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I have a question about the convert key.
First, I have this type of word count in Data Frame.
[Example]
dict = {'forest': 10, 'station': 3, 'office': 7, 'park': 2}
I want to get this result.
[Result]
result = {'name': 'forest', 'value': 10,
'name': 'station', 'value': 3,
'name': 'office', 'value': 7,
'name': 'park', 'value': 2}
Please check this issue.
As Rakesh said:
dict cannot have duplicate keys
The closest way to achieve what you want is to build something like that
my_dict = {'forest': 10, 'station': 3, 'office': 7, 'park': 2}
result = list(map(lambda x: {'name': x[0], 'value': x[1]}, my_dict.items()))
You will get
result = [
{'name': 'forest', 'value': 10},
{'name': 'station', 'value': 3},
{'name': 'office', 'value': 7},
{'name': 'park', 'value': 2},
]
As Rakesh said, You can't have duplicate values in the dictionary
You can simply try this.
dict = {'forest': 10, 'station': 3, 'office': 7, 'park': 2}
result = {}
count = 0;
for key in dict:
result[count] = {'name':key, 'value': dict[key]}
count = count + 1;
print(result)
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
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 a dicts in a list and some dicts are identical. I want to find duplicated ones and want to add to new list or dictionary with how many duplicate they have.
import itertools
myListCombined = list()
for a, b in itertools.combinations(myList, 2):
is_equal = set(a.items()) - set(b.items())
if len(is_equal) == 0:
a.update(count=2)
myListCombined.append(a)
else:
a.update(count=1)
b.update(count=1)
myListCombined.append(a)
myListCombined.append(b)
myListCombined = [i for n, i enumerate(myListCombine) if i not in myListCombine[n + 1:]]
This code is kinda working, but it's just for 2 duplicated dicts in list. a.update(count=2) won't work in this situations.
I'm also deleting duplicated dicts after separete them in last line, but i'm not sure if it's going to work well.
Input:
[{'name': 'Mary', 'age': 25, 'salary': 1000},
{'name': 'John', 'age': 25, 'salary': 2000},
{'name': 'George', 'age': 30, 'salary': 2500},
{'name': 'John', 'age': 25, 'salary': 2000},
{'name': 'John', 'age': 25, 'salary': 2000}]
Desired Output:
[{'name': 'Mary', 'age': 25, 'salary': 1000, 'count':1},
{'name': 'John', 'age': 25, 'salary': 2000, 'count': 3},
{'name': 'George', 'age': 30, 'salary': 2500, 'count' 1}]
You could try the following, which first converts each dictionary to a frozenset of key,value tuples (so that they are hashable as required by collections.Counter).
import collections
a = [{'a':1}, {'a':1},{'b':2}]
print(collections.Counter(map(lambda x: frozenset(x.items()),a)))
Edit to reflect your desired input/output:
from copy import deepcopy
def count_duplicate_dicts(list_of_dicts):
cpy = deepcopy(list_of_dicts)
for d in list_of_dicts:
d['count'] = cpy.count(d)
return list_of_dicts
x = [{'a':1},{'a':1}, {'c':3}]
print(count_duplicate_dicts(x))
If your dict data is well structured and the contents of the dict are simple data types, e.g., numbers and string, and you have following data analysis processing, I would suggest you use pandas, which provide rich functions. Here is a sample code for your case:
In [32]: data = [{'name': 'Mary', 'age': 25, 'salary': 1000},
...: {'name': 'John', 'age': 25, 'salary': 2000},
...: {'name': 'George', 'age': 30, 'salary': 2500},
...: {'name': 'John', 'age': 25, 'salary': 2000},
...: {'name': 'John', 'age': 25, 'salary': 2000}]
...:
...: df = pd.DataFrame(data)
...: df['counts'] = 1
...: df = df.groupby(df.columns.tolist()[:-1]).sum().reset_index(drop=False)
...:
In [33]: df
Out[33]:
age name salary counts
0 25 John 2000 3
1 25 Mary 1000 1
2 30 George 2500 1
In [34]: df.to_dict(orient='records')
Out[34]:
[{'age': 25, 'counts': 3, 'name': 'John', 'salary': 2000},
{'age': 25, 'counts': 1, 'name': 'Mary', 'salary': 1000},
{'age': 30, 'counts': 1, 'name': 'George', 'salary': 2500}]
The logical are:
(1) First build the DataFrame from your data
(2) The groupby function can do aggregate function on each group.
(3) To output back to dict, you can call pd.to_dict
Pandas is a big package, which costs some time to learn it, but it worths to know pandas. It is so powerful that can make your data analysis quite faster and elegant.
Thanks.
You can try this:
import collections
d = [{'name': 'Mary', 'age': 25, 'salary': 1000},
{'name': 'John', 'age': 25, 'salary': 2000},
{'name': 'George', 'age': 30, 'salary': 2500},
{'name': 'John', 'age': 25, 'salary': 2000},
{'name': 'John', 'age': 25, 'salary': 2000}]
count = dict(collections.Counter([i["name"] for i in d]))
a = list(set(map(tuple, [i.items() for i in d])))
final_dict = [dict(list(i)+[("count", count[dict(i)["name"]])]) for i in a]
Output:
[{'salary': 2000, 'count': 3, 'age': 25, 'name': 'John'}, {'salary': 2500, 'count': 1, 'age': 30, 'name': 'George'}, {'salary': 1000, 'count': 1, 'age': 25, 'name': 'Mary'}]
You can take the count values using collections.Counter and then rebuild the dicts after adding the count value from the Counter to each frozenset:
from collections import Counter
l = [dict(d | {('count', c)}) for d, c in Counter(frozenset(d.items())
for d in myList).items()]
print(l)
# [{'salary': 1000, 'name': 'Mary', 'age': 25, 'count': 1},
# {'name': 'John', 'salary': 2000, 'age': 25, 'count': 3},
# {'salary': 2500, 'name': 'George', 'age': 30, 'count': 1}]
This question already has answers here:
Remove duplicate dict in list in Python
(16 answers)
Closed 6 years ago.
I have a list of dictionaries where I want to drop any dictionaries that repeat their id key. What's the best way to do this e.g:
example dict:
product_1={ 'id': 1234, 'price': 234}
List_of_products[product1:, product2,...........]
How can I the list of products so I have non repeating products based on their product['id']
Select one of product dictionaries in which the values with the same id are different. Use itertools.groupby,
import itertools
list_products= [{'id': 12, 'price': 234},
{'id': 34, 'price': 456},
{'id': 12, 'price': 456},
{'id': 34, 'price': 78}]
list_dicts = list()
for name, group in itertools.groupby(sorted(list_products, key=lambda d : d['id']), key=lambda d : d['id']):
list_dicts.append(next(group))
print(list_dicts)
# Output
[{'price': 234, 'id': 12}, {'price': 456, 'id': 34}]
If the product dictionaries with the same id are totally the same, there is an easier way as described in Remove duplicate dict in list in Python. Here is a MWE.
list_products= [{'id': 12, 'price': 234},
{'id': 34, 'price': 456},
{'id': 12, 'price': 234},
{'id': 34, 'price': 456}]
result = [dict(t) for t in set([tuple(d.items()) for d in list_products])]
print(result)
# Output
[{'price': 456, 'id': 34}, {'price': 234, 'id': 12}]
a = [{'id': 124, 'price': 234}, {'id': 125, 'price': 234}, {'id': 1234, 'price': 234}, {'id': 1234, 'price': 234}]
a.sort()
for indx, val in enumerate(a):
if val['id'] == a[indx+1]['id']:
del a[indx]