 1, and then immediately you do for x in d2.values(). This overwrites x with each element of d2.values(). If you want to start from the second item in d2.values(), you need to do create an iterator and skip the first value:

d2_iter = iter(d2.values())
next(d2_iter) # Consume one element
for item in d2_iter: # Iterate over the remaining iterator
    # Do what you want here.

Another problem is that you append the entire list-of-lists to every value in d2. Don’t do that. Instead use zip() to iterate over the list-of-lists and the values in d2 simultaneously

d2_iter = iter(d2.values())
next(d2_iter) # Consume one element
for item_from_dict, item_to_append in zip(d2_iter, list_of_lists): 
    item_from_dict.append(item_to_append)

which leaves you with:

{'Youngstown': ['OH', 4110, 8065, 115436],
 'Yankton': ['SD', 4288, 9739, 12011, [966]],
 'Yakima': ['WA', 4660, 12051, 49826, [1513, 2410]]}

Note that appending like this only works because lists are mutable. If you had an immutable type like tuples as the values of d2, you’d have to create a new tuple and assign it to the key:

d3 = {'Youngstown': ('OH', 4110, 8065, 115436), 'Yankton': ('SD', 4288, 9739, 12011), 'Yakima': ('WA', 4660, 12051, 49826)}

d3_iter = iter(d2.keys())
next(d3_iter) # Consume one element

for key_from_dict, item_to_append in zip(d3_iter, list_of_lists): 
    new_item = d3[key_from_dict] + (item_to_append,) # Create a new item
    d3[key_from_dict] = new_item

and you’d get

{'Youngstown': ('OH', 4110, 8065, 115436),
 'Yankton': ('SD', 4288, 9739, 12011, [966]),
 'Yakima': ('WA', 4660, 12051, 49826, [1513, 2410])}

Method 3

You can use itertools.islice() to skip the first element, and then use zip() to pair each list value with the list to append:

from itertools import islice
for lst_value, lst_to_append in zip(islice(d2.values(), 1, None), list_of_lists):
    lst_value.append(lst_to_append)
    
print(d2)

This outputs:

{
 'Youngstown': ['OH', 4110, 8065, 115436],
 'Yankton': ['SD', 4288, 9739, 12011, [966]],
 'Yakima': ['WA', 4660, 12051, 49826, [1513, 2410]]
}

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0