I have to sort a dict like:
jobs = {'elem_05': {'id': 'fifth'},
'elem_03': {'id': 'third'},
'elem_01': {'id': 'first'},
'elem_00': {'id': 'zeroth'},
'elem_04': {'id': 'fourth'},
'elem_02': {'id': 'second'}}
based on the "id" elements, whose order can be found in a list:
sorting_list = ['zeroth', 'first', 'second', 'third', 'fourth', 'fifth']
The trivial way to solve the problem is to use:
tmp = {}
for x in sorting_list:
for k, v in jobs.items():
if v["id"] == x:
tmp.update({k: v})
but I was trying to figure out a more efficient and pythonic way.
I've been trying sorted and lambda functions as key, but I'm not familiar with that yet, so I was unsuccessful so far.
I would use a dictionary as key for sorted:
order = {k:i for i,k in enumerate(sorting_list)}
# {'zeroth': 0, 'first': 1, 'second': 2, 'third': 3, 'fourth': 4, 'fifth': 5}
out = dict(sorted(jobs.items(), key=lambda x: order.get(x[1].get('id'))))
output:
{'elem_00': {'id': 'zeroth'},
'elem_01': {'id': 'first'},
'elem_02': {'id': 'second'},
'elem_03': {'id': 'third'},
'elem_04': {'id': 'fourth'},
'elem_05': {'id': 'fifth'}}
There is a way to sort the dict using lambda as a sorting key:
jobs = {'elem_05': {'id': 'fifth'},
'elem_03': {'id': 'third'},
'elem_01': {'id': 'first'},
'elem_00': {'id': 'zeroth'},
'elem_04': {'id': 'fourth'},
'elem_02': {'id': 'second'}}
sorting_list = ['zeroth', 'first', 'second', 'third', 'fourth', 'fifth']
sorted_jobs = dict(sorted(jobs.items(), key=lambda x: sorting_list.index(x[1]['id'])))
print(sorted_jobs)
This outputs
{'elem_00': {'id': 'zeroth'}, 'elem_01': {'id': 'first'}, 'elem_02': {'id': 'second'}, 'elem_03': {'id': 'third'}, 'elem_04': {'id': 'fourth'}, 'elem_05': {'id': 'fifth'}}
I have a feeling the sorted expression could be cleaner but I didn't get it to work any other way.
You can use OrderedDict:
from collections import OrderedDict
sorted_jobs = OrderedDict([(el, jobs[key]['id']) for el, key in zip(sorting_list, jobs.keys())])
This creates an OrderedDict object which is pretty similar to dict, and can be converted to dict using dict(sorted_jobs).
Similar to what is already posted, but with error checking in case id doesn't appear in sorting_list
sorting_list = ['zeroth', 'first', 'second', 'third', 'fourth', 'fifth']
jobs = {'elem_05': {'id': 'fifth'},
'elem_03': {'id': 'third'},
'elem_01': {'id': 'first'},
'elem_00': {'id': 'zeroth'},
'elem_04': {'id': 'fourth'},
'elem_02': {'id': 'second'}}
def custom_order(item):
try:
return sorting_list.index(item[1]["id"])
except ValueError:
return len(sorting_list)
jobs_sorted = {k: v for k, v in sorted(jobs.items(), key=custom_order)}
print(jobs_sorted)
The sorted function costs O(n log n) in average time complexity. For a linear time complexity you can instead create a reverse mapping that maps each ID to the corresponding dict entry:
mapping = {d['id']: (k, d) for k, d in jobs.items()}
so that you can then construct a new dict by mapping sorting_list with the ID mapping above:
dict(map(mapping.get, sorting_list))
which, with your sample input, returns:
{'elem_00': {'id': 'zeroth'}, 'elem_01': {'id': 'first'}, 'elem_02': {'id': 'second'}, 'elem_03': {'id': 'third'}, 'elem_04': {'id': 'fourth'}, 'elem_05': {'id': 'fifth'}}
Demo: https://replit.com/#blhsing/WorseChartreuseFonts
Related
I have one list of elements and another list of dictionaries and i want to insert list of elements into each dictionary of list
list_elem = [1,2,3]
dict_ele = [{"Name":"Madhu","Age":25},{"Name":"Raju","Age:24},{""Name":"Mani","Age":12}],
OUTPUT As:
[{"ID":1,"Name":"Madhu","Age":25},{"ID":2,"Name":"Raju","Age:24},{"ID":3,"Name":"Mani","Age":12}]
I have tried this way :
dit = [{"id":item[0]} for item in zip(sam)]
# [{"id":1,"id":2,"id":3}]
dic1 = list(zip(dit,data))
print(dic1)
# [({"id":1},{{"Name":"Madhu","Age":25}},{"id":2},{"Name":"Raju","Age:24},{"id":3},{""Name":"Mani","Age":12})]
What is the most efficient way to do this in Python?
Making an assumption here that the OP's original question has a typo in the definition of dict_ele and also that list_elem isn't really necessary.
dict_ele = [{"Name":"Madhu","Age":25},{"Name":"Raju","Age":24},{"Name":"Mani","Age":12}]
dit = [{'ID': id_, **d} for id_, d in enumerate(dict_ele, 1)]
print(dit)
Output:
[{'ID': 1, 'Name': 'Madhu', 'Age': 25}, {'ID': 2, 'Name': 'Raju', 'Age': 24}, {'ID': 3, 'Name': 'Mani', 'Age': 12}]
dict_ele = [{"Name":"Madhu","Age":25},{"Name":"Raju","Age":24},{"Name":"Mani","Age":12}]
list_elem = [1,2,3]
[{'ID': id, **_dict} for id, _dict in zip(list_elem, dict_ele)]
[{'ID': 1, 'Name': 'Madhu', 'Age': 25}, {'ID': 2, 'Name': 'Raju', 'Age': 24}, {'ID': 3, 'Name': 'Mani', 'Age': 12}]
try this: r = [{'id':e[0], **e[1]} for e in zip(list_elem, dict_ele)]
Not working too much with dictionaries and thinking about it for too long. I'm looking for merging dictionaries in a way I'm describing a little lower. Dictionary might be a bit bigger.
Thanks!
dict1:
{'0': '5251', '1': '5259'}
list:
[{'id': 5259, 'name': 'chair'}, {'id': 5251, 'name': 'table'}]
Result:
{'0': {'id': 5251, 'name': 'table'}, '1': {'id': 5259, 'name': 'chair'}}
This should work:
dict1 = {'0': '5251', '1': '5259'}
ls = [{'id': 5259, 'name': 'chair'}, {'id': 5251, 'name': 'table'}]
idToEntry = dict([x['id'], x] for x in ls)
dict2 = dict([k, idToEntry[int(v)]] for k, v in dict1.items())
print(dict2)
The output:
{'0': {'id': 5251, 'name': 'table'}, '1': {'id': 5259, 'name': 'chair'}}
This solution will be somewhat slow for larger lists, but it is very simple:
result = {}
for key, val in dict1.items():
for dict2 in list1:
if dict2['id'] == val:
result['key'] = dict2
Construct a dictionary with reversed keys and values from your original:
d = {'0': '5251', '1': '5259'}
rev_dict = {v: k for k, v in d.items()}
Now you can use the id from each item of your list as an index into your dict
l = [{'id': 5259, 'name': 'chair'}, {'id': 5251, 'name': 'table'}]
merged_data = {rev_dict[str(x['id'])]: x for x in l}
# {'1': {'id': 5259, 'name': 'chair'}, '0': {'id': 5251, 'name': 'table'}}
d = {'0': '5251', '1': '5259'}
l = [{'id': 5259, 'name': 'chair'}, {'id': 5251, 'name': 'table'}]
# Lets prepare an intermediate dict for faster computation on large data
d1 = {str(x['id']): x for x in l}
merged_dict = {x: d1[y] for x, y in d.items()}
print(merged_dict)
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)
For each item in dictA, I want to search for it in dictB, if dictB has it then I want to pull some other values from dictB and add it to dictA.
An example that is working is here, however it is rather slow as I have 50,000+ items to search through and it will perform this similar function on multiple dicts.
Is there a fast method of performing this search?
dictA = [
{'id': 12345},
{'id': 67890},
{'id': 11111},
{'id': 22222}
]
dictB = [
{'id': 63351, 'name': 'Bob'},
{'id': 12345, 'name': 'Carl'},
{'id': 59933, 'name': 'Amy'},
{'id': 11111, 'name': 'Chris'}
]
for i in dictA:
name = None
for j in dictB:
if i['id'] == j['id']:
name = j['name']
i['name'] = name
The dictA output after this would be:
dictA = [
{'id': 12345, 'name': 'Carl'},
{'id': 67890, 'name': None},
{'id': 11111, 'name': 'Chris'},
{'id': 22222, 'name': None}
]
The given is list of dict. You can create dict from that assuming id is uninque. Converting from list of dict to dict will work for your case.
dictA = [
{'id': 12345},
{'id': 67890},
{'id': 11111},
{'id': 22222}
]
dictB = [
{'id': 63351, 'name': 'Bob'},
{'id': 12345, 'name': 'Carl'},
{'id': 59933, 'name': 'Amy'},
{'id': 11111, 'name': 'Chris'}
]
actual_dictB = dict()
for d in dictB:
actual_dictB[d['id']] = d['name']
for i in dictA:
i['name'] = actual_dictB.pop(i['id'], None) # now search have became O(1) constant. So best time complexity achived O(n) n=length of dictA
print(dictA)
Follow up for additional question:
actual_dictB = dict()
for d in dictB:
id_ = d['id']
d.pop('id')
actual_dictB[id_] = d
tmp = dict([(k,None) for k in dictB[0].keys() if k!='id'])
for i in dictA:
if i['id'] not in actual_dictB:
i.update(tmp)
else:
i.update(actual_dictB[i['id']])
print(dictA)
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}}