I have a dictionary with missing values (the key is there, but the associated value is empty). For example I want the dictionary below:
dct = {'ID': '', 'gender': 'male', 'age': '20', 'weight': '', 'height': '5.7'}
to be changed to this form:
dct = {'ID': {'link': '','value': ''}, 'gender': 'male', 'age': '20', 'weight': {'link': '','value': ''}, 'height': '5.7'}
I want the ID and Weight key should be replaced with nested dictionary if its empty.
How can I write that in the most time-efficient way?
I have tried solutions from below links but didnt work,
def update(orignal, addition):
for k, v in addition.items():
if k not in orignal:
orignal[k] = v
else:
if isinstance(v, dict):
update(orignal[k], v)
elif isinstance(v, list):
for i in range(len(v)):
update(orignal[k][i], v[i])
else:
if not orignal[k]:
orignal[k] = v
Error: TypeError: 'str' object does not support item assignment
Fill missing keys by comparing example json in python
Adding missing keys in dictionary in Python
It seems similar with this issue https://stackoverflow.com/a/3233356/6396981
import collections.abc
def update(d, u):
for k, v in u.items():
if isinstance(v, collections.abc.Mapping):
d[k] = update(d.get(k, {}) or {}, v)
else:
d[k] = v
return d
For example in your case:
>>> dict1 = {'ID':'', 'gender':'male', 'age':'20', 'weight':'', 'height':'5.7'}
>>> dict2 = {'ID': {'link':'','value':''}, 'weight': {'link':'','value':''}}
>>>
>>> update(dict1, dict2)
{'ID': {'link': '', 'value': ''}, 'gender': 'male', 'age': '20', 'weight': {'link': '', 'value': ''}, 'height': '5.7'}
>>>
You can iterate through the list and see if the value is an empty string('') if it is, replace it with the default value. Here's a small snippet which does it -
dct = {'ID':'', 'gender':'male', 'age':'20', 'weight':'', 'height':'5.7'}
def update(d, default):
for k, v in d.items():
if v == '':
d[k] = default.copy()
update(dct, {'link':'','value':''})
print(dct)
Output :
{'ID': {'link': '', 'value': ''}, 'gender': 'male', 'age': '20', 'weight': {'link': '', 'value': ''}, 'height': '5.7'}
Note that the dict is passed by reference to the function, so any updates made there will be reflected in the original dictionary as well as seen in the above example.
If your dict is nested and you want the replacement to be done for nested items as well then you can use this function -
def nested_update(d, default):
for k, v in d.items():
if v == '':
d[k] = default.copy()
if isinstance(v, list):
for item in v:
nested_update(item, default)
if isinstance(v, dict):
nested_update(v, default)
here's a small example with list of dictionaries and nested dictionary -
dct = {'ID':'', 'gender':'male', 'age':'20', 'weight':'', 'height':'5.7', "list_data":[{'empty': ''}, {'non-empty': 'value'}], "nested_dict": {"key1": "val1", "missing_nested": ""}}
nested_update(dct, {'key1': 'val1-added', 'key2': 'val2-added'})
print(dct)
Output :
{'ID': {'key1': 'val1-added', 'key2': 'val2-added'}, 'gender': 'male', 'age': '20', 'weight': {'key1': 'val1-added', 'key2': 'val2-added'}, 'height': '5.7', 'list_data': [{'empty': {'key1': 'val1-added', 'key2': 'val2-added'}}, {'non-empty': 'value'}], 'nested_dict': {'key1': 'val1', 'missing_nested': {'key1': 'val1-added', 'key2': 'val2-added'}}}
For "this default dictionary to only specified keys like ID and Weight and not for other keys", you can update the condition of when we replace the value -
def nested_update(d, default):
for k, v in d.items():
if k in ('ID', 'weight') and v == '':
d[k] = default.copy()
if isinstance(v, list):
for item in v:
nested_update(item, default)
if isinstance(v, dict):
nested_update(v, default)
Related
I have a dictionary with some values that are type list, i need to convert each list in another dictionary and insert this new dictionary at the place of the list.
Basically, I have this dictionary
Dic = {
'name': 'P1',
'srcintf': 'IntA',
'dstintf': 'IntB',
'srcaddr': 'IP1',
'dstaddr': ['IP2', 'IP3', 'IP4'],
'service': ['P_9100', 'SNMP'],
'schedule' : 'always',
}
I need to reemplace the values that are lists
Expected output:
Dic = {
'name': 'P1',
'srcintf': 'IntA',
'dstintf': 'IntB',
'srcaddr': 'IP1',
'dstaddr': [
{'name': 'IP2'},
{'name': 'IP3'},
{'name': 'IP4'}
],
'service': [
{'name': 'P_9100'},
{'name': 'SNMP'}
],
'schedule' : 'always',
}
So far I have come up with this code:
for k,v in Dic.items():
if not isinstance(v, list):
NewDic = [k,v]
print(NewDic)
else:
values = v
keys = ["name"]*len(values)
for item in range(len(values)):
key = keys[item]
value = values[item]
SmallDic = {key : value}
liste.append(SmallDic)
NewDic = [k,liste]
which print this
['name', 'P1']
['srcintf', 'IntA']
['dstintf', 'IntB']
['srcaddr', 'IP1']
['schedule', 'always']
['schedule', 'always']
I think is a problem with the loop for, but so far I haven't been able to figure it out.
You need to re-create the dictionary. With some modifications to your existing code so that it generates a new dictionary & fixing the else clause:
NewDic = {}
for k, v in Dic.items():
if not isinstance(v, list):
NewDic[k] = v
else:
NewDic[k] = [
{"name": e} for e in v # loop through the list values & generate a dict for each
]
print(NewDic)
Result:
{'name': 'P1', 'srcintf': 'IntA', 'dstintf': 'IntB', 'srcaddr': 'IP1', 'dstaddr': [{'name': 'IP2'}, {'name': 'IP3'}, {'name': 'IP4'}], 'service': [{'name': 'P_9100'}, {'name': 'SNMP'}], 'schedule': 'always'}
Hello I need to compare 2 dicts but in the result, I need to know from which dict the result came.
dict1 = {'name': 'Morgan', 'surename': 'Finch'}
dict2 = {'name': 'David', 'surename': 'Finch'}
so if I compare with input_data.items() ^ response_data.items() result will something like this:
{('name','Morgan'),('name', 'David)}
expected result should look something like {'dict1': ('name','Morgan'), dict2: ('name', 'David')}
I don't care what data-structure just that I could know from what dict it came.
dict1 = {'name': 'Morgan', 'surname': 'Finch'}
dict2 = {'name': 'David', 'surname': 'Finch'}
# symmetric difference (exclusive OR)
print(dict1.items() ^ dict2.items())
# {('name', 'Morgan'), ('name', 'David')}
# dictionary subtraction
print({'dict1': dict1.items() - dict2.items(), 'dict2': dict2.items() - dict1.items()})
# {'dict1': {('name', 'Morgan')} 'dict2': {('name', 'David')}}
If you want the answer in the form of dictionary
You can use these steps
dict1 = {'name': 'Morgan', 'surename': 'Finch'}
dict2 = {'name': 'David', 'surename': 'Finch'}
dict3 = {}
for k,v in dict1.items():
if dict1[k] != dict2[k]:
dict3['dict1'] = (k,dict1[k])
dict3['dict2'] = (k,dict2[k])
print(dict3)
Output:
{'dict1': ('name', 'Morgan'), 'dict2': ('name', 'David')}
Edit:
If all values are different and want to store in a single key like {'dict1' : ('name', 'Morgan', 'surname', 'Finc'), ... }
dict1 = {'name': 'Morgan', 'surename': 'Finch'}
dict2 = {'name': 'David', 'surename': 'Finc'}
dict3 = {'dict1':(), 'dict2':()}
for k,v in dict1.items():
if dict1[k] != dict2[k]:
dict3['dict1'] += (k,dict1[k])
dict3['dict2'] += (k,dict2[k])
print(dict3)
Output:
{'dict1': ('name', 'Morgan', 'surename', 'Finch'), 'dict2': ('name', 'David', 'surename', 'Finc')}
I have a list of dictionary where value contains another dicitonary, I need to search for a string in inner dictionary and if its matches need to return the value of outer dictionary first key.
here is list as follow :
lst=[{'value': 'value1', 'Characteristic': {'ID': 'searchstr1'}},
{'value': 'value2', 'Characteristic': {'ID': 'searchstr2'}},
{'value': 'value3', 'Characteristic': {'ID': 'searchstr3'}}
, {'value': 'value4', 'Characteristic': {'ID': 'searchstr4'}}]
and search string is
search_str="searchstr3"
so in that case it should return the
value3 as result .
I tried by looping through the list and then getting each dictionary item but not sure how to travers back once find the desired value.
here is my code what i tired so far :
def find_value():
for dicts in lst:
current = dicts["value"]
for k,v in dicts.items():
#print("{0} : {1}".format(k, v))
if isinstance(v, dict):
if v['ID']==search_str:
break
return current
In your simple case it's not about the 1st key but a particular key 'value' - just return the needed value immediately:
lst = [{'value': 'value1', 'Characteristic': {'ID': 'searchstr1'}},
{'value': 'value2', 'Characteristic': {'ID': 'searchstr2'}},
{'value': 'value3', 'Characteristic': {'ID': 'searchstr3'}},
{'value': 'value4', 'Characteristic': {'ID': 'searchstr4'}}]
search_str = "searchstr3"
def find_value(lst, search_str):
for d in lst:
for k,v in d.items():
if isinstance(v, dict) and v['ID'] == search_str:
return d['value']
return None
print(find_value(lst, search_str)) # value3
I want to replace the values (formated as strings) with the same values as integers, whenever the key is 'current_values'.
d = {'id': '10', 'datastreams': [{'current_value': '5'}, {'current_value': '4'}]}
Desired Output:
d = {'id': '10', 'datastreams': [{'current_value': 5}, {'current_value': 4}]}
The following piece of code replaces (substrings of) values in a dictionary. It works for nested json structures and copes with json, list and string types. You can easily add other types if needed.
def dict_replace_value(d: dict, old: str, new: str) -> dict:
x = {}
for k, v in d.items():
if isinstance(v, dict):
v = dict_replace_value(v, old, new)
elif isinstance(v, list):
v = list_replace_value(v, old, new)
elif isinstance(v, str):
v = v.replace(old, new)
x[k] = v
return x
def list_replace_value(l: list, old: str, new: str) -> list:
x = []
for e in l:
if isinstance(e, list):
e = list_replace_value(e, old, new)
elif isinstance(e, dict):
e = dict_replace_value(e, old, new)
elif isinstance(e, str):
e = e.replace(old, new)
x.append(e)
return x
# See input and output below
output = dict_replace_value(input, 'string', 'something')
Input:
input = {
'key1': 'a string',
'key2': 'another string',
'key3': [
'a string',
'another string',
[1, 2, 3],
{
'key1': 'a string',
'key2': 'another string'
}
],
'key4': {
'key1': 'a string',
'key2': 'another string',
'key3': [
'a string',
'another string',
500,
1000
]
},
'key5': {
'key1': [
{
'key1': 'a string'
}
]
}
}
Output:
print(output)
{
"key1":"a something",
"key2":"another something",
"key3":[
"a something",
"another something",
[
1,
2,
3
],
{
"key1":"a something",
"key2":"another something"
}
],
"key4":{
"key1":"a something",
"key2":"another something",
"key3":[
"a something",
"another something",
500,
1000
]
},
"key5":{
"key1":[
{
"key1":"a something"
}
]
}
}
d = {'id': '10', 'datastreams': [{'current_value': '5'}, {'current_value': '4'}]}
for elem in d['datastreams']: # for each elem in the list datastreams
for k,v in elem.items(): # for key,val in the elem of the list
if 'current_value' in k: # if current_value is in the key
elem[k] = int(v) # Cast it to int
print(d)
OUTPUT:
{'id': '10', 'datastreams': [{'current_value': 5}, {'current_value': 4}]}
A general approach (assuming you don't know in advance which key of the dict is pointing to a list) would be to iterate over the dict and check the type of its values and then iterate again into each value if needed.
In your case, your dictionary may contain a list of dictionaries as values, so it is enough to check if a value is of type list, if so, iterate over the list and change the dicts you need.
It can be done recursively with a function like the following:
def f(d):
for k,v in d.items():
if k == 'current_value':
d[k] = int(v)
elif type(v) is list:
for item in v:
if type(item) is dict:
f(item)
>>> d = {'id': '10', 'datastreams': [{'current_value': '5'}, {'current_value': '4'}]}
>>> f(d)
>>> d
{'id': '10', 'datastreams': [{'current_value': 5}, {'current_value': 4}]}
Can be done with list comprehension:
d['datastreams'] = [{'current_value': int(ds['current_value'])} if ('current_value' in ds) else ds for ds in d['datastreams']]
You can use ast.literal_eval to evaluate the underlying value for items with current_value key in the d['datastreams'] list. Then check whether the type is an int using isinstance for such values. Finally, type cast such values to int.
import ast
d = {'id': '10', 'datastreams': [{'current_value': '5'}, {'current_value': '4'}]}
for i in d['datastreams']:
for k,v in i.items():
if 'current_value' in k and isinstance(ast.literal_eval(v),int):
i[k] = int(v)
#Output:
print(d)
{'id': '10', 'datastreams': [{'current_value': 5}, {'current_value': 4}]}
You could use this method
which would loop through checks for current_value in list and change it to integer by passing the value through int() function:
for value in d.values():
for element in value:
if 'current_value' in element:
element['current_value'] = int(element['current_value'])
Taking alec_djinn's solution little farther to handle also nested dicts:
def f(d):
for k,v in d.items():
if k == 'current_value':
d[k] = int(v)
elif type(v) is list:
for item in v:
if type(item) is dict:
f(item)
if type(v) is dict:
f(v)
I want to flatten a list of dict but having issues,
let's say i have a list of dict as,
d = [{'val': 454,'c': {'name': 'ss'}, 'r': {'name1': 'ff'}},{'val': 'ss', 'c': {'name': 'ww'}, 'r': {'name1': 'ff'}}, {'val': 22,'c': {'name': 'dd'}, 'r': {'name1': 'aa'}}]
And the output I'm trying to get is,
d = [{'val': 454,'name': 'ss', 'name1': 'ff'},{'val': 'ss','name': 'ww', 'name1': 'ff'},{'val': 22, 'name': 'dd', 'name1': 'aa'}]
For which I'm using the following function,
def flatten(structure, key="", flattened=None):
if flattened is None:
flattened = {}
if type(structure) not in(dict, list):
flattened[key] = structure
elif isinstance(structure, list):
for i, item in enumerate(structure):
flatten(item, "%d" % i, flattened)
else:
for new_key, value in structure.items():
flatten(value, new_key, flattened)
return flattened
Now, the issue I have is, it's only generating the first element in the dict
You are probably initializing something in the wrong place. Take a look at the code below:
d = [{'val': 454, 'c': {'name': 'ss'}, 'r': {'name1': 'ff'}}, {'val': 55, 'c': {'name': 'ww'}, 'r': {'name1': 'ff'}}, {'val': 22, 'c': {'name': 'dd'}, 'r': {'name1': 'aa'}}]
# ^ typo here
def flatten(my_dict):
res = []
for sub in my_dict:
print(sub)
dict_ = {}
for k, v in sub.items():
if isinstance(v, dict):
for k_new, v_new in v.items():
dict_[k_new] = v_new
else:
dict_[k] = v
res.append(dict_)
return res
result = flatten(d)
print(result) # [{'name': 'ss', 'name1': 'ff', 'val': 454}, {'name': 'ww', 'name1': 'ff', 'val': 55}, {'name': 'dd', 'name1': 'aa', 'val': 22}]
You should initialize flattened to the same type as structure if it's None, and pass None when recursing at the list case:
def flatten_2(structure, key="", flattened=None):
if flattened is None:
flattened = {} if isinstance(structure, dict) else []
if type(structure) not in(dict, list):
flattened[key] = structure
elif isinstance(structure, list):
for i, item in enumerate(structure):
flattened.append(flatten(item, "%d" % i))
else:
for new_key, value in structure.items():
flatten(value, new_key, flattened)
return flattened
In [13]: flatten_2(d)
Out[13]:
[{'name': 'ss', 'name1': 'ff', 'val': 454},
{'name': 'ww', 'name1': 'ff', 'val': 'ss'},
{'name': 'dd', 'name1': 'aa', 'val': 22}]
This of course only works for a limited type of data.