How to Get List of Values from Dict

How can I get list of values from dict?

dict.values returns a view of the dictionary's values, so you have to wrap it in list:

list(d.values())

Getting a list of values from a list of dicts

Assuming every dict has a value key, you can write (assuming your list is named l)

[d['value'] for d in l]

If value might be missing, you can use

[d['value'] for d in l if 'value' in d]

How can I extract all values from a dictionary in Python?

If you only need the dictionary keys 1, 2, and 3 use: your_dict.keys().

If you only need the dictionary values -0.3246, -0.9185, and -3985 use: your_dict.values().

If you want both keys and values use: your_dict.items() which returns a list of tuples [(key1, value1), (key2, value2), ...].

Dictionary: Get list of values for list of keys

A list comprehension seems to be a good way to do this:

>>> [mydict[x] for x in mykeys]
[3, 1]

How to get multiple dictionary values?

Use a for loop:

keys = ['firstKey', 'secondKey', 'thirdKey']
for key in keys:
    myDictionary.get(key)

or a list comprehension:

[myDictionary.get(key) for key in keys]

How to get values from a list of dictionaries, which themselves contain lists of dictionaries in Python

First: dict's require a key-value association for every element in the dictionary. Your 2nd level data structure though does not include keys: ({[{'tag': 'tag 1'}]}) This is a set. Unlike dict's, set's do not have keys associated with their elements. So your data structure looks like List[Set[List[Dict[str, str]]]].

Second: when I try to run

# python 3.8.8
player_info = [{[{'tag': 'tag 1'}]},
               {[{'tag': 'tag 2'}]}]

I recieve the error TypeError: unhashable type: 'list'. That's because you're code attempts to contain a list inside a set. Set membership in python demands the members to be hashable. However, you will not find a __hash__() function defined on list objects. Even if you resolve this by replacing the list with a tuple, you will find that dict objects are not hashable either. Potential solutions include using immutable objects like frozendict or tuple, but that is another post.

To answer your question, I have reformulated your problem as

player_info = [[[{'tag': 'tag 1'}]],
               [[{'tag': 'tag 2'}]]]

and compared the performance difference with A) explicit loops:

for i in range(len(player_info)):
  print(player_info[i][0][0]['tag'])

against B) list comprehension

[
  print(single_player_info[0][0]['tag']) 
  for single_player_info in player_info
]

Running the above code blocks in jupyter with the %%timeit cell magic, I got:
A) 154 µs ± 14.6 µs per loop (mean ± std. dev. of 7 runs, 10,000 loops each) and
B) 120 µs ± 11 µs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)

Note: This experiment is highly skewed for at least two reasons:

1. I tested both trials using only the data you provided (N=2). It is very likely that we would observe different scaling behaviors than initial conditions suggest.
2. print consumes a lot of time and makes this problem heavily subject to the status of the kernel

I hope this answers your question.

Converting Dictionary to List?

Your problem is that you have key and value in quotes making them strings, i.e. you're setting aKey to contain the string "key" and not the value of the variable key. Also, you're not clearing out the temp list, so you're adding to it each time, instead of just having two items in it.

To fix your code, try something like:

for key, value in dict.iteritems():
    temp = [key,value]
    dictlist.append(temp)

You don't need to copy the loop variables key and value into another variable before using them so I dropped them out. Similarly, you don't need to use append to build up a list, you can just specify it between square brackets as shown above. And we could have done dictlist.append([key,value]) if we wanted to be as brief as possible.

Or just use dict.items() as has been suggested.

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