CODEWITHSUNDEEP
python
Liondancer
Liondancer

Reputation: 16469

displaying list value inside dictionary inside list

I have a data set that is a list of dictionaries that have a value of a list.

ex:

a = 
[
   {'node1':[1,2,3]},...,..
]

I was wondering how can I access the 1 or 2 inside node1? I know that if I did a[0][0] I would display {'node1':[1,2,3]}. I thought a[0][0][0] would display 1 but I am getting the error:

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
KeyError: 0

Upvotes: 2

Views: 103

Answers (5)

Dineshs91
Dineshs91

Reputation: 2204

a = [ {'node1':[1, 2, 3]} ] 

>>> a[0]['node1'][0]

a[0] selects the first element in the list which is the map {'node1': [1, 2, 3]}. Then a[0]['node1'] selects the value of the map with key 'node1'. Finally a[0]['node1'][0] selects the first element from the list which is the value for the key from map. This gives 1.

for getting the value 2

>>> a[0]['node1'[1]

Change the index accordingly.

Upvotes: 1

Danstahr
Danstahr

Reputation: 4319

a[number] is one of your dictionaries, i.e. {'node1': [1, 2, 3], 'node2': [4, 5, 6]}.

Now, a[number]['node1'] returns the list belonging to 'node1', i.e. [1, 2, 3]. You can access the list normally.

So the complete expression would be 

a[0]['node1'][0]
>>> 1

Upvotes: 4

leeladam
leeladam

Reputation: 1758

Try a[0]['node1'][0]. You can refer to an item in dict by its key (which is 'node1' in your case). Indexes like [0] do not make sense in dictionaries, as their items don't have a specified order. Note that your a[0][0] example should also result an error. a[0] is what returns the first dict in your list: {'node1':[1,2,3]}

Upvotes: 3

qwertynl
qwertynl

Reputation: 3933

You can try doing this: 

a[0]['node1'][...]

Upvotes: 0

mdml
mdml

Reputation: 22882

You need to index into the dictionary using the name of the key, in this case 'node1':

>>> a = [{'node1':[1,2,3]}]
>>> a[0]['node1'][0]
1

Upvotes: 3

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