Python Create a Dictionary of Lists

Question

Alright I searched, but I did not find an answer to my specific problem. I am creating a dictionary of lists. I have while loop creating a few lists of values as well as a name to associate with it:

dict = {} 
list = [1, 2]
name = key1
list = [3, 4]
name = key1

If the name is the same, the list will be appended to the existing set of values for the key. Here is the code I have:

if key in dict:
    dict[key].append(list)
else:
    dict[key] = list

This is the output I want:

dictionary = {'key1': [1, 2], [3, 4]}

However I keep getting this output:

dictionary = {'key1': [1, 2, [3, 4]]}

where the second list to the key is being put inside of the first list.

In the output you want you seem to have a key with 2 values. That's not going to happen. should it perhaps be dictionary = `{'key1': [[1, 2], [3, 4]]}` or h 2 values. That's not going to happen. should it perhaps be dictionary = `{'key1': [1, 2, 3, 4]}` — Leon, Apr 19 '15 at 19:51
don't name your variables as `list` and `dict` as they shadow the builtin — Bhargav Rao, Apr 19 '15 at 19:51

score 4 · Accepted Answer · edited May 23 '17 at 12:22

4

This is a very common error. I am fairly certain that you are doing:

list1.append(list2)

Instead, you want to do:

list1.extend(list2)

Here is a very useful resource

However, since you want [[1,2], [3,4]] instead of [1,2,3,4], you should do:

if key in d1:
    d1[key].append(l1)
else:
    d1[key] = [l1]

edited May 23 '17 at 12:22

Community

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1

answered Apr 19 '15 at 19:50

sshashank124

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That's a little better, but extend gives me: dict = {[1, 2, 3, 5]} instead of dict = {[1, 2], [3, 4]} – MarcAlex Apr 19 '15 at 19:58
1

@MarcAlex, What do you mean by `[1,2], [3,4]`? That is not possible. A dictionary cannot have multiple values for the same key. Did you mean `[[1,2], [3,4]]` – sshashank124 Apr 19 '15 at 20:02
Yeah, I guess I did mean to create a list of lists in the dictionary. So I would nee [[1, 2], [3, 4]] in the dictionary. – MarcAlex Apr 19 '15 at 20:04

score 2 · Answer 2 · answered Apr 19 '15 at 19:50

2

That is because you are appending a list to a list each time. You need to use extend instead. Code:

keys = ['name1', 'name2', 'name2', 'name1']
somelist = [1, 2, 3]
d = {}
for k in keys:
   d.setdefault(k, []).extend(somelist)

answered Apr 19 '15 at 19:50

Saksham Varma

2,122
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Lynch · Answer 3 · 2015-04-19T20:11:32.773

You need a list of list in fact, your output will look like this:

dictionary = {'key1': [[1, 2], [3, 4]]}

To have a key associated to multiple values you could use this line:

dictionary.setdefault(key, []).append(a_list)

setdefault will associate the key to the default value [] if the key is not present in your dictionary.

Also you should avoid using dict or list to define your own variables, they are built-in and you are redefining them.

Edit

To make it obvious to the reader maybe this could help, its the output of an interative python session:

>>> d = {}
>>> d.setdefault('key1', []).append([1, 2])
>>> d.setdefault('key1', []).append([3, 4])
>>> d
{'key1': [[1, 2], [3, 4]]}

dictionary = {} .... dictionary.setdefault("key", []).append([1,2]) .... i like this :) — Jose Ricardo Bustos M., Apr 19 '15 at 20:09

Jose Ricardo Bustos M. · Answer 4 · 2015-04-19T20:04:37.683

1

I hope not misunderstand , since dictionary = {'key1': [1, 2], [3, 4]} is poorly expressed

def add2dict(d, l, k):
  if not k in d:
    dict[k] = []
  d[k].append(l)

dict = {} 
add2dict(dict, [1,2], "key1")
add2dict(dict, [3,4], "key1")

result in dict:

{'key1': [[1, 2], [3, 4]]}

edited Apr 19 '15 at 20:04

answered Apr 19 '15 at 19:58

Jose Ricardo Bustos M.

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Python Create a Dictionary of Lists

4 Answers4