This question already has answers here:
How to get a random value from dictionary?
(19 answers)
How to pick one key from a dictionary randomly
(2 answers)
Closed 10 months ago.
The essence of the question is how to derive a random key/value from a JSON dictionary.
def randkey():
with open("file.json") as file:
dict = json.load(file)
for k, v in sorted(dict.items())[-1:]:
randkeyvalue = f"Name: {v['Name']}\n" \
f"Age: {v['Age']}\n"
If I can sort the dictionary in reverse order (sorted()) and take the very first value from it [-1:]. Can I sort it randomly and take the first value and of course it will always be different?
You can use the choice function from random:
https://docs.python.org/3/library/random.html#random.choice
For example:
>>> from random import choice
>>>
>>> d = {'a': 1, 'b': 2, 'c': 3}
>>>
>>> choice(list(d))
'a'
>>> choice(list(d))
'c'
>>> choice(list(d))
'c'
>>> d
{'a': 1, 'b': 2, 'c': 3}
Related
This question already has answers here:
Sort a list with a custom order in Python
(4 answers)
Closed 2 months ago.
I have a list of objects like this:
my_list = [obj1,obj2,obj3, ...]
Each object has some data like:
obj = {
'id':1,
'name':'Some name',
'value': 'Some value',
'category':'A'
}
The category options are: CATEGORY = {'A','B','C','D'}
Is it possible to sort the list somehow to display data in sequence like:
my_sordet_list = ['all in C categ', 'all in B categ', 'all in D categ','all in A categ']
So the question is: How can I explain to the program that I want it to be sorted in a strict custom sequence?
You could create a lookup dictionary for your custom ordering:
>>> category_order = {'C': 0, 'B': 1, 'D': 2, 'A': 3}
>>> my_sorted_list = sorted(mylist, key=lambda o: category_order[o['category']])
my_sorted_list will be sorted, as you desire.
This question already has answers here:
Is it possible to assign the same value to multiple keys in a dict object at once?
(8 answers)
Closed 3 years ago.
Is there a compact way to have alternative keys for the same value in a dictionary?
Taking a dictionary like the following
dict={'A':1,'B':2,'one':1,'two':2}
I can obtain the value 1 using two different keys:
dict['A']
dict['one']
I would like to know if there is a more compact way to write it, something akin to:
dict2={['A','one']:1,['B','two']:2}
You can define the dict with keys of the same value grouped as a tuple first:
d = {('A', 'one'): 1, ('B', 'two'): 2}
so that you can then convert it to the desired dict with:
d = {key: value for keys, value in d.items() for key in keys}
d becomes:
{'A': 1, 'one': 1, 'B': 2, 'two': 2}
This question already has answers here:
How do I merge two dictionaries in a single expression in Python?
(43 answers)
Closed 6 years ago.
In python, I have a dictionary named dict_a :
dict_a = {'a':1}
and I want to get a new dictionary dict_b as the update of dict_a,at the same time I don't want to change the dict_a, so I used this:
dict_b = copy.deepcopy(dict_a).update({'b':2})
However, the return value of dict.update is None, so the above code doesn't work. And I have to use this：
temp = copy.deepcopy(dict_a)
temp.update({'b':2})
dict_b = temp
The question is, how to get my goal in one line? Or some
What about:
>>> a = {'a':1}
>>> update = {'b': 1}
>>> b = dict(a.items() + update.items())
>>> b
{'a': 1, 'b': 1}
update - is your value that need to update(extend) dict a
b - is a new resulting dict
Anfd in this case a stay unchanged
This question already has answers here:
Access nested dictionary items via a list of keys?
(20 answers)
Xpath like query for nested python dictionaries
(11 answers)
Closed 6 years ago.
Is there a recursive itemgetter in python. Let's say you have an object like
d = {'a': (1,2,3), 'b': {1: (5,6)}}
and I wanted to get the first element of the tuple from d['a']? As far as I can tell itemgetter will only do one level, i.e. get me a or b or both.
Is there some clever way of combining itertools with itemgetter to produce the desired result.
So basically what I want to be able to call is
from operator import itemgetter
d = {'a': (1,2), 'b': (4, (5,))}
itemgetter({'a': 0})(d) --> 1
itemgetter({'b': 0})(d) --> 4
itemgetter({'b': {1: 0}})(d) --> 5
d = {'a': {'b': (1,2,3)}}
itemgetter({'a': {'b': 2}})(d) --> 3
I don't like 'clever' ways. Obvious is better.
You can very easily write a getter that iterates along a list of subscripts:
def getter(x, *args):
for k in args:
x = x[k]
return x
>>> d = {'a': (1,2,3), 'b': {1: (5,6)}}
>>> getter(d, 'a', 0)
1
This question already has answers here:
How to keep keys/values in same order as declared?
(13 answers)
Closed 6 years ago.
I am trying to add elements to a dictionary in the following way:
a = {}
a['b'] = 1
a['a'] = 2
Finally a looks like:
{'a': 2, 'b': 1}
But actually I wanted the dictionary to contain keys in the order:
{'b': 1, 'a': 2}
Can anyone explain me this? Why are the keys getting sorted alphabetically when actually dictionaries (hashmaps) don't have any order?
You are correct in that dictionaries are not ordered, but hashed. As a result, the order of a dictionary should not be relied on.
You can use the OrderedDict to help you achieve your goal:
from collections import OrderedDict
a = OrderedDict()
a['b'] = 1
a['a'] = 2
> a
> OrderedDict([('b', 1), ('a', 2)])