What is a good way to get an ordered dictionary from a regular dictionary? I need the keys (and these keys are known ahead of time) to be in a certain order. I will be "dump"ing a list of these dictionaries into a JSON file and need things ordered a certain way.
--- Edited and added the following
For instance i have a dictionary ...
employee = { 'phone': '1234567890', 'department': 'HR', 'country': 'us', 'name': 'Smith' }
when i dump it into JSON format, i would like for it to print out as
{ 'name': 'Smith', 'department': 'HR', 'country': 'us', 'phone': '1234567890'}
Sort your dict items and create an OrderedDict from the sorted elements making sure to pass reverse=True to sort from highest to lowest:
from collections import OrderedDict
order = ("name","department","country","phone")
employee = { 'phone': '1234567890', 'department': 'HR', 'country': 'us', 'name': 'Smith' }
od = OrderedDict((k, employee[k]) for k in order)
But if you dump to a json file and load again the order will not be maintained and you will not get an OrderedDict back, when you dump it will look like:
{"name": "Smith", "department": "HR", "country": "us", "phone": "1234567890"}
But loading will will not be in the same order because normal dicts have no order like below:
{'phone': '1234567890', 'name': 'Smith', 'country': 'us', 'department': 'HR'}
If you are trying to just store the dicts to use again and want to maintain order you can pickle:
import pickle
with open("foo.pkl","wb") as f:
pickle.dump(od,f)
with open("foo.pkl","rb") as f:
d = pickle.load(f)
print(d)
You could do something like the following ... you collect the keys in order in a list of String, traverse through the list and look up in the dictionary, and create an ordered dictionary
def makeOrderedDict(dictToOrder, keyOrderList):
tupleList = []
for key in keyOrderList:
tupleList.append((key, dictToOrder[key]))
return OrderedDict(tupleList)
Related
For a JSON file, on doing json.load its getting converted into list of dictionaries.
sample rows are from list of dictionaries are as below: -
config_items_list = [{'hostname': '"abc2164"', 'Status': '"InUse"', 'source': '"excel"', 'port': '"[445]"', 'tech': '"Others"', 'ID': '"123456"'},
{'hostname': '"xyz2164"', 'Status': '"InUse"', 'source': '"web"', 'port': '"[123]"', 'tech': '"Others"', 'ID': '"456789"'},
{'hostname': '"pqr2164"', 'Status': '"NotInUse"', 'source': '"web"', 'port': '"[777]"', 'tech': '"Others"', 'ID': '"123456"'}]
Requirement is to parse this list of dictionaries and extract all rows having specific value in ID key. For example, 123456 (means two rows from above sample).
try this solution it should solve your problem
def filter_json_with_id(config_items_list, id):
items = []
for item in config_items_list:
if item['ID'] == id:
items.append(item)
return items
select_id = '"123456"'
[d for d in config_items_list if d["ID"] == select_id]
will produce a list with those dictionaries that have given select_id as "ID" in config_items_list.
By using the below solution, it searches for the id_value even if the id_value is surrounded by any symbol or any other number/character as well:
expectedResult = [d for d in config_items_list if "123456" in d['ApplicationInstanceID']]
I need to compare the values of the items in two different dictionaries.
Let's say that dictionary RawData has items that represent phone numbers and number names.
Rawdata for example has items like: {'name': 'Customer Service', 'number': '123987546'} {'name': 'Switchboard', 'number': '48621364'}
Now, I got dictionary FilteredData, which already contains some items from RawData: {'name': 'IT-support', 'number': '32136994'} {'name': 'Company Customer Service', 'number': '123987546'}
As you can see, Customer Service and Company Customer Service both have the same values, but different keys. In my project, there might be hundreds of similar duplicates, and we only want unique numbers to end up in FilteredData.
FilteredData is what we will be using later in the code, and RawData will be discarded.
Their names(keys) can be close duplicates, but not their numbers(values)**
There are two ways to do this.
A. Remove the duplicate items in RawData, before appending them into FilteredData.
B. Append them into FilteredData, and go through the numbers(values) there, removing the duplicates. Can I use a set here to do that? It would work on a list, obviously.
I'm not looking for the most time-efficient solution. I'd like the most simple and easy to learn one, if and when someone takes over my job someday. In my project it's mandatory for the next guy working on the code to get a quick grip of it.
I've already looked at sets, and tried to face the problem by nesting two for loops, but something tells me there gotta be an easier way.
Of course I might have missed the obvious solution here.
Thanks in advance!
I hope I understands your problem here:
data = [{'name': 'Customer Service', 'number': '123987546'}, {'name': 'Switchboard', 'number': '48621364'}]
newdata = [{'name': 'IT-support', 'number': '32136994'}, {'name': 'Company Customer Service', 'number': '123987546'}]
def main():
numbers = set()
for entry in data:
numbers.add(entry['number'])
for entry in newdata:
if entry['number'] not in numbers:
data.append(entry)
print data
main()
Output:
[{'name': 'Customer Service', 'number': '123987546'},
{'name': 'Switchboard', 'number': '48621364'},
{'name': 'IT-support', 'number': '32136994'}]
What you can do is take a dict.values(), create a set of those to remove duplicates and then go through the old dictionary and find the first key with that value and add it to a new one. Keep the set around because when you get the next dict entry, try adding the element to that set and see if the length of the set is longer that before adding it. If it is, it's a unique element and you can add it to the dict.
If you're willing on changing how FilteredData is currently, you can just use a dict and use the number as your key:
RawData = [
{'name': 'Customer Service', 'number': '123987546'},
{'name': 'Switchboard', 'number': '48621364'}
]
# Change how FilteredData is structured
FilteredDataMap = {
'32136994':
{'name': 'IT-support', 'number': '32136994'},
'123987546':
{'name': 'Company Customer Service', 'number': '123987546'}
}
for item in RawData:
number = item.get('number')
if number not in FilteredDataMap:
FilteredDataMap[number] = item
# If you need the list of items
FilteredData = list(FilteredDataMap.values())
You can just pull the actual list from the Map using .values()
I take the numbers are unique. Then, another solution would be taking advantage of the uniqueness of dictionary keys. This means converting each list of dictionary to a dictionary of 'number:name' pairs. Then, you simple need to update RawData with FilteredData.
RawData = [
{'name': 'Customer Service', 'number': '123987546'},
{'name': 'Switchboard', 'number': '48621364'}
]
FilteredData = [
{'name': 'IT-support', 'number': '32136994'},
{'name': 'Company Customer Service', 'number': '123987546'}
]
def convert_list(input_list):
return {item['number']:item['name'] for item in input_list}
def unconvert_dict(input_dict):
return [{'name':val, 'number': key} for key, val in input_dict.items()]
NewRawData = convert_list(RawData)
NewFilteredData = conver_list(FilteredData)
DesiredResultConverted = NewRawData.update(NewFilteredData)
DesuredResult = unconvert_dict(DesiredResultConverted)
In this example, the variables will have the following values:
NewRawData = {'123987546':'Customer Service', '48621364': 'Switchboard'}
NewFilteredData = {'32136994': 'IT-support', '123987546': 'Company Customer Service'}
When you update NewRawData with NewFilteredData, Company Customer Service will overwrite Customer Service as the value associated with the key 123987546. So,
DesiredResultConverted = {'123987546':'Company Customer Service', '48621364': 'Switchboard', '32136994': 'IT-support'}
Then, if you still prefer the original format, you can "unconvert" back.
I've got a json format list with some dictionaries within each list, it looks like the following:
[{"id":13, "name":"Albert", "venue":{"id":123, "town":"Birmingham"}, "month":"February"},
{"id":17, "name":"Alfred", "venue":{"id":456, "town":"London"}, "month":"February"},
{"id":20, "name":"David", "venue":{"id":14, "town":"Southampton"}, "month":"June"},
{"id":17, "name":"Mary", "venue":{"id":56, "town":"London"}, "month":"December"}]
The amount of entries within the list can be up to 100. I plan to present the 'name' for each entry, one result at a time, for those that have London as a town. The rest are of no use to me. I'm a beginner at python so I would appreciate a suggestion in how to go about this efficiently. I initially thought it would be best to remove all entries that don't have London and then I can go through them one by one.
I also wondered if it might be quicker to not filter but to cycle through the entire json and select the names of entries that have the town as London.
You can use filter:
data = [{"id":13, "name":"Albert", "venue":{"id":123, "town":"Birmingham"}, "month":"February"},
{"id":17, "name":"Alfred", "venue":{"id":456, "town":"London"}, "month":"February"},
{"id":20, "name":"David", "venue":{"id":14, "town":"Southampton"}, "month":"June"},
{"id":17, "name":"Mary", "venue":{"id":56, "town":"London"}, "month":"December"}]
london_dicts = filter(lambda d: d['venue']['town'] == 'London', data)
for d in london_dicts:
print(d)
This is as efficient as it can get because:
The loop is written in C (in case of CPython)
filter returns an iterator (in Python 3), which means that the results are loaded to memory one by one as required
One way is to use list comprehension:
>>> data = [{"id":13, "name":"Albert", "venue":{"id":123, "town":"Birmingham"}, "month":"February"},
{"id":17, "name":"Alfred", "venue":{"id":456, "town":"London"}, "month":"February"},
{"id":20, "name":"David", "venue":{"id":14, "town":"Southampton"}, "month":"June"},
{"id":17, "name":"Mary", "venue":{"id":56, "town":"London"}, "month":"December"}]
>>> [d for d in data if d['venue']['town'] == 'London']
[{'id': 17,
'name': 'Alfred',
'venue': {'id': 456, 'town': 'London'},
'month': 'February'},
{'id': 17,
'name': 'Mary',
'venue': {'id': 56, 'town': 'London'},
'month': 'December'}]
I would like to filter items out of one dictionary where that dictionary contains items of another dictionary. So, say that I have two dictionary's dict1 and dict2 where
dict1 = {
1:{'account_id':1234, 'case':1234, 'date': 12/31/15, 'content': 'some content'},
2:{'account_id':1235, 'case':1235, 'date': 12/15/15, 'content': 'some content'}
}
dict2 = {
1:{'account_id':1234, 'case':1234, 'date': 12/31/15, 'content': 'some different content'},
2:{'account_id':4321, 'case':4321, 'date': 6/12/15, 'content': 'some different content'},
3:{'account_id':1235, 'case':1235, 'date': 12/15/15, 'content': 'some different content'}
}
I would like to match on account_id, case and date and have the output be a third dictionary with matched entries from dict2 being 1 and 3.
out = {
1:{'account_id':1234, 'case':1234, 'date': 12/31/15, 'content': 'some different content'},
2:{'account_id':1235, 'case':1235, 'date': 12/15/15, 'content': 'some different content'}
}
How would I accomplish this? I am using Python 3.5
Well then, I believe this is what you are looking for:
from itertools import count
from operator import itemgetter
# Set the criteria for unique entry (prevents us from needing to write this twice)
get_identifier = itemgetter("account_id","case","date")
# Create a set of all unique items.
unique_entries = set(map(get_identifier, dict1.values()))
# Get all entries that match one of the unique entries
matched_entires = (d for d in dict2.values() if get_identifier(d) in unique_entries)
# Recreate a new dict together with a counter for items.
out = dict(zip(count(1), matched_entires))
For more info about count() and itemgetter(), see their respective docs.
Using a set and generator comprehensions ensures efficiency at the highest level.
I have some data like this:
{'cities': [{'abbrev': 'NY', 'name': 'New York'}, {'abbrev': 'BO', 'name': 'Boston'}]}
From my scarce knowledge of Python this looks like a dictionary within a dictionary.
But either way how can I use "NY" as a key to fetch the value "New York"?
It's a dictionary with one key-value pair. The value is a list of dictionaries.
d = {'cities': [{'abbrev': 'NY', 'name': 'New York'}, {'abbrev': 'BO', 'name': 'Boston'}]}
To find the name for an abbreviation you should iterate over the dictionaries in the list and then compare the abbrev-value for a match:
for city in d['cities']: # iterate over the inner list
if city['abbrev'] == 'NY': # check for a match
print(city['name']) # print the matching "name"
Instead of the print you can also save the dictionary containing the abbreviation, or return it.
When you've got a dataset not adapted to your need, instead of using it "as-is", you can build another dictionary from that one, using a dictionary comprehension with key/values as values of your sub-dictionaries, using the fixed keys.
d = {'cities': [{'abbrev': 'NY', 'name': 'New York'}, {'abbrev': 'BO', 'name': 'Boston'}]}
newd = {sd["abbrev"]:sd["name"] for sd in d['cities']}
print(newd)
results in:
{'NY': 'New York', 'BO': 'Boston'}
and of course: print(newd['NY']) yields New York
Once the dictionary is built, you can reuse it as many times as you need with great lookup speed. Build other specialized dictionaries from the original dataset whenever needed.
Use next and filter the sub dictionaries based upon the 'abbrev' key:
d = {'cities': [{'abbrev': 'NY', 'name': 'New York'},
{'abbrev': 'BO', 'name': 'Boston'}]}
city_name = next(city['name'] for city in d['cities']
if city['abbrev'] == 'NY')
print city_name
Output:
New York
I think that I understand your problem.
'NY' is a value, not a key.
Maybe you need something like {'cities':{'NY':'New York','BO':'Boston'}, so you could type: myvar['cities']['NY'] and it will return 'New York'.
If you have to use x = {'cities': [{'abbrev': 'NY', 'name': 'New York'}, {'abbrev': 'BO', 'name': 'Boston'}]} you could create a function:
def search(abbrev):
for cities in x['cities']:
if cities['abbrev'] == abbrev:
return cities['name']
Output:
>>> search('NY')
'New York'
>>> search('BO')
'Boston'
PD: I use python 3.6
Also with this code you could also find abbrev:
def search(s, abbrev):
for cities in x['cities']:
if cities['abbrev'] == abbrev: return cities['name'], cities['abbrev']
if cities['name'] == abbrev: return cities['name'], cities['abbrev']