I am new with the concept of dictionaries and trying to learn them. What I have is a dictionary like this:
{'cars': [{'values': [1, 534],
{'values': [25,32,164]
'bikes': [{'values': [23,12,1]
{'values': [2,4]
{'values': [68,69]
{'values': [4,93]
What I try to achieve is add Ids to all inner values starting from 1
If you want the ID as part of the value group, like this:
{'cars': [{'values': [1, 534], 'sedan': 1, 'count': 2, 'ID': 1},
{'values': [25, 32, 164], 'sedan': 1, 'count': 10, 'ID': 2}],
'bikes': [{'values': [23, 12, 1], 'road': 0, 'count': 9},
...
You can do:
for i in range(len(try_dict['cars'])):
try_dict['cars'][i]['ID'] = i+1
If you want what Phydeaux suggests, you can do:
new_dict = {'cars': {}}
for i in range(len(try_dict['cars'])):
new_dict['cars'][i+1] = try_dict['cars'][i]
Which will give you:
{'cars': {1: {'values': [1, 534], 'sedan': 1, 'count': 2},
2: {'values': [25, 32, 164], 'sedan': 1, 'count': 10}}}
If you want not just cars but also bikes (and maybe trucks, trains, whatever...). Use:
new_dict = {}
for key in try_dict.keys():
new_dict[key] = {}
for i in range(len(try_dict[key])):
new_dict[key][i+1] = try_dict[key][i]
This will give you:
{'cars': {1: {'values': [1, 534], 'sedan': 1, 'count': 2},
2: {'values': [25, 32, 164], 'sedan': 1, 'count': 10}},
'bikes': {1: {'values': [23, 12, 1], 'road': 0, 'count': 9},
2: {'values': [2, 4], 'road': 1, 'count': 24},
3: {'values': [68, 69], 'sedan': 0, 'count': 28},
4: {'values': [4, 93], 'sedan': 0, 'count': 6}}}
You can do this using a simple function:
def idx(dict, key):
dict = dict
dict[key].insert(0, 0)
return dict
Full Code:
def idx(dict, key):
dict = dict
dict[key].insert(0, 0)
return dict
dict = {'cars': [{'values': [1, 534],
'sedan': 1,
'count': 2},
{'values': [25,32,164],
'sedan': 1,
'count': 10}],
'bikes': [{'values': [23,12,1],
'road': 0,
'count': 9},
{'values': [2,4],
'road': 1,
'count': 24},
{'values': [68,69],
'sedan': 0,
'count': 28},
{'values': [4,93],
'sedan': 0,
'count': 6}]}
dict = idx(dict, "cars")
print(dict["cars"][1])
Explanation:
Replace dictionary with a new edited dictionary:
dict = {key: [...,...,...]}
dict = idx(dict, key)
Function is using the .insert method to insert 0 for the value of the first index to the key provided.
Learn more about Python .insert() method at:
[
https://www.w3schools.com/python/ref_list_insert.asp
Related
I have several lists of dictionaries, where each dictionary contains a unique id value that is common among all lists. I'd like to combine them into a single list of dicts, where each dict is joined on that id value.
list1 = [{'id': 1, 'value': 20}, {'id': 2, 'value': 21}]
list2 = [{'id': 1, 'sum': 10}, {'id': 2, 'sum': 11}]
list3 = [{'id': 1, 'total': 30}, {'id': 2, 'total': 32}]
desired_output = [{'id': 1, 'value': 20, 'sum': 10, 'total': 30}, {'id': 2, 'value': 21, 'sum': 11, 'total': 32}]
I tried doing something like the answer found at https://stackoverflow.com/a/42018660/7564393, but I'm getting very confused since I have more than 2 lists. Should I try using a defaultdict approach? More importantly, I am NOT always going to know the other values, only that the id value is present in all dicts.
You can use itertools.groupby():
from itertools import groupby
list1 = [{'id': 1, 'value': 20}, {'id': 2, 'value': 21}]
list2 = [{'id': 1, 'sum': 10}, {'id': 2, 'sum': 11}]
list3 = [{'id': 1, 'total': 30}, {'id': 2, 'total': 32}]
desired_output = []
for _, values in groupby(sorted([*list1, *list2, *list3], key=lambda x: x['id']), key=lambda x: x['id']):
temp = {}
for d in values:
temp.update(d)
desired_output.append(temp)
Result:
[{'id': 1, 'value': 20, 'sum': 10, 'total': 30}, {'id': 2, 'value': 21, 'sum': 11, 'total': 32}]
list1 = [{'id': 1, 'value': 20}, {'id': 2, 'value': 21}]
list2 = [{'id': 1, 'sum': 10}, {'id': 2, 'sum': 11}]
list3 = [{'id': 1, 'total': 30}, {'id': 2, 'total': 32}]
# combine all lists
d = {} # id -> dict
for l in [list1, list2, list3]:
for list_d in l:
if 'id' not in list_d: continue
id = list_d['id']
if id not in d:
d[id] = list_d
else:
d[id].update(list_d)
# dicts with same id are grouped together since id is used as key
res = [v for v in d.values()]
print(res)
You can first build a dict of dicts, then turn it into a list:
from itertools import chain
from collections import defaultdict
list1 = [{'id': 1, 'value': 20}, {'id': 2, 'value': 21}]
list2 = [{'id': 1, 'sum': 10}, {'id': 2, 'sum': 11}]
list3 = [{'id': 1, 'total': 30}, {'id': 2, 'total': 32}]
dict_out = defaultdict(dict)
for d in chain(list1, list2, list3):
dict_out[d['id']].update(d)
out = list(dict_out.values())
print(out)
# [{'id': 1, 'value': 20, 'sum': 10, 'total': 30}, {'id': 2, 'value': 21, 'sum': 11, 'total': 32}]
itertools.chain allows you to iterate on all the dicts contained in the 3 lists. We build a dict dict_out having the id as key, and the corresponding dict being built as value. This way, we can easily update the already built part with the small dict of our current iteration.
Here, I have presented a functional approach without using itertools (which is excellent in rapid development work).
This solution will work for any number of lists as the function takes variable number of arguments and also let user to specify the type of return output (list/dict).
By default it returns list as you want that otherwise it returns dictionary in case if you pass as_list = False.
I preferred dictionary to solve this because its fast and search complexity is also less.
Just have a look at the below get_packed_list() function.
get_packed_list()
def get_packed_list(*dicts_lists, as_list=True):
output = {}
for dicts_list in dicts_lists:
for dictionary in dicts_list:
_id = dictionary.pop("id") # id() is in-built function so preferred _id
if _id not in output:
# Create new id
output[_id] = {"id": _id}
for key in dictionary:
output[_id][key] = dictionary[key]
dictionary["id"] = _id # push back the 'id' after work (call by reference mechanism)
if as_list:
return [output[key] for key in output]
return output # dictionary
Test
list1 = [{'id': 1, 'value': 20}, {'id': 2, 'value': 21}]
list2 = [{'id': 1, 'sum': 10}, {'id': 2, 'sum': 11}]
list3 = [{'id': 1, 'total': 30}, {'id': 2, 'total': 32}]
output = get_packed_list(list1, list2, list3)
print(output)
# [{'id': 1, 'value': 20, 'sum': 10, 'total': 30}, {'id': 2, 'value': 21, 'sum': 11, 'total': 32}]
output = get_packed_list(list1, list2, list3, as_list=False)
print(output)
# {1: {'id': 1, 'value': 20, 'sum': 10, 'total': 30}, 2: {'id': 2, 'value': 21, 'sum': 11, 'total': 32}}
list1 = [{'id': 1, 'value': 20}, {'id': 2, 'value': 21}]
list2 = [{'id': 1, 'sum': 10}, {'id': 2, 'sum': 11}]
list3 = [{'id': 1, 'total': 30}, {'id': 2, 'total': 32}]
print(list1+list2+list3)
list1 = [{'id': 1, 'value': 20}, {'id': 2, 'value': 21}]
list2 = [{'id': 1, 'sum': 10}, {'id': 2, 'sum': 11}]
list3 = [{'id': 1, 'total': 30}, {'id': 2, 'total': 32}]
result = []
for i in range(0,len(list1)):
final_dict = dict(list(list1[i].items()) + list(list2[i].items()) + list(list3[i].items()))
result.append(final_dict)
print(result)
output : [{'id': 1, 'value': 20, 'sum': 10, 'total': 30}, {'id': 2, 'value': 21, 'sum': 11, 'total': 32}]
I have list p1:
p1 = [
{'id': 1, 'area': 5},
{'id': 2, 'area': 6},
{'id': 3, 'area': 10},
{'id': 4, 'area': 6},
{'id': 5, 'area': 6},
{'id': 6, 'area': 6},
{'id': 7, 'area': 4},
{'id': 8, 'area': 4}
]
And I need to separate this list by area value, like this (p2):
p2 = {
4: [
{'id': 7, 'area': 4},
{'id': 8, 'area': 4}
],
5: [
{'id': 1, 'area': 5}
],
6: [
{'id': 2, 'area': 6},
{'id': 4, 'area': 6},
{'id': 5, 'area': 6},
{'id': 6, 'area': 6}
],
10: [
{'id': 3, 'area': 10}
]
}
My solution is:
areas = {x['area'] for x in p1}
p2 = {}
for area in areas:
p2[area] = [x for x in p1 if x['area'] == area]
It seems to work, but is there any better and more "pythonic" solution?
Using groupby you get
>>> import itertools
>>> f = lambda t: t['area']
>>> {i: list(b) for i, b in itertools.groupby(sorted(p1, key=f), key=f)}
Gives
{4: [{'area': 4, 'id': 7},
{'area': 4, 'id': 8}],
5: [{'area': 5, 'id': 1}],
6: [{'area': 6, 'id': 2},
{'area': 6, 'id': 4},
{'area': 6, 'id': 5},
{'area': 6, 'id': 6}],
10: [{'area': 10, 'id': 3}]}
edit: If you don't like using lambdas you can also do, as suggested by bro-grammer
>>> import operator
>>> f = operator.itemgetter('area')
You can simply use defaultdict:
from collections import defaultdict
result = defaultdict(list)
for i in p1:
result[i['area']].append(i)
Yes, use one of the grouping idioms. Using a vanilla dict:
In [15]: p1 = [
...: {'id': 1, 'area': 5},
...: {'id': 2, 'area': 6},
...: {'id': 3, 'area': 10},
...: {'id': 4, 'area': 6},
...: {'id': 5, 'area': 6},
...: {'id': 6, 'area': 6},
...: {'id': 7, 'area': 4},
...: {'id': 8, 'area': 4}
...: ]
In [16]: p2 = {}
In [17]: for d in p1:
...: p2.setdefault(d['area'], []).append(d)
...:
In [18]: p2
Out[18]:
{4: [{'area': 4, 'id': 7}, {'area': 4, 'id': 8}],
5: [{'area': 5, 'id': 1}],
6: [{'area': 6, 'id': 2},
{'area': 6, 'id': 4},
{'area': 6, 'id': 5},
{'area': 6, 'id': 6}],
10: [{'area': 10, 'id': 3}]}
Or more neatly, using a defaultdict:
In [23]: from collections import defaultdict
In [24]: p2 = defaultdict(list)
In [25]: for d in p1:
...: p2[d['area']].append(d)
...:
In [26]: p2
Out[26]:
defaultdict(list,
{4: [{'area': 4, 'id': 7}, {'area': 4, 'id': 8}],
5: [{'area': 5, 'id': 1}],
6: [{'area': 6, 'id': 2},
{'area': 6, 'id': 4},
{'area': 6, 'id': 5},
{'area': 6, 'id': 6}],
10: [{'area': 10, 'id': 3}]})
Before writing a function, I would like to be sure there is no pre-built (optimized) solution (like sorted()) that can:
From a dictionary like this one :
tags = {'pinoyako':{'likes': 119, 'comments': 11, 'count': 1}, 'dii':{'likes': 151, 'comments': 3, 'count': 1},'djiphantom3':{'likes': 127, 'comments': 6, 'count': 1}}
Order the keys based on 'likes', 'comments' or 'count'. If it's based on 'likes', the output should be a list ordered :
output = [['dii',151],['djiphantom3',127],['pinoyako',119]]
Use a generator expression within sorted() function with a proper key function:
In [22]: from operator import itemgetter
In [23]: sorted(((k, v['likes']) for k, v in tags.items()), key=itemgetter(1), reverse=True)
Out[23]: [('dii', 151), ('djiphantom3', 127), ('pinoyako', 119)]
You need more entries to illustrate. Check this:
tags = {'pinoyako':{'likes': 119, 'comments': 11, 'count': 1},
'pinoyako2':{'likes': 120, 'comments': 5, 'count': 2},
'djiphantom32':{'likes': 1275, 'comments': 61, 'count': 15},
'dii2':{'likes': 151, 'comments': 33, 'count': 13},
'dii':{'likes': 151, 'comments': 3, 'count': 1},
'djiphantom3':{'likes': 1275, 'comments': 61, 'count': 1}
}
tags_sorted = sorted(tags.items(), key=lambda x: (x[1]['likes'], x[1]['comments'], x[1]['count']))
tags_sorted
Output:
[('pinoyako', {'comments': 11, 'count': 1, 'likes': 119}),
('pinoyako2', {'comments': 5, 'count': 2, 'likes': 120}),
('dii', {'comments': 3, 'count': 1, 'likes': 151}),
('dii2', {'comments': 33, 'count': 13, 'likes': 151}),
('djiphantom3', {'comments': 61, 'count': 1, 'likes': 1275}),
('djiphantom32', {'comments': 61, 'count': 15, 'likes': 1275})]
Then you can do this:
tags_sorted = [[k, v['likes']] for k,v in tags_sorted]
tags_sorted
Output:
[['pinoyako', 119],
['pinoyako2', 120],
['dii', 151],
['dii2', 151],
['djiphantom3', 1275],
['djiphantom32', 1275]]
I have this list:
list_users= [[{'points': 9, 'values': 1, 'division': 1, 'user_id': 3}], [{'points': 3, 'values': 0, 'division': 1, 'user_id': 1}], [{'points': 2, 'values': 0, 'division': 1, 'user_id': 4}], [{'points': 9, 'values': 0, 'division': 1, 'user_id': 11}], [{'points': 3, 'values': 0, 'division': 1, 'user_id': 10}], [{'points': 100, 'values': 4, 'division': 1, 'user_id': 2}], [{'points': 77, 'values': 2, 'division': 1, 'user_id': 5}], [{'points': 88, 'values': 3, 'division': 1, 'user_id': 6}], [{'points': 66, 'values': 1, 'division': 1, 'user_id': 7}], [{'points': 2, 'values': 0, 'division': 1, 'user_id': 8}]]
I need to sort the list by points and values.
How can I sort it if dict is inside a list inside the main list?
I generated this list by query and than just append to list_users?
Access the dictionary containing points and values by indexing on the inner list:
list_users_sorted = sorted(list_users, key=lambda x: (x[0]['points'], x[0]['values']))
# ^ ^
Sort using a key function for sorted that builds a tuple of points and values for each dict in each list.
def kf(x):
return (x[0]["points"], x[0]["values"])
s = sorted(list_users, key=kf)
print(s)
Output:
[[{'division': 1, 'points': 2, 'user_id': 4, 'values': 0}],
[{'division': 1, 'points': 2, 'user_id': 8, 'values': 0}],
[{'division': 1, 'points': 3, 'user_id': 1, 'values': 0}],
[{'division': 1, 'points': 3, 'user_id': 10, 'values': 0}],
[{'division': 1, 'points': 9, 'user_id': 11, 'values': 0}],
[{'division': 1, 'points': 9, 'user_id': 3, 'values': 1}],
[{'division': 1, 'points': 66, 'user_id': 7, 'values': 1}],
[{'division': 1, 'points': 77, 'user_id': 5, 'values': 2}],
[{'division': 1, 'points': 88, 'user_id': 6, 'values': 3}],
[{'division': 1, 'points': 100, 'user_id': 2, 'values': 4}]]
I have a csv file that I converted into dataframe using Pandas. Here's the dataframe:
Customer ProductID Count
John 1 50
John 2 45
Mary 1 75
Mary 2 10
Mary 5 15
I need an output in the form of a dictionary that looks like this:
{ProductID:1, Count:{John:50, Mary:75}},
{ProductID:2, Count:{John:45, Mary:10}},
{ProductID:5, Count:{John:0, Mary:15}}
I read the following answers:
python pandas dataframe to dictionary
and
Convert dataframe to dictionary
This is the code that I'm having:
df = pd.read_csv('customer.csv')
dict1 = df.set_index('Customer').T.to_dict('dict')
dict2 = df.to_dict(orient='records')
and this is my current output:
dict1 = {'John': {'Count': 45, 'ProductID': 2}, 'Mary': {'Count': 15, 'ProductID': 5}}
dict2 = [{'Count': 50, 'Customer': 'John', 'ProductID': 1},
{'Count': 45, 'Customer': 'John', 'ProductID': 2},
{'Count': 75, 'Customer': 'Mary', 'ProductID': 1},
{'Count': 10, 'Customer': 'Mary', 'ProductID': 2},
{'Count': 15, 'Customer': 'Mary', 'ProductID': 5}]
IIUC you can use:
d = df.groupby('ProductID').apply(lambda x: dict(zip(x.Customer, x.Count)))
.reset_index(name='Count')
.to_dict(orient='records')
print (d)
[{'ProductID': 1, 'Count': {'John': 50, 'Mary': 75}},
{'ProductID': 2, 'Count': {'John': 45, 'Mary': 10}},
{'ProductID': 5, 'Count': {'Mary': 15}}]