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
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 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:
How to Create Nested Dictionary in Python with 3 lists
(7 answers)
Closed 4 years ago.
Suppose I have three lists:
list_a = [1,2,3]
list_b = ['a','b','c']
list_c = [4,5,6]
How do I create a nested dictionary that looks like this:
dict = {1:{'a':4},2:{'b':5},3:{'c':6}
I was thinking of using the defaultdict command from the collections module or creating a class but I don't know how to do that
You can utilize zip and dictionary comprehension to solve this:
list_a = [1,2,3]
list_b = ['a','b','c']
list_c = [4,5,6]
final_dict = {a:{b:c} for a, b, c in zip(list_a, list_b, list_c)}
Output:
{1: {'a': 4}, 2: {'b': 5}, 3: {'c': 6}}
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:
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)])