I'm trying to make a function that would take nested array (dict/list in any order) and a key name as arguments and return all values of that key in a list.
my_key = "Items"
my_dict = [{'z': 0, 'x': 0, 'y': 0, 'Items': [{'Slot': 1, 'id': 'minecraft:rail', 'Count': 1}, {'Slot': 2, 'id': 'minecraft:white_shulker_box', 'tag': {'BlockEntityTag': {'id': 'minecraft:shulker_box', 'Items': [{'Slot': 0, 'Count': 1, 'tag': {'Items': [{'id': 'minecraft:amethyst_shard', 'Count': 1}]}, 'id': 'minecraft:bundle'}]}}, 'Count': 1}]}]
def recursive_lookup(data, key):
if isinstance(data, list):
for i in data:
recursive_lookup(i, key)
elif isinstance(data, dict):
for i, v in data.items():
if i == key:
print(f'{v = }')
if isinstance(v, list) or isinstance(v, dict): recursive_lookup(v, key)
print(recursive_lookup(my_dict, my_key))
Currently it prints out found items at print(f'{v = }'). How can I store those in a list and pass as a function return?
You can use .extend() to concatenate the result of recursive calls to a list.
def recursive_lookup(data, key):
values = []
if isinstance(data, list):
for i in data:
values.extend(recursive_lookup(i, key))
elif isinstance(data, dict):
for i, v in data.items():
if i == key:
values.append(v)
if isinstance(v, list) or isinstance(v, dict):
values.extend(recursive_lookup(v, key))
return values
You can what you want without any explicit recursion at all by making use of the json module in the standard library (assuming your data can be serialized into that format). This is because the JSON decoder supports an object_hook argument which is a function it will call everytime it encounters a dictionary.
The basic idea is to specify a function via this argument that merely "watches" what is being decoded and checks it for the sought-after key.
Here's what I mean:
import json
my_key = "Items"
my_dict = [{'z': 0, 'x': 0, 'y': 0, 'Items': [{'Slot': 1, 'id': 'minecraft:rail', 'Count': 1}, {'Slot': 2, 'id': 'minecraft:white_shulker_box', 'tag': {'BlockEntityTag': {'id': 'minecraft:shulker_box', 'Items': [{'Slot': 0, 'Count': 1, 'tag': {'Items': [{'id': 'minecraft:amethyst_shard', 'Count': 1}]}, 'id': 'minecraft:bundle'}]}}, 'Count': 1}]}]
def lookup(data, key):
results = []
def decode_dict(a_dict):
try:
results.append(a_dict[key])
except KeyError:
pass
return a_dict
json_repr = json.dumps(data) # Convert to JSON format.
json.loads(json_repr, object_hook=decode_dict) # Return value ignored.
return results
from pprint import pprint
pprint(lookup(my_dict, my_key), sort_dicts=False)
Pretty-printed result list:
[[{'id': 'minecraft:amethyst_shard', 'Count': 1}],
[{'Slot': 0,
'Count': 1,
'tag': {'Items': [{'id': 'minecraft:amethyst_shard', 'Count': 1}]},
'id': 'minecraft:bundle'}],
[{'Slot': 1, 'id': 'minecraft:rail', 'Count': 1},
{'Slot': 2,
'id': 'minecraft:white_shulker_box',
'tag': {'BlockEntityTag': {'id': 'minecraft:shulker_box',
'Items': [{'Slot': 0,
'Count': 1,
'tag': {'Items': [{'id': 'minecraft:amethyst_shard',
'Count': 1}]},
'id': 'minecraft:bundle'}]}},
'Count': 1}]]
You can keep a running list:
def recursive_lookup(data, key):
lst = []
if isinstance(data, list):
for i in data:
lst.append(recursive_lookup(i, key))
elif isinstance(data, dict):
for i, v in data.items():
if i == key:
lst.append([v])
if isinstance(v, list) or isinstance(v, dict): lst.append(recursive_lookup(v, key))
return lst
print(recursive_lookup(data, 'Items'))
Related
So I have this params:
p = [{'quantity': 1}, {'args': {'id': 12345678, 'age': 12}}]
And I want to be able to search for quantity and get the value 1 or the args key and get its doctionary ({'id': 12345678, 'age: 12})
This is what I have try:
def search(name: str, params: list[dict]):
try:
return next((x[name] for x in params if x), None)
except KeyError:
return None
I case I search for the value of quantity:
search(name='quantity', params=p)
This return 1
But in case I want to value of args:
search(name='args', params=p)
This return None
I have a set of functions that you could use for this [ getNestedVal(p, 'args') would return {'id': 12345678, 'age: 12} ]...or you can use this generator
def yieldNestedVals(obj, key):
if isinstance(obj, str): return
try: yield obj[key]
except: pass
if isinstance(obj, dict): obj = obj.values()
for i in (obj if hasattr(obj, '__iter__') else []):
for v in yieldNestedVals(i, key): yield v
inside search like
def search(name, params, defaultVal=None):
for v in yieldNestedVals(params, name): return v # return 1st generated value
return defaultVal # if nothing if generated
[ Ofc you can omit both the defaultVal parameter and the last line if you want to just return None when nothing if found, since that is the default behavior when a function doesn't come across a return statement. ]
Now search('args', p) should also return {'id': 12345678, 'age: 12} and you can try it with other keys as well:
# p = [{'quantity': 1}, {'args': {'id': 12345678, 'age': 12}}]
{k: search(k, p) for k in ['quantity', 'args', 'id', 'age', 'dneKey']}
would return
{'quantity': 1,
'args': {'id': 12345678, 'age': 12},
'id': 12345678,
'age': 12,
'dneKey': None}
I have a dictionary which contains some key-value pairs as strings, but some key-values are dictionaries.
The data looks like this:
{'amount': 123,
'baseUnit': 'test',
'currency': {'code': 'EUR'},
'dimensions': {'height': {'iri': 'http://www.example.com/data/measurement-height-12345',
'unitOfMeasure': 'm',
'value': 23},
'length': {'iri': 'http://www.example.com/data/measurement-length-12345',
'unitOfMeasure': 'm',
'value': 8322},
'volume': {'unitOfMeasure': '', 'value': 0},
'weight': {'iri': 'http://www.example.com/data/measurement-weight-12345',
'unitOfMeasure': 'KG',
'value': 23},
'width': {'iri': 'http://www.example.com/data/measurement-width-12345',
'unitOfMeasure': 'm',
'value': 1}},
'exportListNumber': '1234',
'iri': 'http://www.example.com/data/material-12345',
'number': '12345',
'orderUnit': 'sdf',
'producerFormattedPID': '12345',
'producerID': 'example',
'producerNonFormattedPID': '12345',
'stateID': 'm70',
'typeID': 'FERT'}
for the dimensions and price keys, there are some nested dictionaries as values. How can I extract that data so that the final variable is a dictionary with only keys-values as strings. For the price, I would need something like:
{'pricecurrencycode':'EUR','priceamount':123} instead of 'price': {'currency': {'code': 'EUR'}, 'amount': 123}.
and the same happening to dimensions key->to extract all the nested dictionaries so that it could be easier to transform into a final dataframe.
You can define a recursive flatten function that gets called whenever the dictionary value is a dictionary.
Assuming python>=3.9:
def flatten(my_dict, prefix=""):
res = {}
for k, v in my_dict.items():
if isinstance(v, dict):
res |= flatten(v, prefix+k)
else:
res[prefix+k] = v
return res
A slightly more verbose option for older python versions:
def flatten(my_dict, prefix=""):
res = {}
for k, v in my_dict.items():
if isinstance(v, dict):
for k_flat, v_flat in flatten(v, prefix+k).items():
res[k_flat] = v_flat
else:
res[prefix+k] = v
return res
I need to change values in a nested dictionary. Consider this dictionary:
stocks = {
'name': 'stocks',
'IBM': 146.48,
'MSFT': 44.11,
'CSCO': 25.54,
'micro': {'name': 'micro', 'age': 1}
}
I need to loop through all the keys and change the values of all the name keys.
stocks.name
stocks.micro.name
These keys need to be changed. But, I will not know which keys to change before hand. So, I'll need to loop through keys and change the values.
Example
change_keys("name", "test")
Output
{
'name': 'test',
'IBM': 146.48,
'MSFT': 44.11,
'CSCO': 25.54,
'micro': {'name': 'test', 'age': 1}
}
A recursive solution that supports unknown number of nesting levels:
def change_key(d, required_key, new_value):
for k, v in d.items():
if isinstance(v, dict):
change_key(v, required_key, new_value)
if k == required_key:
d[k] = new_value
stocks = {
'name': 'stocks',
'IBM': 146.48,
'MSFT': 44.11,
'CSCO': 25.54,
'micro': {'name': 'micro', 'age': 1}
}
change_key(stocks, 'name', 'new_value')
print(stocks)
# {'name': 'new_value',
# 'MSFT': 44.11,
# 'CSCO': 25.54,
# 'IBM': 146.48,
# 'micro': {'name': 'new_value',
# 'age': 1}
# }
def changeKeys(d, repl):
for k,v in zip(d.keys(),d.values()):
if isinstance(v, dict):
changeKeys(v,repl)
elif k == "name":
d[k]= repl
I am having trouble with the following code. I need to add a key to a python dictionary if it does not exist, and if it does then I need to add to the value.
My output should look like this.
{'STA': {'value':-62**.***, 'count': 4}}.....
But I'm getting this instead.
{'STA': {'value': -1194.14562548, 'count': 0}},
{'STA': {'value': -5122.985396600001, 'count': 0}},
{'STA': {'value': 25.2293, 'count': 0}},
{'STA': {'value': 34.0099, 'count': 0}},
What am I doing wrong?
new_dict = []
for item in sales_orders['progress_output']:
ex = (list(filter(lambda ex:ex['_branch_shortname'].strip() == item['_branch_shortname'].strip(), expenses)))
value = float(item['gross_profit'].strip().replace(',', '')) - (float(item['load_factor_extended'].strip().replace(',', '')) * float(ex[0]['value']))
branch = item['_branch_shortname'].strip()
if branch in new_dict:
new_dict[branch]['value'] += value
new_dict[branch]['count'] += 1
else:
new_dict.append({branch: {'value': value, 'count': 0}})
#print(item['_branch_shortname'], value)
print(new_dict)
You can use setdefault to ensure there is a default value, setdefault returns the value if it already exists for key (first parameter) or adds the default (second parameter) and returns it:
new_dict = {}
for ...:
b = new_dict.setdefault(branch, {'value': 0, 'count': 0})
b['value'] += value
b['count'] += 1
You can also use a defaultdict:
from collections import defaultdict
new_dict = defaultdict(lambda: {'value': 0, 'count': 0})
for ...:
new_dict[branch]['value'] += value
new_dict[branch]['count'] += 1
I have a dictionary that looks like this:
{'items': [{'id': 1}, {'id': 2}, {'id': 3}]}
and I'm looking for a way to directly get the inner dictionary with id = 1.
Is there a way to reach this other than looping the list items and comparing the id?
first_with_id_or_none = \
next((value for value in dictionary['items'] if value['id'] == 1), None)
You will have to loop through the list. The good news is is that you can use a generator expression with next() to do that looping:
yourdict = next(d for d in somedict['items'] if d['id'] == 1)
This can raise a StopIteration exception if there is no such matching dictionary.
Use
yourdict = next((d for d in somedict['items'] if d['id'] == 1), None)
to return a default instead for that edge-case (here None is used, but pick what you need).
Make it into a function:
def get_inner_dict_with_value(D, key, value):
for k, v in D.items():
for d in v:
if d.get(key) == value:
return d
else:
raise ValueError('the dictionary was not found')
With explanation:
def get_inner_dict_with_value(D, key, value):
for k, v in D.items(): # loop the dictionary
# k = 'items'
# v = [{'id': 1}, {'id': 2}, {'id': 3}]
for d in v: # gets each inner dictionary
if d.get(key) == value: # find what you look for
return d # return it
else: # none of the inner dictionaries had what you wanted
raise ValueError('the dictionary was not found') # oh no!
Running it:
>>> get_inner_dict_with_value({'items': [{'id': 1}, {'id': 2}, {'id': 3}]}, 'id', 1)
{'id': 1}
Another method:
def get_inner_dict_with_value2(D, key, value):
try:
return next((d for l in D.values() for d in l if d.get(key) == value))
except StopIteration:
raise ValueError('the dictionary was not found')
>>> get_inner_dict_with_value2({'items': [{'id': 1}, {'id': 2}, {'id': 3}]}, 'id', 1)
{'id': 1}