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
Related
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}
This question already has answers here:
How do I create variable variables?
(17 answers)
Closed 1 year ago.
The question is about to getting one of multiple data in a run time. There are several data such as
dict_a = {'shape':1}
dict_b = {'opt':'sp'}
dict_c = ...
From the function with job of 'a', 'b', 'c', etc. Rather than using just long repetitive coding such as
if job == 'a':
dic = dict_a
elif job == 'b':
dic = dict_b
...
I want to write a more smart coding. I have tried this with exec, but this does not work.
def running(job):
dictname = 'dict_'+job
var = 'dic'
exec("%s = %s" % (var, dictname))
if "dic" in locals():
print(f"True, {dic}")
"dic" seems to be made by exec() but cannot use dic. The error message goes as follows:
print(f"True, {dic}")
NameError: name 'dic' is not defined
dictionary of dictionaries is more efficient but another possible solution to this is the use of eval
>>> dict_a = {'shape':1}
>>> dict_b = {'opt':'sp'}
>>> job = "dict_a"
>>> dic = eval(job)
>>> dic
{'shape': 1}
Using a dictionary of dictionaries would be much easier:
mega_dict = {'a': {'shape':1}, 'b' :{'opt':'sp'}}
print(mega_dict['a']['shape']) # or any other usage...
You can create a dictionary consisting of all dictionaries. Something like this
final_dict = {'dict_a' : dict_a, 'dict_b' : dict_b}
Then in your running method, you can just simply do
dictname = 'dict_' + job
dic = final_dict[dictname]
This question already has answers here:
How do I merge a list of dicts into a single dict?
(11 answers)
How do I merge dictionaries together in Python?
(7 answers)
Closed 1 year ago.
I have a large list of dictionaries, each with exactly one entry and with unique keys, and I want to 'combine' them into a single dict with python 3.8.
So here is an example that actually works:
mylist = [{'a':1}, {'b':2}, {'c':3}]
mydict = {list(x.keys())[0]:list(x.values())[0] for x in mylist}
which gives as result the expected output:
{'a': 1, 'b': 2, 'c': 3}
But it looks ugly and not quite pythonic. Is there a better one-line solution to this problem?
This is similar to the question asked HERE, but in my example I am looking for an answer (1) to merge many dicts together and (2) for a one-line solution. That makes my question different from the question already asked.
mydict = { k:v for elt in mylist for k, v in elt.items()}
Try this out, simple and effective.
mylist = [{'a':1}, {'b':2}, {'c':3}]
result = {}
for d in mylist:
result.update(d)
Result
{'a': 1, 'b': 2, 'c': 3}
This question already has answers here:
Change the name of a key in dictionary
(23 answers)
Closed 6 years ago.
I have got a dictionary in the form of:
{0.1: (0.7298579,0.7987254)}
which corresponds to: {test_size: (train_error, test_error)}.
I would like to change the key value test_size into 1 - test_size. So that we obtain:
{0.9: (0.7298579, 0.7987254)}
How can I do this?
You can do this :
>>> d = {0.1:(0.7298579,0.7987254)}
>>> new_d = {1-k: v for k, v in d.items()}
>>> new_d
{0.9: (0.7298579, 0.7987254)}
If the dict is not horribly huge, you can just create a new dict and copy the old one with new keys.
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)])