Develop Reference

Combining objects in one list

I have a list of dicts that looks like this:
{
"Player_Name":"Byeong-Hun An",
"Tournament":[
{
"Name":"Arnold Palmer Invitational presented by Mastercard",
"Points":"32.80",
"Salary":"10300.00"
}
]
},
{
"Player_Name":"Byeong-Hun An",
"Tournament":[
{
"Name":"Different",
"Points":"18.80",
"Salary":"10400.00"
}
]
}
and I want this:
[
{
"Player_Name":"Byeong-Hun An",
"Tournament":[
{
"Name":"Arnold Palmer Invitational presented by Mastercard",
"Points":"32.80",
"Salary":"10300.00"
},
{
"Name":"Different",
"Points":"18.80",
"Salary":"10400.00"
}
]
}
]
I've tried collections, but it doesn't do exactly what I'm wanting. I essentially want to take every single player and combine all the tournament objects into one so each player has one object instead of each event having its own object.
Here's my code
import json
import numpy as np
import pandas as pd
from collections import Counter
# using json open the player objects file and set it equal to data
with open('PGA_Player_Objects.json') as json_file:
data = json.load(json_file)
points = []
players = []
for a in data:
for b in a['Tournament']:
points.append(int(float(b['Points'])))
for x in data:
players.append(x['Player_Name'])
def Average(lst):
unrounded = sum(lst) / len(lst)
return round(unrounded,2)
result = Counter()
for d in data:
for b in d['Tournament']:
result[d['Player_Name']] += int(float(b['Points']))
How can I do that?

if your list is in l:
l = [{'Player_Name': 'Byeong-Hun An', 'Tournament': [{'Name': 'Arnold Palmer Invitational presented by Mastercard', 'Points': '32.80', 'Salary': '10300.00'}]},
{'Player_Name': 'Byeong-Hun An', 'Tournament': [{'Name': 'Different', 'Points': '18.80', 'Salary': '10400.00'}]},]
Try this:
from itertools import groupby
result = []
for k,g in groupby(sorted(l, key=lambda x:x['Player_Name']), lambda x:x['Player_Name']):
result.append({'Player_Name':k, 'Tournament':[i['Tournament'][0] for i in g]})
Then the result will be:
[{'Player_Name': 'Byeong-Hun An',
'Tournament': [
{'Name': 'Arnold Palmer Invitational presented by Mastercard',
'Points': '32.80',
'Salary': '10300.00'},
{'Name': 'Different',
'Points': '18.80',
'Salary': '10400.00'}]}]

This works as well, and it's a more general solution that works for arbitrary key names:
from collections import defaultdict
d = defaultdict(list)
for dic in lst:
for k, v in dic.items():
if isinstance(v, list):
d[k].extend(v)
else:
d[k] = v
answer = [dict(d)]

Here's my take on a solution.
Create a new list of dictionaries
Iterate through the original list of dictionaries.
Store one copy of the beginning data for each player that is the same into the new list of dictionaries
Append additional Tournament data for each player into that one dictionary into a unified Tournament list.
Untested code below as an example, but should work with some tweaks.
listofDicts = [{'Player_Name': 'Byeong-Hun An', 'Tournament': [{'Name': 'Arnold Palmer Invitational presented by Mastercard', 'Points': '32.80', 'Salary': '10300.00'}]},{'Player_Name': 'Byeong-Hun An', 'Tournament': [{'Name': 'Different', 'Points': '18.80', 'Salary': '10400.00'}]}]
newListOfDicts = []
playerName = " "
playerNo = -1
for dicts in listofDicts:
if playerName == dicts['Player_Name']:
newListOfDicts[playerNo]['Tournament'].append(dicts['Tournament'][0])
else:
newListOfDicts.append(dicts)
playerName = dicts['Player_Name']
playerNo += 1

How to conditionally select elements in a list comprehension?

I couldn't find any examples that match my use case. Still working through my way in python lists and dictionaries.
Problem:
all_cars = {'total_count': 3,'cars': [{'name': 'audi','model': 'S7'}, {'name': 'honda', 'model': 'accord'},{'name': 'jeep', 'model': 'wrangler'} ]}
owners = {'users':[{'owner': 'Nick', 'car': 'audi'},{'owner': 'Jim', 'car': 'ford'},{'owner': 'Mike', 'car': 'mercedes'} ]}
def duplicate():
for c in all_cars['cars']:
if c['name'] == [c['users']for c in owners['users']]:
pass
else:
res = print(c['name'])
return res
output = ['honda', 'jeep', audi']
and
def duplicate():
for c in all_cars['cars']:
if c['name'] == 'audi':
pass
else:
res = print(c['name'])
return res
output - ['honda', 'jeep']
I am trying to find matching values in both dictionaries, using list comprehension, then return non-matching values only.
Solution: Using 'in' rather than '==' operator, I was able to compare values between both lists and skip duplicates.
def duplicate():
for c in all_cars['cars']:
if c['name'] in [c['users']for c in owners['users']]:
pass
else:
res = print(c['name'])
return res

To answer the question in your title, you can conditionally add elements during a list comprehension using the syntax [x for y in z if y == a], where y == a is any condition you need - if the condition evaluates to True, then the element y will be added to the list, otherwise it will not.

I would just keep a dictionary of all of the owner data together:
ownerData = { "Shaft" : {
"carMake" : "Audi",
"carModel" : "A8",
"year" : "2015" },
"JamesBond" : {
"carMake" : "Aston",
"carModel" : "DB8",
"year" : "2012" },
"JeffBezos" : {
"carMake" : "Honda",
"carModel" : "Accord"
"year" : "1989"}
}
Now you can loop through and query it something like this:
for o in ownerData:
if "Audi" in o["carMake"]:
print("Owner %s drives a %s %s %s" % (o, o["year"], o["carMake"], o["carModel"]))
Should output:
"Owner Shaft drives a 2015 Audi A8"
This way you can expand your data set for owners without creating multiple lists.

OK, based on your feedback on the solution above, here is how I would tackle your problem. Drop your common items into lists and then use "set" to print out the diff.
all_cars = {'total_count': 3,'cars': [{'name': 'audi','model': 'S7'},
{'name': 'honda', 'model': 'accord'},{'name': 'jeep', 'model': 'wrangler'} ]}
owners = {'users':[{'owner': 'Nick', 'car': 'audi'},{'owner': 'Jim',
'car': 'ford'},{'owner': 'Mike', 'car': 'mercedes'} ]}
allCarList = []
ownerCarList = []
for auto in all_cars['cars']:
thisCar = auto['name']
if thisCar not in allCarList:
allCarList.append(thisCar)
for o in owners['users']:
thisCar = o['car']
if thisCar not in ownerCarList:
ownerCarList.append(thisCar)
diff = list(set(allCarList) - set(ownerCarList))
print(diff)
I put this in and ran it and came up with this output:
['jeep', 'honda']
Hope that helps!

Better way to reorder list of dictionaries?

So I have a small data like this:
data = [
{"Name":"Arab","Code":"Zl"},
{"Name":"Korea","Code":"Bl"},
{"Name":"China","Code":"Bz"}
]
I want to find a graph so that the x-axis is: "Bl", "Bz", "Zl" (alphabetic order)
and the y-axis is: "Korea", "China", "Arab" (corresponding to the codenames).
I thought of:
new_data = {}
for dic in data:
country_data = dic["Name"]
code_data = dic["Code"]
new_data[code_data] = country_data
code_data = []
for codes in new_data.keys():
code_data.append(codes)
code_data.sort()
name_data = []
for code in code_data:
name_data.append(new_data[code])
Is there a better way to do this?
Perhaps by not creating a new dictionary?

So here's the data:
data = [
{"Name":"Arab","Code":"Zl"},
{"Name":"Korea","Code":"Bl"},
{"Name":"China","Code":"Bz"}
]
To create a new sorted list:
new_list = sorted(data, key=lambda k: k['Code'])
If you don't want to get a new list:
data[:] = sorted(data, key=lambda k: k['Code'])
The result is:
[{'Code': 'Bl', 'Name': 'Korea'}, {'Code': 'Bz', 'Name': 'China'}, {'Code': 'Zl', 'Name': 'Arab'}]
I hope I could help you!

Better way to produce same results:
from operator import itemgetter
data = [
{"Name": "Arab", "Code": "Zl"},
{"Name": "Korea", "Code": "Bl"},
{"Name": "China", "Code": "Bz"}
]
sorted_data = ((d["Code"], d["Name"]) for d in sorted(data, key=itemgetter("Code")))
code_data, name_data = (list(item) for item in zip(*sorted_data))
print(code_data) # -> ['Bl', 'Bz', 'Zl']
print(name_data) # -> ['Korea', 'China', 'Arab']

Here's one way using operator.itemgetter and unpacking via zip:
from operator import itemgetter
_, data_sorted = zip(*sorted(enumerate(data), key=lambda x: x[1]['Code']))
codes, names = zip(*map(itemgetter('Code', 'Name'), data_sorted))
print(codes)
# ('Bl', 'Bz', 'Zl')
print(names)
# ('Korea', 'China', 'Arab')

Dictionary key name from variable

I am trying to create a nested dictionary, whereby the key to each nested dictionary is named from the value from a variable. My end result should look something like this:
data_dict = {
'jane': {'name': 'jane', 'email': 'jane#example.com'},
'jim': {'name': 'jim', 'email': 'jim#example.com'}
}
Here is what I am trying:
data_dict = {}
s = "jane"
data_dict[s][name] = 'jane'
To my surprise, this does not work. Is this possible?

You want something like:
data_dict = {}
s = "jane"
data_dict[s] = {}
data_dict[s]['name'] = s
That should work, though I would recommend instead of a nested dictionary that you use a dictionary of names to either namedtuples or instances of a class.

Try this:
data_dict = {}
s = ["jane", "jim"]
for name in s:
data_dict[name] = {}
data_dict[name]['name'] = name
data_dict[name]['email'] = name + '#example.com'

as #Milad in the comment mentioned, you first need to initialize s as empty dictionary first
data={}
data['Tom']={}
data['Tom']['name'] = 'Tom Marvolo Riddle'
data['Tom']['email'] = 'iamlordvoldermort.com'

For existing dictionaries you can do dict[key] = value although if there is no dict that would raise an error. I think this is the code you want to have:
data_dict = {}
s = "jane"
data_dict[s] = {"name": s, "email": f"{s}#example.com"}
print(data_dict)
I just realized when I got a notification about this question:
data_dict = defaultdict(dict)
data_dict["jane"]["name"] = "jane"
Would be a better answer I think.

Turn a simple dictionary into dictionary with nested lists

Given the following data received from a web form:
for key in request.form.keys():
print key, request.form.getlist(key)
group_name [u'myGroup']
category [u'social group']
creation_date [u'03/07/2013']
notes [u'Here are some notes about the group']
members[0][name] [u'Adam']
members[0][location] [u'London']
members[0][dob] [u'01/01/1981']
members[1][name] [u'Bruce']
members[1][location] [u'Cardiff']
members[1][dob] [u'02/02/1982']
How can I turn it into a dictionary like this? It's eventually going to be used as JSON but as JSON and dictionaries are easily interchanged my goal is just to get to the following structure.
event = {
group_name : 'myGroup',
notes : 'Here are some notes about the group,
category : 'social group',
creation_date : '03/07/2013',
members : [
{
name : 'Adam',
location : 'London',
dob : '01/01/1981'
}
{
name : 'Bruce',
location : 'Cardiff',
dob : '02/02/1982'
}
]
}
Here's what I have managed so far. Using the following list comprehension I can easily make sense of the ordinary fields:
event = [ (key, request.form.getlist(key)[0]) for key in request.form.keys() if key[0:7] != "catches" ]
but I'm struggling with the members list. There can be any number of members. I think I need to separately create a list for them and add that to a dictionary with the non-iterative records. I can get the member data like this:
tmp_members = [(key, request.form.getlist(key)) for key in request.form.keys() if key[0:7]=="members"]
Then I can pull out the list index and field name:
member_arr = []
members_orig = [ (key, request.form.getlist(key)[0]) for key in request.form.keys() if key[0:7] ==
"members" ]
for i in members_orig:
p1 = i[0].index('[')
p2 = i[0].index(']')
members_index = i[0][p1+1:p2]
p1 = i[0].rfind('[')
members_field = i[0][p1+1:-1]
But how do I add this to my data structure. The following won't work because I could be trying to process members[1][name] before members[0][name].
members_arr[int(members_index)] = {members_field : i[1]}
This seems very convoluted. Is there a simper way of doing this, and if not how can I get this working?

You could store the data in a dictionary and then use the json library.
import json
json_data = json.dumps(dict)
print(json_data)
This will print a json string.
Check out the json library here

Yes, convert it to a dictionary, then use json.dumps(), with some optional parameters, to print out the JSON in the format you need:
eventdict = {
'group_name': 'myGroup',
'notes': 'Here are some notes about the group',
'category': 'social group',
'creation_date': '03/07/2013',
'members': [
{'name': 'Adam',
'location': 'London',
'dob': '01/01/1981'},
{'name': 'Bruce',
'location': 'Cardiff',
'dob': '02/02/1982'}
]
}
import json
print json.dumps(eventdict, indent=4)
The order of the key:value pairs is not always consistent, but if you're just looking for pretty-looking JSON that can be parsed by a script, while remaining human-readable, this should work. You can also sort the keys alphabetically, using:
print json.dumps(eventdict, indent=4, sort_keys=True)

The following python functions can be used to create a nested dictionary from the flat dictionary. Just pass in the html form output to decode().
def get_key_name(str):
first_pos = str.find('[')
return str[:first_pos]
def get_subkey_name(str):
'''Used with lists of dictionaries only'''
first_pos = str.rfind('[')
last_pos = str.rfind(']')
return str[first_pos:last_pos+1]
def get_key_index(str):
first_pos = str.find('[')
last_pos = str.find(']')
return str[first_pos:last_pos+1]
def decode(idic):
odic = {} # Initialise an empty dictionary
# Scan all the top level keys
for key in idic:
# Nested entries have [] in their key
if '[' in key and ']' in key:
if key.rfind('[') == key.find('[') and key.rfind(']') == key.find(']'):
print key, 'is a nested list'
key_name = get_key_name(key)
key_index = int(get_key_index(key).replace('[','',1).replace(']','',1))
# Append can't be used because we may not get the list in the correct order.
try:
odic[key_name][key_index] = idic[key][0]
except KeyError: # List doesn't yet exist
odic[key_name] = [None] * (key_index + 1)
odic[key_name][key_index] = idic[key][0]
except IndexError: # List is too short
odic[key_name] = odic[key_name] + ([None] * (key_index - len(odic[key_name]) + 1 ))
# TO DO: This could be a function
odic[key_name][key_index] = idic[key][0]
else:
key_name = get_key_name(key)
key_index = int(get_key_index(key).replace('[','',1).replace(']','',1))
subkey_name = get_subkey_name(key).replace('[','',1).replace(']','',1)
try:
odic[key_name][key_index][subkey_name] = idic[key][0]
except KeyError: # Dictionary doesn't yet exist
print "KeyError"
# The dictionaries must not be bound to the same object
odic[key_name] = [{} for _ in range(key_index+1)]
odic[key_name][key_index][subkey_name] = idic[key][0]
except IndexError: # List is too short
# The dictionaries must not be bound to the same object
odic[key_name] = odic[key_name] + [{} for _ in range(key_index - len(odic[key_name]) + 1)]
odic[key_name][key_index][subkey_name] = idic[key][0]
else:
# This can be added to the output dictionary directly
print key, 'is a simple key value pair'
odic[key] = idic[key][0]
return odic