I have a list with dictionary inside, with the following keys: id, external_id_client, product_id, as below:
[{'id': 3, 'external_id_client': '7298', 'product_id': 13},
{'id': 3, 'external_id_client': '7298', 'product_id': 8},
{'id': 4, 'external_id_client': '3', 'product_id': 12},
{'id': 4, 'external_id_client': '3', 'product_id': 9},
{'id': 5, 'external_id_client': '4', 'product_id': 12}]
I need that if the ID and external_id are repeated and I have information in the product_id key it creates a tuple in the product_id.
Expected output:
[{'id': 3, 'external_id_client': '7298', 'product_id': (13, 8)},
{'id': 4, 'external_id_client': '3', 'product_id': (12, 9)},
{'id': 5, 'external_id_client': '4', 'product_id': (12)}]
How can I do this?
You can use itertools.groupby for this:
from itertools import groupby
lst = [
{'id': 3, 'external_id_client': '7298', 'product_id': 13},
{'id': 3, 'external_id_client': '7298', 'product_id': 8},
{'id': 4, 'external_id_client': '3', 'product_id': 12},
{'id': 4, 'external_id_client': '3', 'product_id': 9},
{'id': 5, 'external_id_client': '4', 'product_id': 12}
]
result = [
{
"id": itemid,
"external_id_client": client,
"product_id": tuple(item["product_id"] for item in group)
}
for (itemid, client), group
in groupby(lst, key = lambda item: (item["id"], item["external_id_client"]))
]
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I have a list of dicts (same format) like this :
L = [
{'id': 1, 'name': 'john', 'age': 34},
{'id': 1, 'name': 'john', 'age': 34},
{'id': 2, 'name': 'hanna', 'age': 30},
{'id': 2, 'name': 'hanna', 'age': 30},
{'id': 3, 'name': 'stack', 'age': 40}
]
I want to remove duplication and get the number of this duplication like this
[
{'id': 1, 'name': 'john', 'age': 34, 'duplication': 2},
{'id': 2, 'name': 'hanna', 'age': 30, 'duplication': 2},
{'id': 3, 'name': 'stack', 'age': 40, 'duplication': 1}
]
I already managed to remove the duplication by using a set.... but I can't get the number of duplications
my code :
no_duplication = [dict(s) for s in set(frozenset(d.items()) for d in L)]
no_duplication = [
{'id': 1, 'name': 'john', 'age': 34},
{'id': 2, 'name': 'hanna', 'age': 30},
{'id': 3, 'name': 'stack', 'age': 40}
]
Here is a solution you can give a try using collections.Counter,
from collections import Counter
print([
{**dict(k), "duplicated": v}
for k, v in Counter(frozenset(i.items()) for i in L).items()
])
[{'age': 34, 'duplicated': 2, 'id': 1, 'name': 'john'},
{'age': 30, 'duplicated': 2, 'id': 2, 'name': 'hanna'},
{'age': 40, 'duplicated': 1, 'id': 3, 'name': 'stack'}]
ar = [
{'id': 1, 'name': 'john', 'age': 34},
{'id': 1, 'name': 'john', 'age': 34},
{'id': 2, 'name': 'hanna', 'age': 30},
{'id': 2, 'name': 'hanna', 'age': 30},
{'id': 3, 'name': 'stack', 'age': 40}
]
br = []
cnt = []
for i in ar:
if i not in br:
br.append(i)
cnt.append(1)
else:
cnt[br.index(i)] += 1
for i in range(len(br)):
br[i]['duplication'] = cnt[i]
The desired output is contained in br as:
[
{'id': 1, 'name': 'john', 'age': 34, 'duplication': 2},
{'id': 2, 'name': 'hanna', 'age': 30, 'duplication': 2},
{'id': 3, 'name': 'stack', 'age': 40, 'duplication': 1}
]
I'm trying to code a faster way to solve the following problem but I don't know how to do it:
I have the following list of dicts and list of identifiers:
list_of_dicts = [{'id': 1, 'name': 'A'}, {'id': 2, 'name': 'B'}, {'id': 3, 'name': 'C'}, {'id': 4, 'name': 'D'}]
list_of_ids = [1, 3, 2, 4, 1, 3, 4]
I'd like to have the following output:
[{'id': 1, 'name': 'A'}, {'id': 3, 'name': 'C'}, {'id': 2, 'name': 'B'}, {'id': 4, 'name': 'D'}, {'id': 1, 'name': 'A'}, {'id': 3, 'name': 'C'}, {'id': 4, 'name': 'D'}]
The way I'm doing it is:
list_of_dict_ids = [d['id'] for d in list_of_dicts]
ordered_list_by_ids = [list_of_dicts[list_of_dict_ids.index(i)] for i in list_of_ids]
Is there any faster way to do it?
You can do like this :
dic = {d["id"]: d for d in list_of_dicts}
dic
>>>{1: {'id': 1}, 2: {'id': 2}, 3: {'id': 3}, 4: {'id': 4}}
lst =[dic[i] for i in list_of_ids]
lst
>>>[{'id': 1}, {'id': 3}, {'id': 2}, {'id': 4}, {'id': 1}, {'id': 3}, {'id': 4}]
I have this :
[
[{ 'position': 1, 'user_id': 2, 'value': 4, 'points': 100}],
[{ 'position': 2, 'user_id': 6, 'value': 3, 'points': 88}],
[{ 'position': 3, 'user_id': 5, 'value': 2, 'points': 77}],
[{ 'position': 4, 'user_id': 7, 'value': 1, 'points': 66}],
[{ 'position': 5, 'user_id': 3, 'value': 1, 'points': 9}],
[{ 'position': 6, 'user_id': 11, 'value': 0, 'points': 9}],
[{ 'position': 7, 'user_id': 1, 'value': 0, 'points': 3}],
[{ 'position': 8, 'user_id': 10, 'value': 0, 'points': 3}],
[{ 'position': 9, 'user_id': 4, 'value': 0, 'points': 2}],
[{ 'position': 10, 'user_id': 8, 'value': 0, 'points': 2}]
]
is organized by points.
The idea is to choose the user_id and generate a new list with the selected 5 users.
Example:
user_id=3:
[{ 'position': 3, 'user_id': 5, 'value': 2, 'points': 77}],
[{ 'position': 4, 'user_id': 7, 'value': 1, 'points': 66}],
[{ 'position': 5, 'user_id': 3, 'value': 1, 'points': 9}],
[{ 'position': 6, 'user_id': 11, 'value': 0, 'points': 9}],
[{ 'position': 7, 'user_id': 1, 'value': 0, 'points': 3}]
It returns user_id 3 in the middle with 2 users hight and 2 users lower
user_id=2
[{ 'position': 1, 'user_id': 2, 'value': 4, 'points': 100}],
[{ 'position': 2, 'user_id': 6, 'value': 3, 'points': 88}],
[{ 'position': 3, 'user_id': 5, 'value': 2, 'points': 77}],
[{ 'position': 4, 'user_id': 7, 'value': 1, 'points': 66}],
[{ 'position': 5, 'user_id': 3, 'value': 1, 'points': 9}],
As user_id hasn't higher users it returns 4 lower users. So is always same logic.
user_id=9:
[{ 'position': 6, 'user_id': 11, 'value': 0, 'points': 9}],
[{ 'position': 7, 'user_id': 1, 'value': 0, 'points': 3}],
[{ 'position': 8, 'user_id': 10, 'value': 0, 'points': 3}],
[{ 'position': 9, 'user_id': 4, 'value': 0, 'points': 2}],
[{ 'position': 10, 'user_id': 8, 'value': 0, 'points': 2}]
on user_id=9 We only have 1 user lower so we add 3 higher users
If for example we just have 2 users in list, it should return that 2 users.
Main rules:
If we have 5 users or more, as to return 5 users.
if we have 4 users, as to return 4 users
How is a good way to do it?
thanks
This is basically only an update of my answer to your original question.
a = [
[{ 'position': 1, 'user_id': 2, 'value': 4, 'points': 100}],
[{ 'position': 2, 'user_id': 6, 'value': 3, 'points': 88}],
[{ 'position': 3, 'user_id': 5, 'value': 2, 'points': 77}],
[{ 'position': 4, 'user_id': 7, 'value': 1, 'points': 66}],
[{ 'position': 5, 'user_id': 3, 'value': 1, 'points': 9}],
[{ 'position': 6, 'user_id': 11, 'value': 0, 'points': 9}],
[{ 'position': 7, 'user_id': 1, 'value': 0, 'points': 3}],
[{ 'position': 8, 'user_id': 10, 'value': 0, 'points': 3}],
[{ 'position': 9, 'user_id': 4, 'value': 0, 'points': 2}],
[{ 'position': 10, 'user_id': 8, 'value': 0, 'points': 2}]
]
# Sort it if not already sorted
# a.sort(key=lambda x: x[0]['position'])
def find_index(l, user_id):
i = 0
while l[i][0]['user_id'] != user_id:
i += 1
return i
def get_subset(l, i):
return l[:(i + 1 + max(2, 4 - i))][-5:]
get_subset(a, find_index(a, 3))
I have below list with nested lists (sort of key,values)
inp1=[{'id': 0, 'name': 98, 'value': 9}, {'id': 1, 'name': 66, 'value': 8}, {'id': 2, 'name': 29, 'value': 5}, {'id': 3, 'name': 99, 'value': 3}, {'id': 4, 'name': 15, 'value': 9}]
Am trying to replace 'name' with 'wid' and 'value' with 'wrt', how can I do it on same list?
My output should be like
inp1=[{'id': 0, 'wid': 98, 'wrt': 9}, {'id': 1, 'wid': 66, 'wrt': 8}, {'id': 2, 'wid': 29, 'wrt': 5}, {'id': 3, 'wid': 99, 'wrt': 3}, {'id': 4, 'wid': 15, 'wrt': 9}]
I tried below, but it doesn't work as list cannot be indexed with string but integer
inp1['name'] = inp1['wid']
inp1['value'] = inp1['wrt']
I tried if I can find any examples, but mostly I found only this for dictionary and not list.
You need to iterate each item, and remove the old entry (dict.pop is handy for this - it removes an entry and return the value) and assign to new keyes:
>>> inp1 = [
... {'id': 0, 'name': 98, 'value': 9},
... {'id': 1, 'name': 66, 'value': 8},
... {'id': 2, 'name': 29, 'value': 5},
... {'id': 3, 'name': 99, 'value': 3},
... {'id': 4, 'name': 15, 'value': 9}
... ]
>>>
>>> for d in inp1:
... d['wid'] = d.pop('name')
... d['wrt'] = d.pop('value')
...
>>> inp1
[{'wid': 98, 'id': 0, 'wrt': 9},
{'wid': 66, 'id': 1, 'wrt': 8},
{'wid': 29, 'id': 2, 'wrt': 5},
{'wid': 99, 'id': 3, 'wrt': 3},
{'wid': 15, 'id': 4, 'wrt': 9}]
def f(item):
if(item.has_key('name') and not item.has_key('wid')):
item['wid']=item.pop('name')
if(item.has_key('value') and not item.has_key('wrt')):
item['wrt']=item.pop('value')
map(f,inp1)
Output:
[{'wrt': 9, 'wid': 98, 'id': 0}, {'wrt': 8, 'wid': 66, 'id': 1}, {'wrt': 5, 'wid': 29, 'id': 2}, {'wrt': 3, 'wid': 99, 'id': 3}, {'wrt': 9, 'wid': 15, 'id': 4}]
I am trying to replace list element value with value looked up in dictionary how do I do that?
list = [1, 3, 2, 10]
d = {'id': 1, 'val': 30},{'id': 2, 'val': 53}, {'id': 3, 'val': 1}, {'id': 4, 'val': 9}, {'id': 5, 'val': 2}, {'id': 6, 'val': 6}, {'id': 7, 'val': 11}, {'id': 8, 'val': 89}, {'id': 9, 'val': 2}, {'id': 10, 'val': 4}
for i in list:
for key, v in d.iteritems():
???
???
so at the end I am expecting:
list = [30, 1, 53, 4]
thank you
D2 = dict((x['id'], x['val']) for x in D)
L2 = [D2[x] for x in L]
td = (
{'val': 30, 'id': 1},
{'val': 53, 'id': 2},
{'val': 1, 'id': 3},
{'val': 9, 'id': 4},
{'val': 2, 'id': 5},
{'val': 6, 'id': 6},
{'val': 11, 'id': 7},
{'val': 89, 'id': 8},
{'val': 2, 'id': 9},
{'val': 4, 'id': 10}
)
source_list = [1, 3, 2, 10]
final_list = []
for item in source_list:
for d in td:
if d['id'] == item:
final_list.append(d['val'])
print('Source : ', source_list)
print('Final : ', final_list)
Result
Source : [1, 3, 2, 10]
Final : [30, 1, 53, 4]