Change one dict/list python structure to another - python

I have the following data
data={
None: [
{'ne': '1', 'na': '1'},
{'ne': '2', 'na': '2'},
{'ne': '3', 'na': '3'},
{'ne': '4', 'na': '4'}
],
'AO': [
{'ne': '2', 'na': '2'},
{'ne': '6', 'na': '6'}
],
'NZ': [
{'ne': '1', 'na': '1'}
]
}
and I want to have a list from it like this:
[
{'ne': '1', 'na': '1', 'country': [None, 'NZ']},
{'ne': '2', 'na': '2', 'country': [None, 'AO']},
{'ne': '3', 'na': '3', 'country': [None]},
{'ne': '4', 'na': '4', 'country': [None]},
{'ne': '6', 'na': '6', 'country': ['AO']}
]
my code is doing it fine but it's far from being "pythonic" because I'm a newbie at python:
data = {None: [{'ne': '1', 'na': '1'}, {'ne': '2', 'na': '2'}, {'ne': '3', 'na': '3'}, {'ne': '4', 'na': '4'}], 'AO': [{'ne': '2', 'na': '2'}, {'ne': '6', 'na': '6'}], 'NZ': [{'ne': '1', 'na': '1'}]}
data_list = []
for k,d in data.items():
for dd in d:
dd['country'] = k
data_list.append(dd)
help_dict = {}
for item in data_list:
help_dict[item['ne']] = False
final_list = []
for idx, val in enumerate(data_list):
if not help_dict[val['ne']]:
val['country'] = [val['country']]
for idx2, val2 in enumerate(data_list):
if idx2 != idx and val['ne'] == val2['ne']:
val['country'].append(val2['country'])
help_dict[val['ne']] = True
final_list.append(val)
print(final_list)
can someone help me with a better way to do this?


new = [x for key,value in data.items() for x in value]
# remove duplicate dictionaries
new = [dict(t) for t in {tuple(d.items()) for d in new}]
for d in new:
d['country'] = [key for key,data in data.items() if d in data]
print(new)
>>> [{'ne': '2', 'na': '2', 'country': [None, 'AO']},
{'ne': '4', 'na': '4', 'country': [None]},
{'ne': '1', 'na': '1', 'country': [None, 'NZ']},
{'ne': '6', 'na': '6', 'country': ['AO']},
{'ne': '3', 'na': '3', 'country': [None]}]
If you want to preserve the order
new = [x for n,(key,value) in enumerate(data.items()) for x in value]
seen = set()
new_l = []
for d in new:
t = tuple(d.items())
if t not in seen:
seen.add(t)
new_l.append(d)
for d in new_l:
d['country'] = [key for key,data in data.items() if d in data]
print(new_l)
>>> [{'ne': '1', 'na': '1', 'country': [None, 'NZ']},
{'ne': '2', 'na': '2', 'country': [None, 'AO']},
{'ne': '3', 'na': '3', 'country': [None]},
{'ne': '4', 'na': '4', 'country': [None]},
{'ne': '6', 'na': '6', 'country': ['AO']}]


This is a really naive approach to solve your problem, due to it requiring that the inner dictionaries are sorted in the same order for it to "match" earlier found dictionaries.
For more complex dictionaries inside of the country this might not give correct results:
data={
None: [
{'ne': '1', 'na': '1'},
{'ne': '2', 'na': '2'},
{'ne': '3', 'na': '3'},
{'ne': '4', 'na': '4'}
],
'AO': [
{'ne': '2', 'na': '2'},
{'ne': '6', 'na': '6'}
],
'NZ': [
{'ne': '1', 'na': '1'}
]
}
d = {}
for country in data:
for dictionary in data[country]:
# Create a key that is a string of the dictionary, and value is dictionary plus country
x = d.setdefault(str(dictionary), dictionary | {"country": []})
# If you're using Python < 3.9, use this instead:
# x = d.setdefault(str(dictionary), {**dictionary, "country": []})
x["country"].append(country)
# pprint only used to represent data better
import pprint
pprint.pp(list(d.values()))
Output:
[{'ne': '1', 'na': '1', 'country': [None, 'NZ']},
{'ne': '2', 'na': '2', 'country': [None, 'AO']},
{'ne': '3', 'na': '3', 'country': [None]},
{'ne': '4', 'na': '4', 'country': [None]},
{'ne': '6', 'na': '6', 'country': ['AO']}]


Firstly, I'm assuming ne and na are always the same.
An optimal intermediate data structure is a dict with ne/na as keys and country lists as values:
{'1': [None, 'NZ'],
'2': [None, 'AO'],
'3': [None],
'4': [None],
'6': ['AO']}
Once you have that goal in mind, it's super simple to do it Pythonically:
inter = {}
for k, dicts in data.items():
for d in dicts:
inter.setdefault(d['ne'], []).append(k)
dict.setdefault() is used to get the value if it exists, or if not, set it to a default, which is an empty list here. It's functionally the same as this:
ne = d['ne']
if ne not in inter:
inter[ne] = []
inter[ne].append(k)
You could also use collections.defaultdict(list) to do the same thing even more easily.
And once you have that dict, you just need to unpack it into a list of dicts:
result = [{'ne': ne, 'na': ne, 'country': c} for ne, c in inter.items()]
Which becomes:
[{'ne': '1', 'na': '1', 'country': [None, 'NZ']},
{'ne': '2', 'na': '2', 'country': [None, 'AO']},
{'ne': '3', 'na': '3', 'country': [None]},
{'ne': '4', 'na': '4', 'country': [None]},
{'ne': '6', 'na': '6', 'country': ['AO']}]

Related

group dict emelents on lists based on key name

I want to group dict keys in list of dicts based on dict keys name. The dict keys alawys starts with number for example :
{'1_account_id': '5', '1_qty_in': '10.01', '1_uom_in': '1', '2_account_id': '23', '2_qty_in': '2.01', '2_uom_in': '1','3_account_id': '23', '3_qty_in': '2.01', '3_uom_in': '1' ,'some_x':1,'some_y':0}
I want to convert that to list grouped by this number like:
[{'1':{account_id': '5','qty_in':10.01,'uom_in':1}},{'2':{account_id': '23','qty_in':2.01,'uom_in':1}}....] etc
This numbers is not static.
I tried multiple solutions but it seems not good for performace:
Like i looped range for 0 to SOMEBIGNUMBER but this is not a good solution.
i also tried to loop throw the elements to extract the numbers first and then loop again to group but the dict myaybe come not clean as it has other not needed keys.
What should i do ?

So given the input d
d = {'1_account_id': '5', '1_qty_in': '10.01', '1_uom_in': '1', '2_account_id': '23', '2_qty_in': '2.01', '2_uom_in': '1','3_account_id': '23', '3_qty_in': '2.01', '3_uom_in': '1' ,'some_x':1,'some_y':0}
ans = {}
for key in d:
n,k = key.split('_', 1)
if n in ans:
ans[n][k] = d[key]
else:
ans[n] = {'account_id': 0, 'qty_in': 0, 'uom_in': 0}
ans[n][k] = d[key]
The above code produces the following output in which the entire output ans is a dictionary with the numbers as keys, and the values for each of the keys is also a dictionary, each containing 3 keys and corresponding values :
{'1': {'account_id': '5', 'qty_in': '10.01', 'uom_in': '1'},
'2': {'account_id': '23', 'qty_in': '2.01', 'uom_in': '1'},
'3': {'account_id': '23', 'qty_in': '2.01', 'uom_in': '1'},
'some': {'account_id': 0, 'qty_in': 0, 'uom_in': 0, 'x': 1, 'y': 0}}
Now to convert it into the list which you have shown in the desired output, use the follwing code. Here we convert the dictionary of dictionaries to a list containing dictionaries.
ans2 = []
for row in ans:
temp = {}
temp[row] = ans[row]
ans2.append(temp)
which produces the desired output :
[{'1': {'account_id': '5', 'qty_in': '10.01', 'uom_in': '1'}},
{'2': {'account_id': '23', 'qty_in': '2.01', 'uom_in': '1'}},
{'3': {'account_id': '23', 'qty_in': '2.01', 'uom_in': '1'}},
{'some': {'account_id': 0, 'qty_in': 0, 'uom_in': 0, 'x': 1, 'y': 0}}]
Hope this helps !

Mostafa.
You can run this code example to solve it:
example_dict = {'1_account_id': '5', '1_qty_in': '10.01', '1_uom_in': '1', '2_account_id': '23', '2_qty_in': '2.01', '2_uom_in': '1','3_account_id': '23', '3_qty_in': '2.01', '3_uom_in': '1' ,'some_x':1,'some_y':0}
desired_keys = ['account_id','qty_in','uom_in']
result = {}
for key,value in example_dict.items():
key_num, key_name = key.split('_', 1)
if key_name in desired_keys:
result.setdefault(key_num, {})[key_name] = value
print(result)
{'1': {'accountid': '5', 'qtyin': '10.01', 'uomin': '1'}, '2': {'accountid': '23', 'qtyin': '2.01', 'uomin': '1'}, '3': {'accountid': '23', 'qtyin': '2.01', 'uomin': '1'}, 'some': {'x': 1, 'y': 0}}
It will print (as you desired):
{'1': {'accountid': '5', 'qtyin': '10.01', 'uomin': '1'}, '2': {'accountid': '23', 'qtyin': '2.01', 'uomin': '1'}, '3': {'accountid': '23', 'qtyin': '2.01', 'uomin': '1'}, 'some': {'x': 1, 'y': 0}}
Tell me if you do not understand something.

arranging dictionary elements in sorted order

In my scinario, I am having data as kind of a sorted list:
sample = ['1:4', '2:2', '3:1', '4:1', '5:1', 'B5S94Y:1', 'DB:4', 'ICEauthority:1', 'JPG:1', 'MZPZ5Y:1', 'Mdg80A:1', 'TAG:1', 'V2XO5Y:1', 'Xauthority:1', 'apmrc:1', 'asc:1', 'bak:1', 'baseA:4', 'baseB:4', 'bash:1', 'bash_history:1', 'bash_logout:1', 'bashrc:1', 'bat:1', 'bau:1', 'bdic:1', 'bin:5', 'c:282', 'cache-6:55', 'cache:103', 'cfg:3', 'conf:15', 'converted-launchers:1', 'cson:2', 'css:34', 'dat:2', 'db-journal:3', 'db-shm:1', 'db-wal:1', 'db:17', 'dbf:1', 'dbt:1', 'deb:3', 'desktop:2', 'dic:1', 'dirs:1', 'dll:42', 'dmrc:1', 'docx:9', 'exc:1', 'exe:20', 'final:1', 'fingerprint:4', 'fmt:1', 'gif:5', 'gitconfig:1', 'gitignore:6', 'gitkeep:2', 'gitmodules:1', 'gpg-agent:1', 'gpg:1', 'gradle:3', 'gvdb:1', 'gz:169', 'h:14', 'htm:1', 'html:43', 'icc:1', 'ico:1', 'ics:3', 'idx:5', 'info:1', 'ini:14', 'ino:1', 'isrunning:1', 'jar:1', 'java:7', 'journal:2', 'jpg:16', 'js-20170612122310:1', 'js-20170816210634:1', 'js:245', 'json:367', 'jsonlz4:39', 'kbx:1', 'keyring:1', 'keystore:1', 'ldb:23', 'list:2', 'little:2', 'locale:1', 'localstorage-journal:71', 'localstorage:71', 'lock:1', 'log:41', 'lst:2', 'lsup7I:1', 'lz4:1', 'm:1', 'map:2', 'md:92', 'md~:1', 'metadata-v2:6', 'metadata:6', 'mozlz4:2', 'name:1', 'nls:1', 'odb:2', 'odt:1', 'old:24', 'orig:1', 'otf:31', 'out:19', 'pack:5', 'pak:1', 'parentlock:2', 'pb:32', 'pdf:18', 'pf2:5', 'php:12', 'pl:1', 'pma:2', 'png:2348', 'pro:1', 'profile:1', 'properties:2', 'pset:42', 'pub:1', 'py:41', 'pyc:4', 'rb:1', 'rcache:2', 'rdf:2', 'reg:3', 'run:1', 'sample:72', 'sbstore:42', 'sdv:1', 'sh:27', 'so:2', 'source:1', 'spec:9', 'sqlite-journal:1', 'sqlite-shm:4', 'sqlite-wal:4', 'sqlite3-journal:1', 'sqlite3:1', 'sqlite:24', 'stamp:2', 'stderr:8', 'stdout:8', 'sth:19', 'sublime_session:1', 'sudo_as_admin_successful:1', 'svg:124', 'swo:1', 'swp:2', 'sys:1', 'tdb:3', 'thm:1', 'trashinfo:3', 'tsv:1', 'ttf:107', 'tvc:1', 'txt:63', 'update-timestamp:1', 'usage:6', 'viminfo:1', 'vxd:1', 'woff:1', 'x86_64-pc-linux-gnu:1', 'xba:1', 'xbel:1', 'xcu:1', 'xinputrc:1', 'xlb:2', 'xlc:2', 'xml:35', 'xpi:9', 'xsession-errors:1', 'yml:1', 'zcompdump:1', 'zip:5', 'zsh-template:1', 'zsh-theme:143', 'zsh-update:1', 'zsh:235', 'zsh_history:1', 'zshrc:1']
So after converting this above list to a dictionary with the following code:
dict_val = {a:b for a, b in [i.split(":") for i in sample]}
The output will be:
{'baseA': '4', 'baseB': '4', 'cache-6': '55', 'Xauthority': '1', 'gitmodules': '1', 'apmrc': '1', 'gz': '169', 'dbf': '1', 'kbx': '1', 'sqlite-shm': '4', 'dbt': '1', 'gitignore': '6', 'xml': '35', 'sbstore': '42', 'cache': '103', 'jar': '1', 'desktop': '2', 'source': '1', 'sqlite3-journal': '1', 'TAG': '1', '4': '1', 'tsv': '1', 'spec': '9', 'bin': '5', 'docx': '9', 'woff': '1', 'db': '17', 'V2XO5Y': '1', 'dat': '2', 'fingerprint': '4', 'lz4': '1', 'name': '1', 'bat': '1', 'bau': '1', 'converted-launchers': '1', 'h': '14', 'list': '2', 'lst': '2', 'gradle': '3', 'zsh-update': '1', 'mozlz4': '2', 'stderr': '8', 'sublime_session': '1', 'bak': '1', 'isrunning': '1', 'locale': '1', 'cfg': '3', 'htm': '1', 'odt': '1', 'xlb': '2', 'md~': '1', 'pma': '2', 'sqlite-journal': '1', 'odb': '2', 'dic': '1', 'tvc': '1', 'out': '19', 'ico': '1', 'icc': '1', 'dll': '42', '3': '1', 'rb': '1', 'ics': '3', 'py': '41', 'journal': '2', 'metadata': '6', 'dirs': '1', 'run': '1', 'tdb': '3', 'DB': '4', 'zshrc': '1', 'xpi': '9', 'pub': '1', 'js': '245', 'asc': '1', 'ldb': '23', 'xlc': '2', 'xbel': '1', 'properties': '2', 'bash': '1', 'sys': '1', 'c': '282', 'zip': '5', 'idx': '5', 'lsup7I': '1', 'zcompdump': '1', 'rdf': '2', 'dmrc': '1', 'Mdg80A': '1', 'pdf': '18', 'reg': '3', 'jsonlz4': '39', 'bashrc': '1', 'db-journal': '3', 'pf2': '5', 'localstorage': '71', 'old': '24', 'txt': '63', 'orig': '1', 'gvdb': '1', 'little': '2', 'pyc': '4', 'java': '7', 'log': '41', 'swo': '1', 'stamp': '2', 'vxd': '1', 'fmt': '1', 'gpg': '1', 'zsh-template': '1', 'pb': '32', 'gif': '5', 'json': '367', '2': '2', 'js-20170612122310': '1', 'swp': '2', 'bash_logout': '1', 'final': '1', 'pl': '1', 'gpg-agent': '1', 'sdv': '1', 'x86_64-pc-linux-gnu': '1', 'parentlock': '2', 'cson': '2', 'rcache': '2', 'otf': '31', 'usage': '6', 'bash_history': '1', 'localstorage-journal': '71', 'update-timestamp': '1', 'png': '2348', 'exc': '1', 'info': '1', 'md': '92', 'js-20170816210634': '1', 'sth': '19', 'yml': '1', 'sqlite-wal': '4', 'deb': '3', 'zsh': '235', 'pack': '5', 'zsh_history': '1', 'sqlite': '24', 'stdout': '8', 'lock': '1', 'pro': '1', 'gitkeep': '2', 'jpg': '16', 'sample': '72', 'ino': '1', 'pset': '42', 'ini': '14', 'conf': '15', 'xcu': '1', 'sudo_as_admin_successful': '1', 'xsession-errors': '1', 'keystore': '1', 'nls': '1', 'sh': '27', 'bdic': '1', '1': '4', 'html': '43', '5': '1', 'MZPZ5Y': '1', 'sqlite3': '1', 'pak': '1', 'ttf': '107', 'css': '34', 'profile': '1', 'map': '2', 'metadata-v2': '6', 'm': '1', 'zsh-theme': '143', 'trashinfo': '3', 'ICEauthority': '1', 'php': '12', 'B5S94Y': '1', 'viminfo': '1', 'exe': '20', 'db-shm': '1', 'xinputrc': '1', 'svg': '124', 'keyring': '1', 'JPG': '1', 'thm': '1', 'gitconfig': '1', 'so': '2', 'xba': '1', 'db-wal': '1'}
So I want to know why the order of the output data has changed compared to the input data and how to get the output in the same sorted order as the input is given!!

dicts in python versions older than 3.6 do not have any notion of ordering. If you want a dictionary that maintains order, you'll want the collections.OrderedDict data structure.
from collections import OrderedDict
mapping = OrderedDict(i.split(":") for i in sample)
mapping is an OrderedDict, a subclass of dict which supports all the basic dict functionality, plus the ordering.
Also, do not use dict as a variable name, it shadows the builtin class dict.

creating dictionary from the list with each elements as key:value pairs in python [duplicate]

This question already has answers here:
Python: how to build a dict from plain list of keys and values
(5 answers)
Closed 5 years ago.
I have a list where the content of list are key:value pairs something like shown below
sample =['ldb:21', 'baseB:4', 'cache-6:55', 'Xauthority:1', 'baseA:4',
'apmrc:1', 'gz:169', 'dbf:1', 'lst:2', 'sqlite-shm:4', 'ttf:107',
'gitignore:6', 'xml:35', 'sbstore:42', 'cache:103', 'jar:1',
'desktop:2', 'source:1', 'sqlite3-journal:1', 'TAG:1', '4:1',
'usage:6', 'yml:1', 'bin:5', 'docx:9', 'woff:1', 'db:17',
'gpg-agent:1', 'V2XO5Y:1', 'dat:2', 'fingerprint:4', 'lz4:1',
'cson:2', 'name:1', 'bat:1', 'bau:1', 'converted-launchers:1',
'h:14', 'list:2', 'xlb:2', 'dic:1', 'zsh-update:1',
'mozlz4:2', 'stderr:8', 'sublime_session:1', 'bak:1', 'dll:42',
'old:24', 'locale:1', 'cfg:3', 'htm:1', 'odt:1', 'keyring:1',
'md~:1', 'pma:2', 'sqlite-journal:1', 'odb:2', 'gradle:3', 'tvc:1',
'out:19', 'ico:1', 'icc:1', 'gpg:1', 'dbt:1', '3:1', 'rb:1',
'ics:3', 'reg:3', 'metadata:6', 'dirs:1', 'run:1', 'tdb:3',
'journal:2', 'zshrc:1', 'little:2', 'pub:1', 'js:245',
'asc:1', 'xbel:1', 'properties:2', 'bash:1', 'sys:1', 'c:282',
'zip:5', 'idx:5', 'lsup7I:1', 'zcompdump:1', 'rdf:2', 'dmrc:1',
'Mdg80A:1', 'pdf:18', 'xlc:2', 'jsonlz4:39', 'bashrc:1',
'db-journal:3', 'pf2:5', 'localstorage:71', 'isrunning:1',
'txt:63', 'orig:1', 'gvdb:1', 'xpi:9', 'php:12',
'gitmodules:1', 'log:41', 'swo:1', 'stamp:2', 'vxd:1',
'fmt:1', 'py:41']
I want to convert the above list into dictionary key:value pairs.
Then I want to convert it into json format.

You can try this:
sample =['ldb:21', 'baseB:4', 'cache-6:55', 'Xauthority:1', 'baseA:4', 'apmrc:1', 'gz:169', 'dbf:1', 'lst:2', 'sqlite-shm:4', 'ttf:107', 'gitignore:6', 'xml:35', 'sbstore:42', 'cache:103', 'jar:1', 'desktop:2', 'source:1', 'sqlite3-journal:1', 'TAG:1', '4:1', 'usage:6', 'yml:1', 'bin:5', 'docx:9', 'woff:1', 'db:17', 'gpg-agent:1', 'V2XO5Y:1', 'dat:2', 'fingerprint:4', 'lz4:1', 'cson:2', 'name:1', 'bat:1', 'bau:1', 'converted-launchers:1', 'h:14', 'list:2', 'xlb:2', 'dic:1', 'zsh-update:1', 'mozlz4:2', 'stderr:8', 'sublime_session:1', 'bak:1', 'dll:42', 'old:24', 'locale:1', 'cfg:3', 'htm:1', 'odt:1', 'keyring:1', 'md~:1', 'pma:2', 'sqlite-journal:1', 'odb:2', 'gradle:3', 'tvc:1', 'out:19', 'ico:1', 'icc:1', 'gpg:1', 'dbt:1', '3:1', 'rb:1', 'ics:3', 'reg:3', 'metadata:6', 'dirs:1', 'run:1', 'tdb:3', 'journal:2', 'zshrc:1', 'little:2', 'pub:1', 'js:245', 'asc:1', 'xbel:1', 'properties:2', 'bash:1', 'sys:1', 'c:282', 'zip:5', 'idx:5', 'lsup7I:1', 'zcompdump:1', 'rdf:2', 'dmrc:1', 'Mdg80A:1', 'pdf:18', 'xlc:2', 'jsonlz4:39', 'bashrc:1', 'db-journal:3', 'pf2:5', 'localstorage:71', 'isrunning:1', 'txt:63', 'orig:1', 'gvdb:1', 'xpi:9', 'php:12', 'gitmodules:1', 'log:41', 'swo:1', 'stamp:2', 'vxd:1', 'fmt:1', 'py:41']
final_data = {a:b for a, b in [i.split(":") for i in sample]}
print(final_data)
Output:
{'ldb': '21', 'baseB': '4', 'cache-6': '55', 'Xauthority': '1', 'baseA': '4', 'apmrc': '1', 'gz': '169', 'dbf': '1', 'lst': '2', 'dll': '42', 'ttf': '107', 'gitignore': '6', 'xml': '35', 'sbstore': '42', 'cache': '103', 'jar': '1', 'desktop': '2', 'source': '1', 'sqlite3-journal': '1', 'TAG': '1', '4': '1', 'usage': '6', 'yml': '1', 'bin': '5', 'docx': '9', 'woff': '1', 'dbt': '1', 'V2XO5Y': '1', 'dat': '2', 'fingerprint': '4', 'lz4': '1', 'name': '1', 'bat': '1', 'bau': '1', 'converted-launchers': '1', 'h': '14', 'list': '2', 'xlb': '2', 'gradle': '3', 'zsh-update': '1', 'stderr': '8', 'sublime_session': '1', 'bak': '1', 'old': '24', 'locale': '1', 'cfg': '3', 'htm': '1', 'odt': '1', 'md~': '1', 'pma': '2', 'sqlite-journal': '1', 'odb': '2', 'dic': '1', 'tvc': '1', 'out': '19', 'ico': '1', 'icc': '1', 'sqlite-shm': '4', '3': '1', 'rb': '1', 'ics': '3', 'py': '41', 'reg': '3', 'metadata': '6', 'dirs': '1', 'run': '1', 'tdb': '3', 'journal': '2', 'zshrc': '1', 'xpi': '9', 'pub': '1', 'js': '245', 'asc': '1', 'xbel': '1', 'properties': '2', 'bash': '1', 'c': '282', 'swo': '1', 'idx': '5', 'lsup7I': '1', 'rdf': '2', 'dmrc': '1', 'Mdg80A': '1', 'pdf': '18', 'xlc': '2', 'jsonlz4': '39', 'bashrc': '1', 'db-journal': '3', 'pf2': '5', 'localstorage': '71', 'isrunning': '1', 'txt': '63', 'orig': '1', 'gvdb': '1', 'little': '2', 'gitmodules': '1', 'log': '41', 'zip': '5', 'stamp': '2', 'vxd': '1', 'fmt': '1', 'gpg': '1', 'gpg-agent': '1', 'cson': '2', 'zcompdump': '1', 'mozlz4': '2', 'db': '17', 'sys': '1', 'php': '12', 'keyring': '1'}

Just use dict, splitting on the colons.
dict(item.split(':') for item in sample)
Or the equivalent functional approach,
from operator import methodcaller
dict(map(methodcaller('split', ':'), sample))
From there json.dumps to get a JSON formatted string.

sample =['ldb:21', 'baseB:4', 'cache-6:55', 'Xauthority:1', 'baseA:4', 'apmrc:1', 'gz:169', 'dbf:1', 'lst:2', 'sqlite-shm:4', 'ttf:107', 'gitignore:6', 'xml:35', 'sbstore:42', 'cache:103', 'jar:1', 'desktop:2', 'source:1', 'sqlite3-journal:1', 'TAG:1', '4:1', 'usage:6', 'yml:1', 'bin:5', 'docx:9', 'woff:1', 'db:17', 'gpg-agent:1', 'V2XO5Y:1', 'dat:2', 'fingerprint:4', 'lz4:1', 'cson:2', 'name:1', 'bat:1', 'bau:1', 'converted-launchers:1', 'h:14', 'list:2', 'xlb:2', 'dic:1', 'zsh-update:1', 'mozlz4:2', 'stderr:8', 'sublime_session:1', 'bak:1', 'dll:42', 'old:24', 'locale:1', 'cfg:3', 'htm:1', 'odt:1', 'keyring:1', 'md~:1', 'pma:2', 'sqlite-journal:1', 'odb:2', 'gradle:3', 'tvc:1', 'out:19', 'ico:1', 'icc:1', 'gpg:1', 'dbt:1', '3:1', 'rb:1', 'ics:3', 'reg:3', 'metadata:6', 'dirs:1', 'run:1', 'tdb:3', 'journal:2', 'zshrc:1', 'little:2', 'pub:1', 'js:245', 'asc:1', 'xbel:1', 'properties:2', 'bash:1', 'sys:1', 'c:282', 'zip:5', 'idx:5', 'lsup7I:1', 'zcompdump:1', 'rdf:2', 'dmrc:1', 'Mdg80A:1', 'pdf:18', 'xlc:2', 'jsonlz4:39', 'bashrc:1', 'db-journal:3', 'pf2:5', 'localstorage:71', 'isrunning:1', 'txt:63', 'orig:1', 'gvdb:1', 'xpi:9', 'php:12', 'gitmodules:1', 'log:41', 'swo:1', 'stamp:2', 'vxd:1', 'fmt:1', 'py:41']
dict_data = {}
for data in sample:
item = data.split(':')
dict_data[item[0]] = item[1]
print dict_data
output:
{'ldb': '21', 'baseB': '4', 'cache-6': '55', 'Xauthority': '1', 'baseA': '4', 'apmrc': '1', 'gz': '169', 'dbf': '1', 'lst': '2', 'dll': '42', 'ttf': '107', 'gitignore': '6', 'xml': '35', 'sbstore': '42', 'cache': '103', 'jar': '1', 'desktop': '2', 'source': '1', 'sqlite3-journal': '1', 'TAG': '1', '4': '1', 'usage': '6', 'yml': '1', 'bin': '5', 'docx': '9', 'woff': '1', 'dbt': '1', 'V2XO5Y': '1', 'dat': '2', 'fingerprint': '4', 'lz4': '1', 'name': '1', 'bat': '1', 'bau': '1', 'converted-launchers': '1', 'h': '14', 'list': '2', 'xlb': '2', 'gradle': '3', 'zsh-update': '1', 'stderr': '8', 'sublime_session': '1', 'bak': '1', 'old': '24', 'locale': '1', 'cfg': '3', 'htm': '1', 'odt': '1', 'md~': '1', 'pma': '2', 'sqlite-journal': '1', 'odb': '2', 'dic': '1', 'tvc': '1', 'out': '19', 'ico': '1', 'icc': '1', 'sqlite-shm': '4', '3': '1', 'rb': '1', 'ics': '3', 'py': '41', 'reg': '3', 'metadata': '6', 'dirs': '1', 'run': '1', 'tdb': '3', 'journal': '2', 'zshrc': '1', 'xpi': '9', 'pub': '1', 'js': '245', 'asc': '1', 'xbel': '1', 'properties': '2', 'bash': '1', 'c': '282', 'swo': '1', 'idx': '5', 'lsup7I': '1', 'rdf': '2', 'dmrc': '1', 'Mdg80A': '1', 'pdf': '18', 'xlc': '2', 'jsonlz4': '39', 'bashrc': '1', 'db-journal': '3', 'pf2': '5', 'localstorage': '71', 'isrunning': '1', 'txt': '63', 'orig': '1', 'gvdb': '1', 'little': '2', 'gitmodules': '1', 'log': '41', 'zip': '5', 'stamp': '2', 'vxd': '1', 'fmt': '1', 'gpg': '1', 'gpg-agent': '1', 'cson': '2', 'zcompdump': '1', 'mozlz4': '2', 'db': '17', 'sys': '1', 'php': '12', 'keyring': '1'}

Iterate over your list and split the pair into two to form a dictionary
dicta = {a:b for a,b in [pair.split(":") for pair in sample]}
[pair.split(":") for pair in sample] gives you the list of lists:
[["lib", "1"], ["baseB", "4"], ... ]
You can put key as string and value as int, as it seems to be count.
dicta = {a:int(b) for a,b in [pair.split(":") for pair in sample]}

Merging Python list of dicts grouped by key value

I want to merge two dicts on the id. Dict x contains many distinct ids and different row counts per each id. Dict y contains multiple key values, and always has less rows than dict x.
x = [{'costgroup': '1', 'POC1': '2', 'post': '5','id': '1'},
{'costgroup': '2', 'POC1': '1', 'post': '4','id': '1'},
{'costgroup': '3', 'POC1': '5', 'post': '2', 'id': '1'},
{'costgroup': '1', 'POC1': '2', 'post': '5','id': '2'},
{'costgroup': '2', 'POC1': '1', 'post': '4','id': '2'},
{'costgroup': '3', 'POC1': '5', 'post': '2', 'id': '2'},
{'costgroup': '3', 'POC1': '5', 'post': '2', 'id': '2'}]
y = [{'id': '1', 'laminate': 'D'},
{ 'id':'2', 'laminate': T'}]
The output that I want is the following:
z = [{'costgroup': '1', 'POC1': '2', 'post': '5','id': '1','laminate':'D'},
{'costgroup': '2', 'POC1': '1', 'post': '4','id': '1','laminate': 'D'},
{'costgroup': '3', 'POC1': '5', 'post': '2', 'id': '1','laminate': 'D'},
{'costgroup': '1', 'POC1': '2', 'post': '5','id': '2','laminate': 'T'},
{'costgroup': '2', 'POC1': '1', 'post': '4','id': '2','laminate': 'T'},
{'costgroup': '3', 'POC1': '5', 'post': '2', 'id': '2','laminate': 'T'},
{'costgroup': '3', 'POC1': '5', 'post': '2', 'id': '2','laminate': 'T'}]
This is easy to achieve using pandas
dfx = pd.DataFrame(x)
dfy = pd.DataFrame(y)
pd.merge(dfx,dfy, how ='left', left_on = 'id', right_on = 'id' )
But, I am going to apply this using an AWS Lambda function and I don't want to have the overhead of pandas and the output needs to be a dict. I tried the code below which gets me closer, but then i had to add something to find the distinct values of the ID and iterate through them.But,still don't have the output that I need.
valuelist = ['1']
def copyf(dictlist, key, valuelist):
return [d for d in dictlist if d[key] in valuelist]
y1 = copyf(y, 'id', valuelist)
x1 = copyf(x, 'id', valuelist)
y1.append(x1)
The above provides this output, which is interesting but not what I need.
[{'distance': '2', 'id': '1', 'laminate': 'D'},
[{'POC1': '2', 'costgroup': '1', 'id': '1', 'post': '5'},
{'POC1': '1', 'costgroup': '2', 'id': '1', 'post': '4'},
{'POC1': '5', 'costgroup': '3', 'id': '1', 'post': '2'}]]

def merge(d1, d2):
"""Given two dicts, merge them into a new dict as a shallow copy."""
result = d1.copy()
result.update(d2)
return result
result = [merge(d1, d2) for d1 in x for d2 in y if d1["id"] == d2["id"]]
print(result)
Gives
[{'POC1': '2', 'costgroup': '1', 'id': '1', 'laminate': 'D', 'post': '5'},
{'POC1': '1', 'costgroup': '2', 'id': '1', 'laminate': 'D', 'post': '4'},
{'POC1': '5', 'costgroup': '3', 'id': '1', 'laminate': 'D', 'post': '2'},
{'POC1': '2', 'costgroup': '1', 'id': '2', 'laminate': 'T', 'post': '5'},
{'POC1': '1', 'costgroup': '2', 'id': '2', 'laminate': 'T', 'post': '4'},
{'POC1': '5', 'costgroup': '3', 'id': '2', 'laminate': 'T', 'post': '2'},
{'POC1': '5', 'costgroup': '3', 'id': '2', 'laminate': 'T', 'post': '2'}]
Merge function from here: How to merge two Python dictionaries in a single expression?
There's a more concise syntax in Python 3.5 but you're on 2.7.

import copy
x = [{'costgroup': '1', 'POC1': '2', 'post': '5','id': '1'},
{'costgroup': '2', 'POC1': '1', 'post': '4','id': '1'},
{'costgroup': '3', 'POC1': '5', 'post': '2', 'id': '1'},
{'costgroup': '1', 'POC1': '2', 'post': '5','id': '2'},
{'costgroup': '2', 'POC1': '1', 'post': '4','id': '2'},
{'costgroup': '3', 'POC1': '5', 'post': '2', 'id': '2'},
{'costgroup': '3', 'POC1': '5', 'post': '2', 'id': '2'}]
y = [{'id': '1', 'laminate': 'D'},
{ 'id':'2', 'laminate': 'T'}
]
#create the id->laminate mapping
m = { d['id']: d['laminate'] for d in y }
#create the final output
z = []
for d in x:
#make a copy of the dictionary from x so that the input data
#is not overwritten
item = dict(d) #copy.deepcopy(d)
item.update({'laminate': m[d['id']]})
z.append(item)
print(z)
this produces
[
{'laminate': 'D', 'post': '5', 'POC1': '2', 'id': '1', 'costgroup': '1'},
{'laminate': 'D', 'post': '4', 'POC1': '1', 'id': '1', 'costgroup': '2'},
{'laminate': 'D', 'post': '2', 'POC1': '5', 'id': '1', 'costgroup': '3'},
{'laminate': 'T', 'post': '5', 'POC1': '2', 'id': '2', 'costgroup': '1'},
{'laminate': 'T', 'post': '4', 'POC1': '1', 'id': '2', 'costgroup': '2'},
{'laminate': 'T', 'post': '2', 'POC1': '5', 'id': '2', 'costgroup': '3'},
{'laminate': 'T', 'post': '2', 'POC1': '5', 'id': '2', 'costgroup': '3'}]

z = []
for dx in x:
for dy in y:
if dx['id'] == dy['id']:
z.append(dict(dx.items () + dy.items()))
print z

How to convert list into a dictionary by python

old_list = [ ['ID0', 'ID1'], ['4', '8'], ['5', '6'] ]
I want convert list to new list
key = ['id', 'frame', 'length']
new_list = [{'id': 'ID0', 'frame': '4', 'length': '5'}, {'id': 'ID1', 'frame': '8', 'length': '6'}]

Here's a one-line approach:
>>> [{'id':x, 'frame':y, 'length':z} for x,y,z in zip(*old_list)]
[{'length': '5', 'frame': '4', 'id': 'ID0'}, {'length': '6', 'frame': '8', 'id': 'ID1'}]

new_list=[]
for x,y,z in zip(old_list[0],old_list[1], old_list[2]):
dict = {'id' : x, 'frame' : y, 'length': z}
new_list.append(dict)

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