I have built a dictionary to collect orders according to the client, the products that have been ordered and I need to order it alphabetically by client to use it in a qweb report, nothing I have tried so far has worked for me. Any ideas?
reparto_data = {
'Cliente 1': {
'Pastel manzana': 12,
'Bomba crema': 8,
},
'Cliente 2': {
'Cake calabaza': 18,
'Bombon chocolate': 8,
},
#...
}
reparto_data2 = {}
for i in sorted(reparto_data[key]):
reparto_data2[i] = reparto_data[i]
You need to sort the inner dictionaries.
So, iterate across the keys and values of the outer dict, and for each value (which is an inner dict), sort it.
from pprint import pprint
reparto_data = {
'Cliente 1': {
'Pastel manzana': 12,
'Bomba crema': 8,
},
'Cliente 2': {
'Cake calabaza': 18,
'Bombon chocolate': 8,
},
}
def sort_dict(d):
return {k: d[k] for k in sorted(d.keys())}
reparto_data2 = {k: sort_dict(v) for k,v in reparto_data.items()}
pprint(reparto_data2, width=1)
Output
{'Cliente 1': {'Bomba crema': 8,
'Pastel manzana': 12},
'Cliente 2': {'Bombon chocolate': 8,
'Cake calabaza': 18}}
You can try this too.
reparto_data = {
'Cliente 1': {
'Pastel manzana': 12,
'Bomba crema': 8,
},
'Cliente 2': {
'Cake calabaza': 18,
'Bombon chocolate': 8,
},
}
sorted_clients = sorted(reparto_data.keys(), key=lambda x:x.lower())
reparto_data2 = {i: reparto_data[i] for i in sorted_clients}
Output
{'Cliente 1': {'Pastel manzana': 12, 'Bomba crema': 8}, 'Cliente 2': {'Cake calabaza': 18, 'Bombon chocolate': 8}}
If want to use your own code just replace this with your loop
reparto_data2 = {i: reparto_data[i] for i in sorted(reparto_data[key])}
Thanks for your help but I have posed the question wrong since the dictionary contains objects.
The code that has worked for me is the following:
reparto_data2 = sorted(reparto_data.items(), key=lambda x: x[0].lower())
Related
Hi I am looking to extract list of all keys from a List of dictionaries and their corresponding counts using Python. Used below code but counden't do it can some one help me in this regard
The Data looks like this,
people = [{'name': "Tom", 'age': 10, "city" : "NewYork"},
{'name': "Mark", 'age': 5, "country" : "Japan"},
{'name': "Pam", 'age': 7, "city" : "London"},
{'name': "Tom", 'hight': 163, "city" : "California"},
{'name': "Lena", 'weight': 45, "country" : "Italy"},
{'name': "Ben", 'age': 17, "city" : "Colombo"},
{'name': "Lena", 'gender': "Female", "country" : "Italy"},
{'name': "Ben", 'gender': "Male", "city" : "Colombo"}]
def search(name):
p_count = 0
for p in people:
if p['name'] == name:
p_count = p_count+1
return p, p_count
search("Pam")
def search1(list_dict):
for i in list_dict:
def getList(i):
return i.keys()
search1(people)
I want out put like the following,
name: 8, age: 4, City:3, Country: 2 etc..
I am new to this can some one help me in this
Just create a temporary dictionary to hold the count result, and for each key in each dictionary of the list, increase the the count by 1 using dict.get with default value as 0, finally return this dictionary.
def getKeyCount(lst):
out = {}
for d in lst:
for k in d.keys():
out[k] = out.get(k, 0) + 1
return out
getKeyCount(people)
#output
{'name': 8, 'age': 4, 'city': 5, 'country': 3, 'hight': 1, 'weight': 1, 'gender': 2}
You can make a hash table, and iterate over the keys of each
hash = {}
for d in list_dict:
for k in d.keys():
if k not in hash:
hash[k] = 0
hash[k] += 1
print(hash)
This question already has answers here:
How to convert list to dict
(3 answers)
Closed 1 year ago.
studentdictionary = {}
tuplist = [("Charles",8,85),("Bea",9,95),("Coral",8,75)]
def students(studentlist):
n = 0
for t in tuplist:
n+=1
for name,age,grade in t:
studentdictionary[n]["name"] = name
studentdictionary[n]["age"] = age
studentdictionary[n]["class grade"] = grade
return studentdictionary
print(students(tuplist))
#goal output: {1:{"name": "Charles", "age": 8, "class grade": 85},2:{"name": "Bea", "age": 9, "class grade": 95},3:{"name": "Coral", "age": 8, "class grade": 75}}
I'm just a beginner and have looked all over the place to help. Please don't be brutal!
Here are the errors:
Traceback (most recent call last):
File "<string>", line 81, in <module>
File "<string>", line 76, in students
ValueError: too many values to unpack (expected 3)
You can use dictionary-comprehension with enumerate():
tuplist = [("Charles", 8, 85), ("Bea", 9, 95), ("Coral", 8, 75)]
studentdictionary = {
i: {"name": name, "age": age, "grade": grade}
for i, (name, age, grade) in enumerate(tuplist, 1)
}
print(studentdictionary)
Prints:
{1: {'name': 'Charles', 'age': 8, 'grade': 85}, 2: {'name': 'Bea', 'age': 9, 'grade': 95}, 3: {'name': 'Coral', 'age': 8, 'grade': 75}}
i have these two dictionaries and i would like to add the data dict1 in dict2. how can i do that with python ?
dict1 = {
213688169:
{'stationCode': '6003',
'station_id': 213688169,
'num_bikes_available': 10,
'numBikesAvailable': 10,
'num_bikes_available_types': [{'mechanical': 10}, {'ebike': 0}],
'num_docks_available': 10,
'numDocksAvailable': 10,
'is_installed': 1,
'is_returning': 1,
'is_renting': 1,
'last_reported': 1619207955}
}
dict2 = {
213688169:
{'station_id': 213688169,
'name': 'Benjamin Godard - Victor Hugo',
'lat': 48.865983,
'lon': 2.275725,
'capacity': 35,
'stationCode': '16107'}
}
i tried this but it's too long and complicated :
donnees=[]
for i in stations:
for j in velib_description :
if i['station_id'] == j['station_id']:
List={}
List['name'] = i['name']
List['lat'] = i['lat']
List['lon'] = i['lon']
List['capacity'] = i['capacity']
List['num_bikes_available'] = j['num_bikes_available']
List['num_bikes_available_types'] = j['num_bikes_available_types']
List['last_reported'] = j['last_reported']
donnees.append(List)
I want to add in dict_2 = {num_bikes_available', 'num_bikes_available_types', last_reported': 1619207955 }
thank you
Perhaps you want dict.update?
dict1 = {
213688169: {
"stationCode": "6003",
"station_id": 213688169,
"num_bikes_available": 10,
"numBikesAvailable": 10,
"num_bikes_available_types": [{"mechanical": 10}, {"ebike": 0}],
"num_docks_available": 10,
"numDocksAvailable": 10,
"is_installed": 1,
"is_returning": 1,
"is_renting": 1,
"last_reported": 1619207955,
}
}
dict2 = {
213688169: {
"station_id": 213688169,
"name": "Benjamin Godard - Victor Hugo",
"lat": 48.865983,
"lon": 2.275725,
"capacity": 35,
"stationCode": "16107",
}
}
dict1_item = dict1[213688169]
dict2[213688169].update({k: v for k, v in dict1_item.items() if k not in dict2})
dict2
This prints
{213688169: {'station_id': 213688169,
'name': 'Benjamin Godard - Victor Hugo',
'lat': 48.865983,
'lon': 2.275725,
'capacity': 35,
'stationCode': '6003',
'num_bikes_available': 10,
'numBikesAvailable': 10,
'num_bikes_available_types': [{'mechanical': 10}, {'ebike': 0}],
'num_docks_available': 10,
'numDocksAvailable': 10,
'is_installed': 1,
'is_returning': 1,
'is_renting': 1,
'last_reported': 1619207955}}
We're using a comprehension to get all the key/values from the first dictionary that aren't in the second, then updating the second dictionary. We have to do a little munging with the 213688169 key because your dictionary is nested.
You can use a dictionary comprehension with a dictionary merge trick:
dict3 = {k: {**v, **dict2[k]} for k, v in dict1.items()}
or for Python 3.9 and later:
dict3 = {k: v | dict2[k] for k, v in dict1.items()}
I have a list of dictionaries, which contain a key and a value of dictionaries as such:
Array = [
{'Example1': {'Time Taken': 56, 'Type': 'Quiz'} },
{'Example1': {'Time Taken': 58, 'Type': 'Exam'} },
{'Example2': {'Time Taken': 40, 'Type': 'Quiz'} } ]
I want to iterate through the list and obtain only unique keys, with the values merged together into one list as such:
{ 'Example1': [
{ 'Time Taken': 56, 'Type': 'Quiz' },
{ 'Time Taken': 58, 'Type': 'Exam' } ] }
{ 'Example2': [ { 'Time Taken': 40, 'Type': 'Quiz' } ] }
Any idea on how to go about this? I've tried a lot of different things, but can't seem to get an efficient way to write this code. Any feedback appreciated.
Using collections.defaultdict
Ex:
from collections import defaultdict
Array = [ {"Example1": {"Time Taken": 56, "Type": "Quiz"} }, {"Example1": {"Time Taken": 58, "Type": "Exam"} }, {"Example2": {"Time Taken": 40, "Type": "Quiz"} } ]
result = defaultdict(list)
for ar in Array: #Iterate each element in list
for k, v in ar.items(): #Iterate your dict
result[k].append(v) #Create key-list.
print(result)
Output:
defaultdict(<class 'list'>, {'Example1': [{'Time Taken': 56, 'Type': 'Quiz'}, {'Time Taken': 58, 'Type': 'Exam'}], 'Example2': [{'Time Taken': 40, 'Type': 'Quiz'}]})
You can iterate over the dicts and use dict.setdefault to set the default as empty list if key is missing and append to the list otherwise:
for dct in Array:
for k, v in dct.items():
out.setdefault(k, []).append(v)
out is the desired output dict.
Example:
In [1208]: arr = [ {'Example1': {'Time Taken': 56, 'Type': 'Quiz'} }, {'Example1': {'Time Taken': 58, 'Type': 'Exam'} }, {'Example2': {'Time Taken': 40, 'Type': 'Quiz'} } ]
In [1209]: out = {}
In [1210]: for dct in arr:
...: for k, v in dct.items():
...: out.setdefault(k, []).append(v)
...:
In [1211]: out
Out[1211]:
{'Example1': [{'Time Taken': 56, 'Type': 'Quiz'},
{'Time Taken': 58, 'Type': 'Exam'}],
'Example2': [{'Time Taken': 40, 'Type': 'Quiz'}]}
Try following.
from collections import defaultdict
# l is list of dictionaries
d = defaultdict(list)
for x in l:
for y in x:
d[y].append(x[y])
print(d)
I have created following without any external module
Array = [{'Example1': {'Time Taken': 56, 'Type': 'Quiz'} },
{'Example1': {'Time Taken': 58, 'Type': 'Exam'} },
{'Example2': {'Time Taken': 40, 'Type': 'Quiz'} }]
result = {}
for i in range(len(Array)):
dict = Array[i]
for key in dict :
if key in result:
result[key].append(dict[key])
else:
result[key]=[dict[key]]
print(result)
Output:
{
'Example1': [{'Time Taken': 56, 'Type': 'Quiz'},
{'Time Taken': 58, 'Type': 'Exam'}],
'Example2': [{'Time Taken': 40, 'Type': 'Quiz'}]
}
I have a list of dictionaries and I wanted to group the data. I used the following:
group_list = []
for key, items in itertools.groupby(res, operator.itemgetter('dept')):
group_list.append({key:list(items)})
For data that looks like this
[{'dept':1, 'age':10, 'name':'Sam'},
{'dept':1, 'age':12, 'name':'John'},
.
.
.
{'dept':2,'age':20, 'name':'Mary'},
{'dept':2,'age':11, 'name':'Mark'},
{'dept':2,'age':11, 'name':'Tom'}]
the output would be:
[{1:[{'dept':1, 'age':10, 'name':'Sam'},
{'dept':1, 'age':12, 'name':'John'}],
{2:[{'dept':2,'age':20, 'name':'Mary'},
{'dept':2,'age':11, 'name':'Mark'},
{'dept':2,'age':11, 'name':'Tom'}]
...
]
Now if I want to group using multiple keys say 'dept' and 'age', the above mentioned method returns
[{(2, 20): [{'age': 20, 'dept': 2, 'name': 'Mary'}]},
{(2, 11): [{'age': 11, 'dept': 2, 'name': 'Mark'},
{'age': 11, 'dept': 2, 'name': 'Tom'}]},
{(1, 10): [{'age': 10, 'dept': 1, 'name': 'Sam'}]},
{(1, 12): [{'age': 12, 'dept': 1, 'name': 'John'}]}]
The desired output would be:
[
{
2: {
20: [
{
'age': 20,
'dept': 2,
'name': 'Mary'
}
]
},
{
11: [
{
'age': 11,
'dept': 2,
'name': 'Mark'
},
{
'age': 11,
'dept': 2,
'name': 'Tom'
}
]
}
},
{
1: {
10: [
{
'age': 10,
'dept': 1,
'name': 'Sam'
}
]
},
{
12: [
{
'age': 12,
'dept': 1,
'name': 'John'
}
]
}
}
]
Can it be done with itertools? Or do I need to write that code myself?
Absolutely. You just need to apply itertools.groupby() for the second criterion first, then the other.
You would need to write a (probably recursive) bit of code to do this yourself - itertools doesn't have a tree-builder in it.
Thanks everyone for your help. Here is how I did it:
import itertools, operator
l = [{'dept':1, 'age':10, 'name':'Sam'},
{'dept':1, 'age':12, 'name':'John'},
{'dept':2,'age':20, 'name':'Mary'},
{'dept':2,'age':11, 'name':'Mark'},
{'dept':2,'age':11, 'name':'Tom'}]
groups = ['dept', 'age', 'name']
groups.reverse()
def hierachical_data(data, groups):
g = groups[-1]
g_list = []
for key, items in itertools.groupby(data, operator.itemgetter(g)):
g_list.append({key:list(items)})
groups = groups[0:-1]
if(len(groups) != 0):
for e in g_list:
for k,v in e.items():
e[k] = hierachical_data(v, groups)
return g_list
print hierachical_data(l, groups)