I have a list of dictionary like this
[
{'id':1, 'name': 'name1', 'education':{'university':'university1', 'subject': 'abc1'}},
{'id':2, 'name': 'name2', 'education':{'university':'university2', 'subject': 'abc2'}},
{'id':3, 'name': 'name3', 'education':{'university':'university3', 'subject': 'abc3'}},
]
and I want to convert it like
[
{'id':1, 'name': 'name1', 'university':'university1', 'subject': 'abc1'},
{'id':2, 'name': 'name2', 'university':'university2', 'subject': 'abc2'},
{'id':3, 'name': 'name3', 'university':'university3', 'subject': 'abc3'},
]
is there any pythonic way to solve this.
You could simply do the following:
l = [...]
for d in l:
d.update(d.pop('education', {}))
# l
[{'id': 1, 'name': 'name1', 'subject': 'abc1', 'university': 'university1'},
{'id': 2, 'name': 'name2', 'subject': 'abc2', 'university': 'university2'},
{'id': 3, 'name': 'name3', 'subject': 'abc3', 'university': 'university3'}]
Depending if you want to transform the original list or if you want to return a new one you could go for one of these two approaches:
l = [
{'id':1, 'name': 'name1', 'education':{'university':'university1', 'subject': 'abc1'}},
{'id':2, 'name': 'name2', 'education':{'university':'university2', 'subject': 'abc2'}},
{'id':3, 'name': 'name3', 'education':{'university':'university3', 'subject': 'abc3'}},
]
def flattenReturn(input):
output = {key: value for key, value in input.items() if type(value) != dict}
for value in input.values():
if type(value) == dict:
output.update(value)
return output
def flattenTransform(d):
for key, value in list(d.items()):
if isinstance(value, dict):
d.update(d.pop(key))
print(list(map(flattenReturn, l)))
print(l)
print("-"*80)
map(flattenTransform, l)
print(l)
As you can see flattenReturn generates a new dict filtering the values which are dictionaries and then updates it with their key-values to flatten it while the second option modifies the dict in place. If the size of the data is big, a solution including generators should be prefered.
Related
I have a list of dictionaries:
mydict = [
{'name': 'test1', 'value': '1_1'},
{'name': 'test2', 'value': '2_1'},
{'name': 'test1', 'value': '1_2'},
{'name': 'test1', 'value': '1_3'},
{'name': 'test3', 'value': '3_1'},
{'name': 'test4', 'value': '4_1'},
{'name': 'test4', 'value': '4_2'},
]
I would like to use it to create a dictionary where the values are lists or single values depending of number of their occurrences in the list above.
Expected output:
outputdict = {
'test1': ['1_1', '1_2', '1_3'],
'test2': '2_1',
'test3': '3_1',
'test4': ['4_1', '4_2'],
}
I tried to do it the way below but it always returns a list, even when there is just one value element.
outputdict = {}
outputdict.setdefault(mydict.get('name'), []).append(mydict.get('value'))
The current output is:
outputdict = {
'test1': ['1_1', '1_2', '1_3'],
'test2': ['2_1'],
'test3': ['3_1'],
'test4': ['4_1', '4_2'],
}
Do what you have already done, and then convert single-element lists afterwards:
outputdict = {
name: (value if len(value) > 1 else value[0])
for name, value in outputdict.items()
}
You can use a couple of the built-in functions mainly itertools.groupby:
from itertools import groupby
from operator import itemgetter
mydict = [
{'name': 'test1', 'value': '1_1'},
{'name': 'test2', 'value': '2_1'},
{'name': 'test1', 'value': '1_2'},
{'name': 'test1', 'value': '1_3'},
{'name': 'test3', 'value': '3_1'},
{'name': 'test4', 'value': '4_1'},
{'name': 'test4', 'value': '4_2'},
]
def keyFunc(x):
return x['name']
outputdict = {}
# groupby groups all the items that matches the returned value from keyFunc
# in our case it will use the names
for name, groups in groupby(mydict, keyFunc):
# groups will contains an iterator of all the items that have the matched name
values = list(map(itemgetter('value'), groups))
if len(values) == 1:
outputdict[name] = values[0]
else:
outputdict[name] = values
print(outputdict)
I want to compare below dictionaries. Name key in the dictionary is common in both dictionaries.
If Name matched in both the dictionaries, i wanted to do some other stuff with the data.
PerfData = [
{'Name': 'abc', 'Type': 'Ex1', 'Access': 'N1', 'perfStatus':'Latest Perf', 'Comments': '07/12/2017 S/W Version'},
{'Name': 'xyz', 'Type': 'Ex1', 'Access': 'N2', 'perfStatus':'Latest Perf', 'Comments': '11/12/2017 S/W Version upgrade failed'},
{'Name': 'efg', 'Type': 'Cust1', 'Access': 'A1', 'perfStatus':'Old Perf', 'Comments': '11/10/2017 S/W Version upgrade failed, test data is active'}
]
beatData = [
{'Name': 'efg', 'Status': 'Latest', 'rcvd-timestamp': '1516756202.632'},
{'Name': 'abc', 'Status': 'Latest', 'rcvd-timestamp': '1516756202.896'}
]
Thanks
Rajeev
l = [{'name': 'abc'}, {'name': 'xyz'}]
k = [{'name': 'a'}, {'name': 'abc'}]
[i['name'] for i in l for f in k if i['name'] == f['name']]
Hope above logic work for you.
The answer provided didn't assign the result to any variable. If you want to print it, add the following would work:
result = [i['name'] for i in l for f in k if i['name'] == f['name']]
print(result)
Here‘s a simplified example of some data I have:
{"id": "1234565", "fields": {"name": "john", "email":"john#example.com", "country": "uk"}}
The wholeo nested dictionary is a bigger list of address data. The goal is to create pairs of people from the list with randomized partners where partners from the same country should be preferd. So my first real issue is to find a good way to group them by that country value.
I‘m sure there‘s a smarter way to do this than iterating through the dict and writing all records out to some new list/dict?
I think this is close to what you need:
result = {key:[i for i in value] for key, value in itertools.groupby(people, lambda item: item["fields"]["country"])}
What this does is use itertools.groupby to group all people in the people list by their specified country. The resulting dictionary has countries as keys, and the unpacked groupings (matching people) as values. Input is expected as a list of dictionaries like the one in your example:
people = [{"id": "1234565", "fields": {"name": "john", "email":"john#example.com", "country": "uk"}},
{"id": "654321", "fields": {"name": "sam", "email":"sam#example.com", "country": "uk"}}]
Sample output:
>>> print(result)
>>> {'uk': [{'fields': {'name': 'john', 'email': 'john#example.com', 'country': 'uk'}, 'id': '1234565'}, {'fields': {'name': 'sam', 'email': 'sam#example.com', 'country': 'uk'}, 'id': '654321'}]}
For a cleaner result, the looping construct can be tweaked so that only the ID of each person is included in the result dict:
result = {key:[i["id"] for i in value] for key, value in itertools.groupby(people, lambda item: item["fields"]["country"])}
>>> print(result)
>>> {'uk': ['1234565', '654321']}
EDIT: Sorry, I forgot about the sorting. Simply sort the list of people by country before putting it through groupby. It should now work properly:
sort = sorted(people, key=lambda item: item["fields"]["country"])
Here is another one that uses defaultdict:
import collections
def make_groups(nested_dicts, nested_key):
default = collections.defaultdict(list)
for nested_dict in nested_dicts:
for value in nested_dict.values():
try:
default[value[nested_key]].append(nested_dict)
except TypeError:
pass
return default
To test the results:
import random
COUNTRY = {'af', 'br', 'fr', 'mx', 'uk'}
people = [{'id': i, 'fields': {
'name': 'name'+str(i),
'email': str(i)+'#email',
'country': random.sample(COUNTRY, 1)[0]}}
for i in range(10)]
country_groups = make_groups(people, 'country')
for country, persons in country_groups.items():
print(country, persons)
Random output:
fr [{'id': 0, 'fields': {'name': 'name0', 'email': '0#email', 'country': 'fr'}}, {'id': 1, 'fields': {'name': 'name1', 'email': '1#email', 'country': 'fr'}}, {'id': 4, 'fields': {'name': 'name4', 'email': '4#email', 'country': 'fr'}}]
br [{'id': 2, 'fields': {'name': 'name2', 'email': '2#email', 'country': 'br'}}, {'id': 8, 'fields': {'name': 'name8', 'email': '8#email', 'country': 'br'}}]
uk [{'id': 3, 'fields': {'name': 'name3', 'email': '3#email', 'country': 'uk'}}, {'id': 7, 'fields': {'name': 'name7', 'email': '7#email', 'country': 'uk'}}]
af [{'id': 5, 'fields': {'name': 'name5', 'email': '5#email', 'country': 'af'}}, {'id': 9, 'fields': {'name': 'name9', 'email': '9#email', 'country': 'af'}}]
mx [{'id': 6, 'fields': {'name': 'name6', 'email': '6#email', 'country': 'mx'}}]
If I have a list of contact dictionaries like this:
{'name': 'Rob', 'phoneNumbers': [{'phone': '123-3214', 'type': 'home'}, {'phone': '456-3216', 'type': 'work'}]}
how could I update this dictionary to remove the dashes from the phone numbers in a list of contact dictionaries pythonically?
You could just nest loops:
for contact_dict in list_of_dicts:
for phone_dict in contact_dict['phoneNumbers']:
phone_dict['phone'] = phone_dict['phone'].replace('-', '')
This alters the values in-place.
Or you could create a whole new copy of the structure, with the alterations made:
[dict(contact, phoneNumbers=[
dict(phone_dict, phone=phone_dict['phone'].replace('-', ''))
for phone_dict in contact['phoneNumbers']])
for contact in list_of_dicts]
This creates a semi-shallow copy; only the phoneNumbers key is explicitly copied, but any other mutable values are just referenced by the new dictionaries.
Demo:
>>> list_of_dicts = [{'name': 'Rob', 'phoneNumbers': [{'phone': '123-3214', 'type': 'home'}, {'phone': '456-3216', 'type': 'work'}]}]
>>> [dict(contact, phoneNumbers=[
... dict(phone_dict, phone=phone_dict['phone'].replace('-', ''))
... for phone_dict in contact['phoneNumbers']])
... for contact in list_of_dicts]
[{'phoneNumbers': [{'phone': '1233214', 'type': 'home'}, {'phone': '4563216', 'type': 'work'}], 'name': 'Rob'}]
>>> for contact_dict in list_of_dicts:
... for phone_dict in contact_dict['phoneNumbers']:
... phone_dict['phone'] = phone_dict['phone'].replace('-', '')
...
>>> list_of_dicts
[{'phoneNumbers': [{'phone': '1233214', 'type': 'home'}, {'phone': '4563216', 'type': 'work'}], 'name': 'Rob'}]
Just str.replace the -
d ={'name': "Rob", 'phoneNumbers': [{'phone': '123-3214', 'type': 'home'}, {'phone': '456-3216', 'type': 'work'}]}
for dct in d["phoneNumbers"]:
dct['phone'] = dct['phone'].replace("-","",1)
Which gives you:
{'phoneNumbers': [{'phone': '1233214', 'type': 'home'}, {'phone': '4563216', 'type': 'work'}], 'name': 'Rob'}
I have a python dictionary and I would like to find and replace part of the characters in the values of the dictionary. I'm using python 2.7.
My dictionary is
data1 = {'customer_order': {'id': '20'},
'patient':
{'birthdate': None,
'medical_proc': None,
'medical_ref': 'HG_CTRL12',
'name': 'Patient_96',
'sex': None},
'physician_name': 'John Doe'
}
I would like to change the underscore to backslash underscore only in the values of the dictionary, in this case only for Patient_96 and HG_CTRL12.
I would like to change it to the following:
data1 = {'customer_order': {'id': '20'},
'patient':
{'birthdate': None,
'medical_proc': None,
'medical_ref': 'HG\_CTRL12',
'name': 'Patient\_96',
'sex': None},
'physician_name': 'John Doe'
}
Thank you for your help
This function recursively replaces the underscore in the values of the dictionary with replace_char:
def replace_underscores(a_dict, replace_char):
for k, v in a_dict.items():
if not isinstance(v, dict):
if v and '_' in v:
a_dict[k] = v.replace('_', replace_char)
else:
replace_underscores(v, replace_char)
More on isinstance() here.
>>> for i in data1:
... if type(data1[i]) is str:
... if data1[i]:
... data1[i] = data1[i].replace('_','\_')
... elif type(data1[i]) is dict:
... for j in data1[i]:
... if data1[i][j]:
... data1[i][j] = data1[i][j].replace('_','\_')
...
>>>
>>>
>>> data1
{'physician_name': 'John Doe', 'customer_order': {'id': '20'}, 'patient': {'medical_ref': 'HG\\_CTRL12', 'medical_proc': None, 'name': 'Patient\\_96', 'birthdate': None, 'sex': None}}