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How do I access values from dictionary randomly in python3? [duplicate]

How can I get a random pair from a dict? I'm making a game where you need to guess a capital of a country and I need questions to appear randomly.
The dict looks like {'VENEZUELA':'CARACAS'}
How can I do this?

One way would be:
import random
d = {'VENEZUELA':'CARACAS', 'CANADA':'OTTAWA'}
random.choice(list(d.values()))
EDIT: The question was changed a couple years after the original post, and now asks for a pair, rather than a single item. The final line should now be:
country, capital = random.choice(list(d.items()))

I wrote this trying to solve the same problem:
https://github.com/robtandy/randomdict
It has O(1) random access to keys, values, and items.

If you don't want to use the random module, you can also try popitem():
>> d = {'a': 1, 'b': 5, 'c': 7}
>>> d.popitem()
('a', 1)
>>> d
{'c': 7, 'b': 5}
>>> d.popitem()
('c', 7)
Since the dict doesn't preserve order, by using popitem you get items in an arbitrary (but not strictly random) order from it.
Also keep in mind that popitem removes the key-value pair from dictionary, as stated in the docs.
popitem() is useful to destructively iterate over a dictionary

>>> import random
>>> d = dict(Venezuela = 1, Spain = 2, USA = 3, Italy = 4)
>>> random.choice(d.keys())
'Venezuela'
>>> random.choice(d.keys())
'USA'
By calling random.choice on the keys of the dictionary (the countries).

Try this:
import random
a = dict(....) # a is some dictionary
random_key = random.sample(a, 1)[0]
This definitely works.

This works in Python 2 and Python 3:
A random key:
random.choice(list(d.keys()))
A random value
random.choice(list(d.values()))
A random key and value
random.choice(list(d.items()))

Since the original post wanted the pair:
import random
d = {'VENEZUELA':'CARACAS', 'CANADA':'TORONTO'}
country, capital = random.choice(list(d.items()))
(python 3 style)

If you don't want to use random.choice() you can try this way:
>>> list(myDictionary)[i]
'VENEZUELA'
>>> myDictionary = {'VENEZUELA':'CARACAS', 'IRAN' : 'TEHRAN'}
>>> import random
>>> i = random.randint(0, len(myDictionary) - 1)
>>> myDictionary[list(myDictionary)[i]]
'TEHRAN'
>>> list(myDictionary)[i]
'IRAN'

When they ask for a random pair here they mean a key and value.
For such a dict where the key:values are country:city,
use random.choice().
Pass the dictionary keys to this function as follows:
import random
keys = list(my_dict)
country = random.choice(keys)
You may wish to track the keys that were already called in a round and when getting a fresh country, loop until the random selection is not in the list of those already "drawn"... as long as the drawn list is shorter than the keys list.

Since this is homework:
Check out random.sample() which will select and return a random element from an list. You can get a list of dictionary keys with dict.keys() and a list of dictionary values with dict.values().

I am assuming that you are making a quiz kind of application. For this kind of application I have written a function which is as follows:
def shuffle(q):
"""
The input of the function will
be the dictionary of the question
and answers. The output will
be a random question with answer
"""
selected_keys = []
i = 0
while i < len(q):
current_selection = random.choice(q.keys())
if current_selection not in selected_keys:
selected_keys.append(current_selection)
i = i+1
print(current_selection+'? '+str(q[current_selection]))
If I will give the input of questions = {'VENEZUELA':'CARACAS', 'CANADA':'TORONTO'} and call the function shuffle(questions) Then the output will be as follows:
VENEZUELA? CARACAS
CANADA? TORONTO
You can extend this further more by shuffling the options also

With modern versions of Python(since 3), the objects returned by methods dict.keys(), dict.values() and dict.items() are view objects*. And hey can be iterated, so using directly random.choice is not possible as now they are not a list or set.
One option is to use list comprehension to do the job with random.choice:
import random
colors = {
'purple': '#7A4198',
'turquoise':'#9ACBC9',
'orange': '#EF5C35',
'blue': '#19457D',
'green': '#5AF9B5',
'red': ' #E04160',
'yellow': '#F9F985'
}
color=random.choice([hex_color for color_value in colors.values()]
print(f'The new color is: {color}')
References:
*Python 3.8: Standard Library Documentation - Built-in types: Dictionary view objects
Python 3.8: Data Structures - List Comprehensions:

I just stumbled across a similar problem and designed the following solution (relevant function is pick_random_item_from_dict; other functions are just for completeness).
import random
def pick_random_key_from_dict(d: dict):
"""Grab a random key from a dictionary."""
keys = list(d.keys())
random_key = random.choice(keys)
return random_key
def pick_random_item_from_dict(d: dict):
"""Grab a random item from a dictionary."""
random_key = pick_random_key_from_dict(d)
random_item = random_key, d[random_key]
return random_item
def pick_random_value_from_dict(d: dict):
"""Grab a random value from a dictionary."""
_, random_value = pick_random_item_from_dict(d)
return random_value
# Usage
d = {...}
random_item = pick_random_item_from_dict(d)
The main difference from previous answers is in the way we handle the dictionary copy with list(d.items()). We can partially circumvent that by only making a copy of d.keys() and using the random key to pick its associated value and create our random item.

Try this (using random.choice from items)
import random
a={ "str" : "sda" , "number" : 123, 55 : "num"}
random.choice(list(a.items()))
# ('str', 'sda')
random.choice(list(a.items()))[1] # getting a value
# 'num'

To select 50 random key values from a dictionary set dict_data:
sample = random.sample(set(dict_data.keys()), 50)

I needed to iterate through ranges of keys in a dict without sorting it each time and found the Sorted Containers library. I discovered that this library enables random access to dictionary items by index which solves this problem intuitively and without iterating through the entire dict each time:
>>> import sortedcontainers
>>> import random
>>> d = sortedcontainers.SortedDict({1: 'a', 2: 'b', 3: 'c'})
>>> random.choice(d.items())
(1, 'a')
>>> random.sample(d.keys(), k=2)
[1, 3]

I found this post by looking for a rather comparable solution. For picking multiple elements out of a dict, this can be used:
idx_picks = np.random.choice(len(d), num_of_picks, replace=False) #(Don't pick the same element twice)
result = dict ()
c_keys = [d.keys()] #not so efficient - unfortunately .keys() returns a non-indexable object because dicts are unordered
for i in idx_picks:
result[c_keys[i]] = d[i]

Here is a little Python code for a dictionary class that can return random keys in O(1) time. (I included MyPy types in this code for readability):
from typing import TypeVar, Generic, Dict, List
import random
K = TypeVar('K')
V = TypeVar('V')
class IndexableDict(Generic[K, V]):
def __init__(self) -> None:
self.keys: List[K] = []
self.vals: List[V] = []
self.dict: Dict[K, int] = {}
def __getitem__(self, key: K) -> V:
return self.vals[self.dict[key]]
def __setitem__(self, key: K, val: V) -> None:
if key in self.dict:
index = self.dict[key]
self.vals[index] = val
else:
self.dict[key] = len(self.keys)
self.keys.append(key)
self.vals.append(val)
def __contains__(self, key: K) -> bool:
return key in self.dict
def __len__(self) -> int:
return len(self.keys)
def random_key(self) -> K:
return self.keys[random.randrange(len(self.keys))]

b = { 'video':0, 'music':23,"picture":12 }
random.choice(tuple(b.items())) ('music', 23)
random.choice(tuple(b.items())) ('music', 23)
random.choice(tuple(b.items())) ('picture', 12)
random.choice(tuple(b.items())) ('video', 0)

Dividing dictionary into nested dictionaries, based on the key's name on Python 3.4

I have the following dictionary (short version, real data is much larger):
dict = {'C-STD-B&M-SUM:-1': 0, 'C-STD-B&M-SUM:-10': 4.520475, 'H-NSW-BAC-ART:-9': 0.33784000000000003, 'H-NSW-BAC-ART:0': 0, 'H-NSW-BAC-ENG:-59': 0.020309999999999998, 'H-NSW-BAC-ENG:-6': 0,}
I want to divide it into smaller nested dictionaries, depending on a part of the key name.
Expected output would be:
# fixed closing brackets
dict1 = {'C-STD-B&M-SUM: {'-1': 0, '-10': 4.520475}}
dict2 = {'H-NSW-BAC-ART: {'-9': 0.33784000000000003, '0': 0}}
dict3 = {'H-NSW-BAC-ENG: {'-59': 0.020309999999999998, '-6': 0}}
Logic behind is:
dict1: if the part of the key name is 'C-STD-B&M-SUM', add to dict1.
dict2: if the part of the key name is 'H-NSW-BAC-ART', add to dict2.
dict3: if the part of the key name is 'H-NSW-BAC-ENG', add to dict3.
Partial code so far:
def divide_dictionaries(dict):
c_std_bem_sum = {}
for k, v in dict.items():
if k[0:13] == 'C-STD-B&M-SUM':
c_std_bem_sum = k[14:17], v
What I'm trying to do is to create the nested dictionaries that I need and then I'll create the dictionary and add the nested one to it, but I'm not sure if it's a good way to do it.
When I run the code above, the variable c_std_bem_sum becomes a tuple, with only two values that are changed at each iteration. How can I make it be a dictionary, so I can later create another dictionary, and use this one as the value for one of the keys?

One way to approach it would be to do something like
d = {'C-STD-B&M-SUM:-1': 0, 'C-STD-B&M-SUM:-10': 4.520475, 'H-NSW-BAC-ART:-9': 0.33784000000000003, 'H-NSW-BAC-ART:0': 0, 'H-NSW-BAC-ENG:-59': 0.020309999999999998, 'H-NSW-BAC-ENG:-6': 0,}
def divide_dictionaries(somedict):
out = {}
for k,v in somedict.items():
head, tail = k.split(":")
subdict = out.setdefault(head, {})
subdict[tail] = v
return out
which gives
>>> dnew = divide_dictionaries(d)
>>> import pprint
>>> pprint.pprint(dnew)
{'C-STD-B&M-SUM': {'-1': 0, '-10': 4.520475},
'H-NSW-BAC-ART': {'-9': 0.33784000000000003, '0': 0},
'H-NSW-BAC-ENG': {'-59': 0.020309999999999998, '-6': 0}}
A few notes:
(1) We're using nested dictionaries instead of creating separate named dictionaries, which aren't convenient.
(2) We used setdefault, which is a handy way to say "give me the value in the dictionary, but if there isn't one, add this to the dictionary and return it instead.". Saves an if.
(3) We can use .split(":") instead of hardcoding the width, which isn't very robust -- at least assuming that's the delimiter, anyway!
(4) It's a bad idea to use dict, the name of a builtin type, as a variable name.

That's because you're setting your dictionary and overriding it with a tuple:
>>> a = 1, 2
>>> print a
>>> (1,2)
Now for your example:
>>> def divide_dictionaries(dict):
>>> c_std_bem_sum = {}
>>> for k, v in dict.items():
>>> if k[0:13] == 'C-STD-B&M-SUM':
>>> new_key = k[14:17] # sure you don't want [14:], open ended?
>>> c_std_bem_sum[new_key] = v
Basically, this grabs the rest of the key (or 3 characters, as you have it, the [14:None] or [14:] would get the rest of the string) and then uses that as the new key for the dict.

Get name of dictionary

I find myself needing to iterate over a list made of dictionaries and I need, for every iteration, the name of which dictionary I'm iterating on.
Here's an MRE (Minimal Reproducible Example).
Contents of the dicts are irrelevant:
dict1 = {...}
dicta = {...}
dict666 = {...}
dict_list = [dict1, dicta, dict666]
for dc in dict_list:
# Insert command that should replace ???
print 'The name of the dictionary is: ', ???
If I just use dc where ??? is, it will print the entire contents of the dictionary. How can I get the name of the dictionary being used?

Don't use a dict_list, use a dict_dict if you need their names. In reality, though, you should really NOT be doing this. Don't embed meaningful information in variable names. It's tough to get.
dict_dict = {'dict1':dict1, 'dicta':dicta, 'dict666':dict666}
for name,dict_ in dict_dict.items():
print 'the name of the dictionary is ', name
print 'the dictionary looks like ', dict_
Alternatively make a dict_set and iterate over locals() but this is uglier than sin.
dict_set = {dict1,dicta,dict666}
for name,value in locals().items():
if value in dict_set:
print 'the name of the dictionary is ', name
print 'the dictionary looks like ', value
Again: uglier than sin, but it DOES work.

You should also consider adding a "name" key to each dictionary.
The names would be:
for dc in dict_list:
# Insert command that should replace ???
print 'The name of the dictionary is: ', dc['name']

Objects don't have names in Python, a name is an identifier that can be assigned to an object, and multiple names could be assigned to the same one.
However, an object-oriented way to do what you want would be to subclass the built-in dict dictionary class and add a name property to it. Instances of it would behave exactly like normal dictionaries and could be used virtually anywhere a normal one could be.
class NamedDict(dict):
def __init__(self, *args, **kwargs):
try:
self._name = kwargs.pop('name')
except KeyError:
raise KeyError('a "name" keyword argument must be supplied')
super(NamedDict, self).__init__(*args, **kwargs)
#classmethod
def fromkeys(cls, name, seq, value=None):
return cls(dict.fromkeys(seq, value), name=name)
#property
def name(self):
return self._name
dict_list = [NamedDict.fromkeys('dict1', range(1,4)),
NamedDict.fromkeys('dicta', range(1,4), 'a'),
NamedDict.fromkeys('dict666', range(1,4), 666)]
for dc in dict_list:
print 'the name of the dictionary is ', dc.name
print 'the dictionary looks like ', dc
Output:
the name of the dictionary is dict1
the dictionary looks like {1: None, 2: None, 3: None}
the name of the dictionary is dicta
the dictionary looks like {1: 'a', 2: 'a', 3: 'a'}
the name of the dictionary is dict666
the dictionary looks like {1: 666, 2: 666, 3: 666}

If you want to read name and value
dictionary={"name1":"value1","name2":"value2","name3":"value3","name4":"value4"}
for name,value in dictionary.items():
print(name)
print(value)
If you want to read name only
dictionary={"name1":"value1","name2":"value2","name3":"value3","name4":"value4"}
for name in dictionary:
print(name)
If you want to read value only
dictionary={"name1":"value1","name2":"value2","name3":"value3","name4":"value4"}
for values in dictionary.values():
print(values)
Here is your answer
dic1 = {"dic":1}
dic2 = {"dic":2}
dic3 = {"dic":3}
dictionaries = [dic1,dic2,dic3]
for i in range(len(dictionaries)):
my_var_name = [ k for k,v in locals().items() if v == dictionaries[i]][0]
print(my_var_name)

The following doesn't work on standard dictionaries, but does work just fine with collections dictionaries and counters:
from collections import Counter
# instantiate Counter ditionary
test= Counter()
# create an empty name attribute field
test.name = lambda: None
# set the "name" attribute field to "name" = "test"
setattr(test.name, 'name', 'test')
# access the nested name field
print(test.name.name)
It's not the prettiest solution, but it is easy to implement and access.

Here's my solution for a descriptive error message.
def dict_key_needed(dictionary,key,msg='dictionary'):
try:
value = dictionary[key]
return value
except KeyError:
raise KeyError(f"{msg} is missing key '{key}'")

Setting a value in a nested Python dictionary given a list of indices and value

I'm trying to programmatically set a value in a dictionary, potentially nested, given a list of indices and a value.
So for example, let's say my list of indices is:
['person', 'address', 'city']
and the value is
'New York'
I want as a result a dictionary object like:
{ 'Person': { 'address': { 'city': 'New York' } }
Basically, the list represents a 'path' into a nested dictionary.
I think I can construct the dictionary itself, but where I'm stumbling is how to set the value. Obviously if I was just writing code for this manually it would be:
dict['Person']['address']['city'] = 'New York'
But how do I index into the dictionary and set the value like that programmatically if I just have a list of the indices and the value?
Python

Something like this could help:
def nested_set(dic, keys, value):
for key in keys[:-1]:
dic = dic.setdefault(key, {})
dic[keys[-1]] = value
And you can use it like this:
>>> d = {}
>>> nested_set(d, ['person', 'address', 'city'], 'New York')
>>> d
{'person': {'address': {'city': 'New York'}}}

I took the freedom to extend the code from the answer of Bakuriu. Therefore upvotes on this are optional, as his code is in and of itself a witty solution, which I wouldn't have thought of.
def nested_set(dic, keys, value, create_missing=True):
d = dic
for key in keys[:-1]:
if key in d:
d = d[key]
elif create_missing:
d = d.setdefault(key, {})
else:
return dic
if keys[-1] in d or create_missing:
d[keys[-1]] = value
return dic
When setting create_missing to True, you're making sure to only set already existing values:
# Trying to set a value of a nonexistent key DOES NOT create a new value
print(nested_set({"A": {"B": 1}}, ["A", "8"], 2, False))
>>> {'A': {'B': 1}}
# Trying to set a value of an existent key DOES create a new value
print(nested_set({"A": {"B": 1}}, ["A", "8"], 2, True))
>>> {'A': {'B': 1, '8': 2}}
# Set the value of an existing key
print(nested_set({"A": {"B": 1}}, ["A", "B"], 2))
>>> {'A': {'B': 2}}

Here's another option:
from collections import defaultdict
recursivedict = lambda: defaultdict(recursivedict)
mydict = recursivedict()
I originally got this from here: Set nested dict value and create intermediate keys.
It is quite clever and elegant if you ask me.

First off, you probably want to look at setdefault.
As a function I'd write it as
def get_leaf_dict(dct, key_list):
res=dct
for key in key_list:
res=res.setdefault(key, {})
return res
This would be used as:
get_leaf_dict( dict, ['Person', 'address', 'city']) = 'New York'
This could be cleaned up with error handling and such. Also using *args rather than a single key-list argument might be nice; but the idea is that
you can iterate over the keys, pulling up the appropriate dictionary at each level.

Here is my simple solution: just write
terms = ['person', 'address', 'city']
result = nested_dict(3, str)
result[terms] = 'New York' # as easy as it can be
You can even do:
terms = ['John', 'Tinkoff', '1094535332'] # account in Tinkoff Bank
result = nested_dict(3, float)
result[terms] += 2375.30
Now the backstage:
from collections import defaultdict
class nesteddict(defaultdict):
def __getitem__(self, key):
if isinstance(key, list):
d = self
for i in key:
d = defaultdict.__getitem__(d, i)
return d
else:
return defaultdict.__getitem__(self, key)
def __setitem__(self, key, value):
if isinstance(key, list):
d = self[key[:-1]]
defaultdict.__setitem__(d, key[-1], value)
else:
defaultdict.__setitem__(self, key, value)
def nested_dict(n, type):
if n == 1:
return nesteddict(type)
else:
return nesteddict(lambda: nested_dict(n-1, type))

The dotty_dict library for Python 3 can do this. See documentation, Dotty Dict for more clarity.
from dotty_dict import dotty
dot = dotty()
string = '.'.join(['person', 'address', 'city'])
dot[string] = 'New York'
print(dot)
Output:
{'person': {'address': {'city': 'New York'}}}

Use these pair of methods
def gattr(d, *attrs):
"""
This method receives a dict and list of attributes to return the innermost value of the give dict
"""
try:
for at in attrs:
d = d[at]
return d
except:
return None
def sattr(d, *attrs):
"""
Adds "val" to dict in the hierarchy mentioned via *attrs
For ex:
sattr(animals, "cat", "leg","fingers", 4) is equivalent to animals["cat"]["leg"]["fingers"]=4
This method creates necessary objects until it reaches the final depth
This behaviour is also known as autovivification and plenty of implementation are around
This implementation addresses the corner case of replacing existing primitives
https://gist.github.com/hrldcpr/2012250#gistcomment-1779319
"""
for attr in attrs[:-2]:
# If such key is not found or the value is primitive supply an empty dict
if d.get(attr) is None or isinstance(d.get(attr), dict):
d[attr] = {}
d = d[attr]
d[attrs[-2]] = attrs[-1]

Here's a variant of Bakuriu's answer that doesn't rely on a separate function:
keys = ['Person', 'address', 'city']
value = 'New York'
nested_dict = {}
# Build nested dictionary up until 2nd to last key
# (Effectively nested_dict['Person']['address'] = {})
sub_dict = nested_dict
for key_ind, key in enumerate(keys[:-1]):
if not key_ind:
# Point to newly added piece of dictionary
sub_dict = nested_dict.setdefault(key, {})
else:
# Point to newly added piece of sub-dictionary
# that is also added to original dictionary
sub_dict = sub_dict.setdefault(key, {})
# Add value to last key of nested structure of keys
# (Effectively nested_dict['Person']['address']['city'] = value)
sub_dict[keys[-1]] = value
print(nested_dict)
>>> {'Person': {'address': {'city': 'New York'}}}

This is a pretty good use case for a recursive function. So you can do something like this:
def parse(l: list, v: str) -> dict:
copy = dict()
k, *s = l
if len(s) > 0:
copy[k] = parse(s, v)
else:
copy[k] = v
return copy
This effectively pops off the first value of the passed list l as a key for the dict copy that we initialize, then runs the remaining list through the same function, creating a new key under that key until there's nothing left in the list, whereupon it assigns the last value to the v param.

This is much easier in Perl:
my %hash;
$hash{"aaa"}{"bbb"}{"ccc"}=1; # auto creates each of the intermediate levels
# of the hash (aka: dict or associated array)

How to get a random value from dictionary?

How can I get a random pair from a dict? I'm making a game where you need to guess a capital of a country and I need questions to appear randomly.
The dict looks like {'VENEZUELA':'CARACAS'}
How can I do this?

One way would be:
import random
d = {'VENEZUELA':'CARACAS', 'CANADA':'OTTAWA'}
random.choice(list(d.values()))
EDIT: The question was changed a couple years after the original post, and now asks for a pair, rather than a single item. The final line should now be:
country, capital = random.choice(list(d.items()))

I wrote this trying to solve the same problem:
https://github.com/robtandy/randomdict
It has O(1) random access to keys, values, and items.

If you don't want to use the random module, you can also try popitem():
>> d = {'a': 1, 'b': 5, 'c': 7}
>>> d.popitem()
('a', 1)
>>> d
{'c': 7, 'b': 5}
>>> d.popitem()
('c', 7)
Since the dict doesn't preserve order, by using popitem you get items in an arbitrary (but not strictly random) order from it.
Also keep in mind that popitem removes the key-value pair from dictionary, as stated in the docs.
popitem() is useful to destructively iterate over a dictionary

>>> import random
>>> d = dict(Venezuela = 1, Spain = 2, USA = 3, Italy = 4)
>>> random.choice(d.keys())
'Venezuela'
>>> random.choice(d.keys())
'USA'
By calling random.choice on the keys of the dictionary (the countries).

Try this:
import random
a = dict(....) # a is some dictionary
random_key = random.sample(a, 1)[0]
This definitely works.

This works in Python 2 and Python 3:
A random key:
random.choice(list(d.keys()))
A random value
random.choice(list(d.values()))
A random key and value
random.choice(list(d.items()))

Since the original post wanted the pair:
import random
d = {'VENEZUELA':'CARACAS', 'CANADA':'TORONTO'}
country, capital = random.choice(list(d.items()))
(python 3 style)

If you don't want to use random.choice() you can try this way:
>>> list(myDictionary)[i]
'VENEZUELA'
>>> myDictionary = {'VENEZUELA':'CARACAS', 'IRAN' : 'TEHRAN'}
>>> import random
>>> i = random.randint(0, len(myDictionary) - 1)
>>> myDictionary[list(myDictionary)[i]]
'TEHRAN'
>>> list(myDictionary)[i]
'IRAN'

When they ask for a random pair here they mean a key and value.
For such a dict where the key:values are country:city,
use random.choice().
Pass the dictionary keys to this function as follows:
import random
keys = list(my_dict)
country = random.choice(keys)
You may wish to track the keys that were already called in a round and when getting a fresh country, loop until the random selection is not in the list of those already "drawn"... as long as the drawn list is shorter than the keys list.

Since this is homework:
Check out random.sample() which will select and return a random element from an list. You can get a list of dictionary keys with dict.keys() and a list of dictionary values with dict.values().

I am assuming that you are making a quiz kind of application. For this kind of application I have written a function which is as follows:
def shuffle(q):
"""
The input of the function will
be the dictionary of the question
and answers. The output will
be a random question with answer
"""
selected_keys = []
i = 0
while i < len(q):
current_selection = random.choice(q.keys())
if current_selection not in selected_keys:
selected_keys.append(current_selection)
i = i+1
print(current_selection+'? '+str(q[current_selection]))
If I will give the input of questions = {'VENEZUELA':'CARACAS', 'CANADA':'TORONTO'} and call the function shuffle(questions) Then the output will be as follows:
VENEZUELA? CARACAS
CANADA? TORONTO
You can extend this further more by shuffling the options also

With modern versions of Python(since 3), the objects returned by methods dict.keys(), dict.values() and dict.items() are view objects*. And hey can be iterated, so using directly random.choice is not possible as now they are not a list or set.
One option is to use list comprehension to do the job with random.choice:
import random
colors = {
'purple': '#7A4198',
'turquoise':'#9ACBC9',
'orange': '#EF5C35',
'blue': '#19457D',
'green': '#5AF9B5',
'red': ' #E04160',
'yellow': '#F9F985'
}
color=random.choice([hex_color for color_value in colors.values()]
print(f'The new color is: {color}')
References:
*Python 3.8: Standard Library Documentation - Built-in types: Dictionary view objects
Python 3.8: Data Structures - List Comprehensions:

I just stumbled across a similar problem and designed the following solution (relevant function is pick_random_item_from_dict; other functions are just for completeness).
import random
def pick_random_key_from_dict(d: dict):
"""Grab a random key from a dictionary."""
keys = list(d.keys())
random_key = random.choice(keys)
return random_key
def pick_random_item_from_dict(d: dict):
"""Grab a random item from a dictionary."""
random_key = pick_random_key_from_dict(d)
random_item = random_key, d[random_key]
return random_item
def pick_random_value_from_dict(d: dict):
"""Grab a random value from a dictionary."""
_, random_value = pick_random_item_from_dict(d)
return random_value
# Usage
d = {...}
random_item = pick_random_item_from_dict(d)
The main difference from previous answers is in the way we handle the dictionary copy with list(d.items()). We can partially circumvent that by only making a copy of d.keys() and using the random key to pick its associated value and create our random item.

Try this (using random.choice from items)
import random
a={ "str" : "sda" , "number" : 123, 55 : "num"}
random.choice(list(a.items()))
# ('str', 'sda')
random.choice(list(a.items()))[1] # getting a value
# 'num'

To select 50 random key values from a dictionary set dict_data:
sample = random.sample(set(dict_data.keys()), 50)

I needed to iterate through ranges of keys in a dict without sorting it each time and found the Sorted Containers library. I discovered that this library enables random access to dictionary items by index which solves this problem intuitively and without iterating through the entire dict each time:
>>> import sortedcontainers
>>> import random
>>> d = sortedcontainers.SortedDict({1: 'a', 2: 'b', 3: 'c'})
>>> random.choice(d.items())
(1, 'a')
>>> random.sample(d.keys(), k=2)
[1, 3]

I found this post by looking for a rather comparable solution. For picking multiple elements out of a dict, this can be used:
idx_picks = np.random.choice(len(d), num_of_picks, replace=False) #(Don't pick the same element twice)
result = dict ()
c_keys = [d.keys()] #not so efficient - unfortunately .keys() returns a non-indexable object because dicts are unordered
for i in idx_picks:
result[c_keys[i]] = d[i]

Here is a little Python code for a dictionary class that can return random keys in O(1) time. (I included MyPy types in this code for readability):
from typing import TypeVar, Generic, Dict, List
import random
K = TypeVar('K')
V = TypeVar('V')
class IndexableDict(Generic[K, V]):
def __init__(self) -> None:
self.keys: List[K] = []
self.vals: List[V] = []
self.dict: Dict[K, int] = {}
def __getitem__(self, key: K) -> V:
return self.vals[self.dict[key]]
def __setitem__(self, key: K, val: V) -> None:
if key in self.dict:
index = self.dict[key]
self.vals[index] = val
else:
self.dict[key] = len(self.keys)
self.keys.append(key)
self.vals.append(val)
def __contains__(self, key: K) -> bool:
return key in self.dict
def __len__(self) -> int:
return len(self.keys)
def random_key(self) -> K:
return self.keys[random.randrange(len(self.keys))]

b = { 'video':0, 'music':23,"picture":12 }
random.choice(tuple(b.items())) ('music', 23)
random.choice(tuple(b.items())) ('music', 23)
random.choice(tuple(b.items())) ('picture', 12)
random.choice(tuple(b.items())) ('video', 0)