I have a dictionary setup like this:
company = {'Honda': {}
,'Toyota':{}
,'Ford':{}
}
I have a list containing data like this:
years = ['Year 1', 'Year 2', 'Year 3']
Finally, I also have a list of lists containing data like this:
sales = [[55,9,90],[44,22,67],[83,13,91]]
I am trying to achieve a final result that looks like this:
{'Honda': {'Year 1':55,'Year 2':9,'Year 3':90}
,'Toyota':{'Year 1':44,'Year 2':22,'Year 3':67}
,'Ford':{'Year 1':83,'Year 2':13,'Year 3':91}
}
I can access the sub-list if sales like this:
for i in sales:
for j in i:
#this would give me a flat list of all sales
I can't seem to wrap my head around constructing the final dictionary that would tie everything together.
Any help is appreciated!
You can use a dict comprehension with zip.
res = {k : dict(zip(years, sale)) for k, sale in zip(company, sales)}
You can use zip to pair corresponding information together. First, zip the brand names with the values in sales. Next, zip years with a particular brand's sales numbers.
company = {brand: dict(zip(years, sales_nums))
for brand, sales_nums in zip(["Honda", "Toyota", "Ford"], sales)}
You can use zip and a double for-loop to zip all 3 lists. Here you are:
final_dict = {}
for i, (name, sub_dict) in enumerate(company.items()):
for year, sale in zip(years, sales[i]):
sub_dict[year] = sale
final_dict[name] = sub_dict
Related
Trying hard to create a dictionary by comparing values and to add greater values more than 90 into a new dictionary. I have an actual data set working on with pandas but to simplify, i will try to explain with lists
list_names = ['annie','betty', 'charlie','debie', 'elf', 'frank', 'goe', 'hayri']
list_ages = [19,23,44,31,55,65,15,40]
The ages are corresponding to each name in the list and the target is to create a key:value pair that contains values only older than 30.
Thank you for any help.
Try this
list_names = ['annie','betty', 'charlie','debie', 'elf', 'frank', 'goe', 'hayri']
list_ages = [19,23,44,31,55,65,15,40]
{name: age for name, age in zip(list_names,list_ages) if age>30}
Try taking example from this code:
list_names = name((k, v) for k, v in list_names.items() if v > 30)
I'm trying to create an event manager in which a dictionary stores the events like this
my_dict = {'2020':
{'9': {'8': ['School ']},
'11': {'13': ['Doctors ']},
'8': {'31': ['Interview']}
},
'2021': {}}
In which the outer key is the year the middle key is a month and the most inner key is a date which leads to a list of events.
I'm trying to first sort it so that the months are in order then sort it again so that the days are in order. Thanks in advance
Use-case
DevOrangeCrush wishes to sort on keys in a nested dictionary where the nesting occurs on multiple levels
Solution
Normalize the data so that the dates match ISO8601 format, for easier sorting
In plain English, this means make sure you always use two digits for month and date, and always use four digits for year
Re-normalize the original dictionary data structure into a single list of dictionaries, where each dictionary represents a row, and the list represents an outer containing table
this is known as an Array of Hashes in perl-speak
this is known as a list of objects in JSON-speak
Once your data is restructured you are solving a much more well-known, well-documented, and more obvious problem, how to sort a simple list of dictionaries (which is already documented in the See also section of this answer).
Example
import pprint
## original data is formatted as a nested dictionary, which is clumsy
my_dict = {'2020':
{'9': {'8': ['School ']}, '11':
{'13': ['Doctors ']},'8':
{'31': ['Interview']}}, '2021': {}
}
## we want the data formatted as a standard table (aka list of dictionary)
## this is the most common format for this kind of data as you would see in
## databases and spreadsheets
mydata_table = []
ddtemp = dict()
for year in my_dict:
for month in my_dict[year].keys():
ddtemp['month'] = '{0:02d}'.format(*[int(month)])
ddtemp['year'] = year
for day in my_dict[year][month].keys():
ddtemp['day'] = '{0:02d}'.format(*[int(day)])
mydata_row = dict()
mydata_row['year'] = '{year}'.format(**ddtemp)
mydata_row['month'] = '{month}'.format(**ddtemp)
mydata_row['day'] = '{day}'.format(**ddtemp)
mydata_row['task_list'] = my_dict[year][month][day]
mydata_row['date'] = '{year}-{month}-{day}'.format(**ddtemp)
mydata_table.append(mydata_row)
pass
pass
pass
## output result is now easily sorted and there is no data loss
## you will have to modify this if you want to deal with years that
## do not have any associated task_list data
pprint.pprint(mydata_table)
'''
## now we have something that can be sorted using well-known python idioms
## and easily manipulated using data-table semantics
## (search, sort, filter-by, group-by, select, project ... etc)
[
{'date': '2020-09-08','day': '08',
'month': '09','task_list': ['School '],'year': '2020'},
{'date': '2020-11-13','day': '13',
'month': '11','task_list': ['Doctors '],'year': '2020'},
{'date': '2020-08-31','day': '31',
'month': '08','task_list': ['Interview'],'year': '2020'},
]
'''
See also
How to sort a python list-of-dictionary
How to sort objects by multiple keys
Why you should use ISO8601 date format
ISO8601 vs timestamp
To get sorted events data, you can do something like this:
def sort_events(my_dict):
new_events_data = dict()
for year, month_data in my_dict.items():
new_month_data = dict()
for month, day_data in month_data.items():
sorted_day_data = sorted(day_data.items(), key=lambda kv: int(kv[0]))
new_month_data[month] = OrderedDict(sorted_day_data)
sorted_months_data = sorted(new_month_data.items(), key=lambda kv: int(kv[0]))
new_events_data[year] = OrderedDict(sorted_months_data)
return new_events_data
Output:
{'2020': OrderedDict([('8', OrderedDict([('31', ['Interview'])])),
('9', OrderedDict([('8', ['School '])])),
('11', OrderedDict([('13', ['Doctors '])]))]),
'2021': OrderedDict()}
A simple dict can't be ordered, you could do it using a OrderedDict but if you simply need to get it sorted while iterating on it do like this
for year in sorted(map(int, my_dict)):
year_dict = my_dict[str(year)]
for month in sorted(map(int, year_dict)):
month_dict = year_dict[str(month)]
for day in sorted(map(int, month_dict)):
events = month_dict[str(day)]
for event in events:
print(year, month, day, event)
Online Demo
The conversion to int is to ensure right ordering between the numbers, without you'll get 1, 10, 11, .., 2, 20, 21
A dictionary in Python does not have an order, you might want to try the OrderedDict class from the collections Module which remembers the order of insertion.
Of course you would have to sort and reinsert the elements whenever you insert a new element which should be placed before any of the existing elements.
If you care about order maybe a different data structure works better. For example a list of lists.
I have two lists of dictionaries and json list and I need to grab a value of a specific key based on the value of a key from another dictionary. My data looks like this:
opps = [{'Product2Id': '100','Price': '1645'}, {'Product2Id': '101','Price': '5478'}]
products = [{'Id': '100', 'Name': 'Insertion'}, [{'Id': '101', 'Name': 'Print'}]
sales_json = {'Insertion': {'name': 'BAZ', 'id': '95'}, 'Print': {'name': 'BIC', 'id': '105'}
I need to loop through opps and assign a value to a new variable from sales_json. But for a specific Id that are stored in products and in opps
I tried the following:
for index, my_dict in enumerate(opps):
new_name = sales_json[products[my_dict["Product2Id"]]["Name"]]["name"]
Gives me an error.
The desired output is:
print(new_name)
BAZ,
BIC
You are trying to use the list products as a dictionary. Instead, you should first build a product number to name dictionary from it:
prod_num_to_name = {d['Id']: d['Name'] for d in products}
Then, you can run the loop you wanted, modified like this:
for index, my_dict in enumerate(opps):
new_name = sales_json[prod_num_to_name[my_dict["Product2Id"]]]["name"]
print new_name
To return a list of names that match the criteria, using a List Comprehension:
names = [ sales_json[product['Name']]['name'] for opp in opps for product in products if product['Id'] == opp['Product2Id']]
print (names)
Prints the list of names:
['BAZ', 'BIC']
import csv
def partytoyear():
party_in_power = {}
with open("presidents.txt") as f:
reader = csv.reader(f)
for row in reader:
party = row[1]
for year in row[2:]:
party_in_power[year] = party
print(party_in_power)
return party_in_power
partytoyear()
def statistics():
with open("BLS_private.csv") as f:
statistics = {}
reader = csv.DictReader(f)
for row in reader:
statistics = row
print(statistics)
return statistics
statistics()
These two functions return two dictionaries.
Here is a sample of the first dictionary:
'Democrat', '1981': 'Republican', '1982': 'Republican', '1983'
Sample of the second dictionary:
'2012', '110470', '110724', '110871', '110956', '111072', '111135', '111298', '111432', '111560', '111744'
The first dictionary associates a year and the political party. The next dictionary associates the year with job statistics.
I need to combine these two dictionaries, so I can have the party inside the dictionary with the job statistics.
I would like the dictioary to look like this:
'Democrat, '2012','110470', '110724', '110871', '110956', '111072', '111135', '111298', '111432', '111560', '111744'
How would I go about doing this? I've looked at the syntax for update() but that didn't work for my program
You can’t have a dictionary in that manor in python it’s syntactically wrong but you can have each value be a collection such as a list. Here’s a comprehension that does just that using dict lookups:
first_dict = {'Democrat': '1981': 'Republican': '1982': 'Republican': '1983', ...}
second_dict = {'2012': ['110470', '110724', '110871', '110956', '111072', '111135', '111298', '111432', '111560', '111744'], ...}
result = {party: [year, *second_dict[year] for party, year in first_dict.items()}
Pseudo result dict structure:
{'Party Name': [year, stats, ...], ...}
I'm trying to create a dictionary of dictionaries like this:
food = {"Broccoli": {"Taste": "Bad", "Smell": "Bad"},
"Strawberry": {"Taste": "Good", "Smell": "Good"}}
But I am populating it from an SQL table. So I've pulled the SQL table into an SQL object called "result". And then I got the column names like this:
nutCol = [i[0] for i in result.description]
The table has about 40 characteristics, so it is quite long.
I can do this...
foodList = {}
for id, food in enumerate(result):
addMe = {str(food[1]): {nutCol[id + 2]: food[2], nulCol[idx + 3]:
food[3] ...}}
foodList.update(addMe)
But this of course would look horrible and take a while to write. And I'm still working out how I want to build this whole thing so it's possible I'll need to change it a few times...which could get extremely tedious.
Is there a DRY way of doing this?
In order to make solution position independent you can make use of dict1.update(dict2). This simply merges dict2 with dict1.
In our case since we have dict of dict, we can use dict['key'] as dict1 and simply add any additional key,value pair as dict2.
Here is an example.
food = {"Broccoli": {"Taste": "Bad", "Smell": "Bad"},
"Strawberry": {"Taste": "Good", "Smell": "Good"}}
addthis = {'foo':'bar'}
Suppose you want to add addthis dict to food['strawberry'] , we can simply use,
food["Strawberry"].update(addthis)
Getting result:
>>> food
{'Strawberry': {'Taste': 'Good', 'foo': 'bar', 'Smell': 'Good'},'Broccoli': {'Taste': 'Bad', 'Smell': 'Bad'}}
>>>
Assuming that column 0 is what you wish to use as your key, and you do wish to build a dictionary of dictionaries, then its:
detail_names = [col[0] for col in result.description[1:]]
foodList = {row[0]: dict(zip(detail_names, row[1:]))
for row in result}
Generalising, if column k is your identity then its:
foodList = {row[k]: {col[0]: row[i]
for i, col in enumerate(result.description) if i != k}
for row in result}
(Here each sub dictionary is all columns other than column k)
addMe = {str(food[1]):dict(zip(nutCol[2:],food[2:]))}
zip will take two (or more) lists of items and pair the elements, then you can pass the result to dict to turn the pairs into a dictionary.