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How do I remove the first item from a list?
(12 answers)
Closed 5 years ago.
I wanna remove a dictionary from a nested dictionary and I don't know how.
From this dictionary:
dict = {
'user': [
{
'firstName': 'john',
'lastName': 'doe',
'movieList': []
},
{
'firstName': 'sarah',
'lastName': 'doe',
'movieList': []
},
{
'firstName': 'john',
'lastName': 'smith',
'movieList': []
},
{
'firstName': 'sarah',
'lastName': 'smith',
'movieList': []
}
], 'movie': []
}
I want to remove:
{
'firstName': 'john',
'lastName': 'doe',
'movieList': []
}
which has the index 0
I tried using delete but i get this error:
dict['user'][userId] TypeError: list indices must be integers or slices, not str
First, I wouldn't name the dict "dict", use "d" or something else.
dict['user'].pop(0)
I would write a comment, but my reputation is too low. It seems like your variable userId might be a string instead of an integer. If this is the case, try converting userId to an int:
userId = int(userId)
See this question for more info on del pop and remove Difference between del, remove and pop on lists
If you do not want to use the dictionary you are removing, I would use del, as below:
d = {'user': [{'firstName': 'john', 'lastName': 'doe', 'movieList': []},
{'firstName': 'sarah', 'lastName': 'doe', 'movieList': []},
{'firstName': 'john', 'lastName': 'smith', 'movieList': []},
{'firstName': 'sarah', 'lastName': 'smith', 'movieList': []}], 'movie': []}
del d['user'][0]
This [0] is deleting the first index of the list stored as the value of the 'user' key
If you are repeatedly deleting the first index of your list and your list is long, consider using a deque() instead, which has fast pops from either end:
https://docs.python.org/3/library/collections.html#collections.deque
Also, don't call your variables the same name as their types, aka don't call your dictionary variable dict
You can try this:
dict1 = {'user': [{'firstName': 'john', 'lastName': 'doe', 'movieList': []}, {'firstName': 'sarah', 'lastName': 'doe', 'movieList': []}, {'firstName': 'john', 'lastName': 'smith', 'movieList': []}, {'firstName': 'sarah', 'lastName': 'smith', 'movieList': []}], 'movie': []}
new_dict = {a:b[1:] if b else b for a, b in dict1.items()}
Output:
{'movie': [], 'user': [{'lastName': 'doe', 'movieList': [], 'firstName': 'sarah'}, {'lastName': 'smith', 'movieList': [], 'firstName': 'john'}, {'lastName': 'smith', 'movieList': [], 'firstName': 'sarah'}]}
Related
I have this list of dictionary and I would like to get those with the same exact value of 'name' and 'school' into a new list and also getting their 'age' merged into a list as well and the rest of the dictionary that is not identical to just add into the list as per usual..
Here is an example of the list of dictionary
[{'name': 'Jane', 'age':12, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'SMU'},{'name': 'Jane', 'age':14, 'school': 'SIT'}, {'name': 'Jane', 'age':16, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'NUS'}]
and I would like it to make it into something like this..
[{'name': 'Jane', 'age': [12,14,16], 'school': 'SIT'}, {'name': 'John', 'age': 13, 'school': 'SMU'}, {'name': 'John', 'age':13, 'school': 'NUS'}]
using Python.. please help!
tried using counter, loops but still can't get it to work..
You could use itertools.groupby().
Example:
import itertools
from pprint import pprint
data = [{'name': 'Jane', 'age':12, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'SMU'},{'name': 'Jane', 'age':14, 'school': 'SIT'}, {'name': 'Jane', 'age':16, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'NUS'}]
keyfunc = lambda x: (x["name"], x["school"])
# needs to be sorted to use groupby
data.sort(key=keyfunc)
output = []
for k,v in itertools.groupby(data, key=keyfunc):
this_group = {
"name": k[0],
"school": k[1],
"age": [i["age"] for i in v],
}
output.append(this_group)
pprint(output)
The output is:
[{'age': [12, 14, 16], 'name': 'Jane', 'school': 'SIT'},
{'age': [13], 'name': 'John', 'school': 'NUS'},
{'age': [13], 'name': 'John', 'school': 'SMU'}]
If you wish to go with the solution based on a buffer dictionary, please check out the dict.setdefault() method.
Example:
buffer = {}
for i in data:
buffer.setdefault((i["name"], i["school"]), []).append(i["age"])
For reference:
https://docs.python.org/3/library/itertools.html#itertools.groupby
https://docs.python.org/3/library/stdtypes.html#dict.setdefault
x = [{'name': 'Jane', 'age':12, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'SMU'},{'name': 'Jane', 'age':14, 'school': 'SIT'}, {'name': 'Jane', 'age':16, 'school': 'SIT'}, {'name': 'John', 'age':13, 'school': 'NUS'}]
new_x = {}
for r in x:
if r['name'] in new_x.keys():
if not isinstance(new_x[r['name']]['age'], list):
new_x[r['name']]['age'] = [new_x[r['name']]['age']]
if r['age'] not in new_x[r['name']]['age']:
new_x[r['name']]['age'].append(r['age'])
else:
new_x[r['name']] = {'age': r['age'], 'school': r['school']}
z = [v.update(name=k) for k, v in new_x.items()]
z = [v for k, v in new_x.items()]
Here is a universal solution to your problem. Only name and school are considered "special". All other keys, like age are converted to list when a new value has to be added.
l = [
{"name": "Jane", "age": 12, "school": "SIT"},
{"name": "John", "age": 13, "school": "SMU"},
{"name": "Jane", "age": 14, "school": "SIT"},
{"name": "Jane", "age": 16, "school": "SIT"},
{"name": "John", "age": 13, "school": "NUS"},
]
r = {}
for x in l:
id = f"{x['name']}-{x['school']}"
if id in r:
for k,v in x.items():
if k not in ["name", "school"]:
if k in r[id]:
if isinstance(r[id][k], list):
r[id][k].append(v)
else:
r[id][k] = [r[id][k], v]
else:
r[id][k] = v
else:
r[id] = x
result = [x for x in r.values()]
I am trying to figure out how to filter for the dictionaries that have a status of "awaiting_delivery". I am not sure how to do this (or if it is impossible). I am new to python and programming. I am using Python 3.8.5 on VS Code on Ubuntu 20.04. The data below is sample data that I created that resembles json data from an API. Any help on how to filter for "status" would be great. Thank you.
nested_dict = {
'list_data': [
{
'id': 189530,
'total': 40.05,
'user_data': {
'id': 1001,
'first_name': 'jane',
'last_name': 'doe'
},
'status': 'future_delivery'
},
{
'id': 286524,
'total': 264.89,
'user_data': {
'id': 1002,
'first_name': 'john',
'last_name': 'doe'
},
'status': 'awaiting_delivery'
},
{
'id': 368725,
'total': 1054.98,
'user_data': {
'id': 1003,
'first_name': 'chris',
'last_name': 'nobody'
},
'status': 'awaiting_delivery'
},
{
'id': 422955,
'total': 4892.78,
'user_data': {
'id': 1004,
'first_name': 'mary',
'last_name': 'madeup'
},
'status': 'future_delivery'
}
],
'current_page': 1,
'total': 2,
'first': 1,
'last': 5,
'per_page': 20
}
#confirm that nested_dict is a dictionary
print(type(nested_dict))
#create a list(int_list) from the nested_dict dictionary
int_list = nested_dict['list_data']
#confirm that int_list is a list
print(type(int_list))
#create the int_dict dictionary from the int_list list
for int_dict in int_list:
print(int_dict)
#this is my attempt at filtering the int_dict dictionar for all orders with a status of awaiting_delivery
for order in int_dict:
int_dict.get('status')
print(order)
Output from Terminal Follows:
<class 'dict'>
<class 'list'>
{'id': 189530, 'total': 40.05, 'user_data': {'id': 1001, 'first_name': 'jane', 'last_name': 'doe'}, 'status': 'future_delivery'}
{'id': 286524, 'total': 264.89, 'user_data': {'id': 1002, 'first_name': 'john', 'last_name': 'doe'}, 'status': 'awaiting_delivery'}
{'id': 368725, 'total': 1054.98, 'user_data': {'id': 1003, 'first_name': 'chris', 'last_name': 'nobody'}, 'status': 'awaiting_delivery'}
{'id': 422955, 'total': 4892.78, 'user_data': {'id': 1004, 'first_name': 'mary', 'last_name': 'madeup'}, 'status': 'future_delivery'}
id
total
user_data
status
You can obtain a filtered list of dicts by doing conditional list comprehension on your list of dicts:
# filter the data
list_data_filtered = [entry for entry in nested_dict['list_data']
if entry['status'] == 'awaiting_delivery']
# print out the results
for entry in list_data_filtered:
print(entry)
# results
# {'id': 286524, 'total': 264.89, 'user_data': {'id': 1002, 'first_name': 'john', 'last_name': 'doe'}, 'status': 'awaiting_delivery'}
# {'id': 368725, 'total': 1054.98, 'user_data': {'id': 1003, 'first_name': 'chris', 'last_name': 'nobody'}, 'status': 'awaiting_delivery'}
Attempting to return a specific value of a key that's in a list within a dict. I've tried iterating over items, searching specific ID's and not quite getting the result I'm after.
{
'activeAccount': True,
'country': 'USA',
'state': 'CA',
'users': [
{
'id': 'A',
'firstName': 'Tom',
'lastName': 'Cruise',
'creditScore': '713',
'balance': '65897.22',
'debts': '12414.12',
'savings': '15231.23'
},
{
'id': 'B',
'firstName': 'Jon',
'lastName': 'Snow',
'creditScore': '648',
'balance': '12366.23',
'debts': '522',
'savings': '121588'
},
{
'id': 'C',
'firstName': 'Rick',
'lastName': 'Sanchez',
'creditScore': '655',
'balance': '556425.33',
'debts': '0',
'savings': '125122.23'
},
{
'id': 'D',
'firstName': 'Monty',
'lastName': 'Python',
'creditScore': '815',
'balance': '4512699.13',
'debts': '4.25',
'savings': '5499865.12'
}
]
}
How would I retrieve a specific value of, for example, the credit score of users[2] to return the value?
655
I've tried the solution of cScore = (dict['users'][2]['creditScore']), but this isn't ideal in situations of the objects changing order, etc.
You have to iterate through the list of dicts, looking for one with an id of 2. Then you return the credit score from that dict. Use this starter code:
for account in data_base:
if account['id'] == 2:
score = account['creditScore']
break
Your underlying problem is that you chose a data representation that doesn't match your data access needs. If you always reference your accounts by id, then change your list to a dict, keyed by that value. If you look up accounts through various means, then you need to investigate some data base format that suits your use cases.
The dictionary value of 'users' is a list. So you have to iterate through the list and search for the specific id.
Here's how you can do it.
for i in cust['users']:
if i['id'] == '2':
print (i['creditScore'])
I searched for 'id' == 'D'. The output was 815
The dictionary is defined as :
cust = {
'activeAccount': True,
'country': 'USA',
'state': 'CA',
'users': [
{
'id': 'A',
'firstName': 'Tom',
'lastName': 'Cruise',
'creditScore': '713',
'balance': '65897.22',
'debts': '12414.12',
'savings': '15231.23'
},
{
'id': 'B',
'firstName': 'Jon',
'lastName': 'Snow',
'creditScore': '648',
'balance': '12366.23',
'debts': '522',
'savings': '121588'
},
{
'id': 'C',
'firstName': 'Rick',
'lastName': 'Sanchez',
'creditScore': '655',
'balance': '556425.33',
'debts': '0',
'savings': '125122.23'
},
{
'id': 'D',
'firstName': 'Monty',
'lastName': 'Python',
'creditScore': '815',
'balance': '4512699.13',
'debts': '4.25',
'savings': '5499865.12'
}
]
}
If you need to find the user by ID within of a list of users you can use a filter function:
dict = {
'activeAccount': True,
'country': 'USA',
'state': 'CA',
'users': [
{
'id': 'A',
'firstName': 'Tom',
'lastName': 'Cruise',
'creditScore': '713',
'balance': '65897.22',
'debts': '12414.12',
'savings': '15231.23'
},
{
'id': 'B',
'firstName': 'Jon',
'lastName': 'Snow',
'creditScore': '648',
'balance': '12366.23',
'debts': '522',
'savings': '121588'
},
{
'id': 'C',
'firstName': 'Rick',
'lastName': 'Sanchez',
'creditScore': '655',
'balance': '556425.33',
'debts': '0',
'savings': '125122.23'
},
{
'id': 'D',
'firstName': 'Monty',
'lastName': 'Python',
'creditScore': '815',
'balance': '4512699.13',
'debts': '4.25',
'savings': '5499865.12'
}
]
}
# get the user with ID:"D"
user = list(filter(lambda u: u["id"] == "D", dict["users"]))[0]
# get the value
cScore = user["creditScore"] # <-- Output: 815
Description
lambda u: u["id"] == "D" means: return True if the property "id" of a given element is "D".
It could be rewritten as a plain function:
def check_D(user):
return user["id"] == "D"
user = list(filter(check_D, dict["users"]))[0]
filter( <condition> , dict["users"]) means: to iterate through the list dict["users"] and to return all elements that meet the condition (the "lambda" in this case, or it could be any function that return True or False).
user = list( ... )[0] since a filter function returns an object, it's need to convert the object into a list and to get a first element of the list.
Another option
You can built a list comprehension with a condition (id == "D") and take a first element of the list:
creditScore = [usr["creditScore"] for usr in dict["users"] if usr["id"] == "D"][0]
print(creditScore) # <-- Output: 815
I have a dictionary as follows:
a = {'name': 'Test', 'lastName': 'Test', 'scores': ['1', '2'], 'subjects': ['te','re'] }
I have tried nested loops, but I'm not sure if that's the best approach.
As an output I need a list of dictionaries for each score and subject :
result1 = { 'name':'Test', 'lastName': 'Test', 'score': '1', 'subjects': 'te'}
result2 = { 'name':'Test', 'lastName': 'Test', 'score': '2', 'subjects': 're'}
How to best iterate through the lists and create such dictionary? The number of scores and subjects will always match.
Any help would be appreciated.
Here is a function which unzip your dictionary.
We first use next to find some list value in the dictionary, its length is the expected output size.
Note that this will fail if the dictionary contains no list at all.
def unzip_dict(d):
# Find one of the list in the dictionary to read its length
length = len(next(value for value in d.values() if isinstance(value, list)))
output = []
# Unzip the dictionary
for i in range(length):
output.append({k: v[i] if isinstance(v, list) else v for k, v in d.items()})
return output
a = {'name': 'Test', 'lastName': 'Test', 'scores': ['1', '2'], 'subjects': ['te', 're']}
print(unzip_dict(a))
Output
[{'lastName': 'Test', 'name': 'Test', 'scores': '1', 'subjects': 'te'},
{'lastName': 'Test', 'name': 'Test', 'scores': '2', 'subjects': 're'}]
Try this:
# setup data
a = {'name': 'Test', 'lastName': 'Test', 'scores': ['1', '2'], 'subjects': ['te','re'] }
# create list of dictionary
out_list = []
for value in a.get('scores'):
for item in a.get('subjects'):
out_list.append({'name': 'Test', 'lastName': 'Test', 'scores':value, 'subjects':item})
Output:
{'name': 'Test', 'lastName': 'Test', 'scores': '1', 'subjects': 'te'}
{'name': 'Test', 'lastName': 'Test', 'scores': '1', 'subjects': 're'}
{'name': 'Test', 'lastName': 'Test', 'scores': '2', 'subjects': 'te'}
{'name': 'Test', 'lastName': 'Test', 'scores': '2', 'subjects': 're'}
You don't need nested for loops, a single for loop is sufficient:
def foo(a):
finarr=[]
for i in range(len(a['scores'])):
fnarr={}
fnarr['name']=a['name']
fnarr['lastName']=a['lastName']
fnarr['score']=a['scores'][i]
fnarr['subject']=a['subjects'][i]
finarr.append(fnarr)
return finarr
Output:
[{'name': 'Test', 'lastName': 'Test', 'score': '1', 'subject': 'te'},
{'name': 'Test', 'lastName': 'Test', 'score': '2', 'subject': 're'}]
you can try this:
res1 = {}
res2= {}
for k,v in a.items():
if(k == "scores"):
res1[k] = v[0]
res2[k] = v[1]
elif(k=="subjects"):
res1[k] = v[0]
res2[k] = v[1]
else:
res1[k] = v
res2[k] = v
print(res1)
print(res2)
you can also take a look to defaultdict i think that would help you on your Task
You can utilize zip to attach scores and subjects to each other and then add it to a new list.
d = {'name': 'Test', 'lastName': 'Test', 'scores': ['1', '2'], 'subjects': ['te','re'] }
template = {'name': d['name'], 'lastName': d['lastName']}
res = []
for subject, score in zip(d['subjects'], d['scores']):
template.update({'subjects': subject, 'scores': score})
res.append(template)
print(res)
I have the following three dictionaries in an array:
items = [
{
'FirstName': 'David',
'LastName': 'Smith',
'Language': set(['en'])
},
{
'FirstName': 'David',
'LastName': 'Smith',
'Language': set(['fr'])
},
{
'FirstName': 'Bob',
'LastName': 'Jones',
'Language': set(['en'])
} ]
I want to merge together these dictionaries if the two dictionaries are the same minus the specified key: and add that key together. If using the "Language" key it would merge the array into the following:
[ {
'FirstName': 'David',
'LastName': 'Smith',
'Language': set(['en','fr'])
},{
'FirstName': 'Bob',
'LastName': 'Jones',
'Language': set(['en'])
} ]
Here is what I'm currently doing:
from copy import deepcopy
def _merge_items_on_field(items, field):
'''Given an array of dicts, merge the
dicts together if they are the same except for the 'field'.
If merging dicts, add the unique values of that field together.'''
items = deepcopy(items)
items_merged_on_field = []
for num, item in enumerate(items):
# Remove that key/value from the dict
field_value = item.pop(field)
# Get an array of items *without* that field to compare against
items_without_field = deepcopy(items_merged_on_field)
map(lambda d: d.pop(field), items_without_field)
# If the dict item is found ("else"), add the fields together
# If not ("except"), then add in the dict item to the array
try:
index = items_without_field.index(item)
except ValueError:
item[field] = field_value
items_merged_on_field.append(item)
else:
items_merged_on_field[index][field] = items_merged_on_field[index][field].union(field_value)
return items_merged_on_field
>>> items = [{'LastName': 'Smith', 'Language': set(['en']), 'FirstName': 'David'}, {'LastName': 'Smith', 'Language': set(['fr']), 'FirstName': 'David'}, {'LastName': 'Jones', 'Language': set(['en']), 'FirstName': 'Bob'}]
>>> _merge_items_on_field(items, 'Language')
[{'LastName': 'Smith', 'Language': set(['fr', 'en']), 'FirstName': 'David'}, {'LastName': 'Jones', 'Language': set(['en']), 'FirstName': 'Bob'}]
This seems a bit complicated -- is there a better way to do this?
There are a couple of ways of doing this. The most painless method to my knowledge utilises the pandas library—in particular, a groupby + apply.
import pandas as pd
merged = (
pd.DataFrame(items)
.groupby(['FirstName', 'LastName'], sort=False)
.Language
.apply(lambda x: set.union(*x))
.reset_index()
.to_dict(orient='records')
)
print(merged)
[
{'FirstName': 'David', 'LastName': 'Smith', 'Language': {'en', 'fr'}},
{'FirstName': 'Bob', 'LastName': 'Jones', 'Language': {'en'}}
]
The other method (that I mentioned) uses itertools.groupby, but seeing as you have 30 columns to group on, I'd just recommend sticking to pandas.
If you want to turn this into a function,
def merge(items, field):
df = pd.DataFrame(items)
columns = df.columns.difference([field]).tolist()
return (
df.groupby(columns, sort=False)[field]
.apply(lambda x: set.union(*x))
.reset_index()
.to_dict(orient='records')
)
merged = merge(items, 'Language')
print(merged)
[
{'FirstName': 'David', 'LastName': 'Smith', 'Language': {'en', 'fr'}},
{'FirstName': 'Bob', 'LastName': 'Jones', 'Language': {'en'}}
]
You can use itertools.groupby:
import itertools
d = [{'FirstName': 'David', 'LastName': 'Smith', 'Language': {'en'}}, {'FirstName': 'David', 'LastName': 'Smith', 'Language': {'fr'}}, {'FirstName': 'Bob', 'LastName': 'Jones', 'Language': {'en'}}]
v = [[a, list(b)] for a, b in itertools.groupby(sorted(d, key=lambda x:x['FirstName']), key=lambda x:x['FirstName'])]
final_dict = [{**{'FirstName':a}, **{'LastName':(lambda x:[list(set(x)), x[0]][len(set(x)) == 1])([i['LastName'] for i in b])}, **{'Language':set([list(i['Language'])[0] for i in b])}} for a, b in v]
Output:
[{'FirstName': 'Bob', 'LastName': 'Jones', 'Language': {'en'}}, {'FirstName': 'David', 'LastName': 'Smith', 'Language': {'en', 'fr'}}]
If pandas is not an option:
from itertools import groupby
from functools import reduce
arr = [
{'FirstName': 'David', 'LastName': 'Smith', 'Language': set(['en'])},
{'FirstName': 'David', 'LastName': 'Smith', 'Language': set(['fr'])},
{'FirstName': 'David', 'LastName': 'Jones', 'Language': set(['sp'])}
]
def reduce_field(items, field, op=set.union, sort=False):
def _key(d):
return tuple((k, v) for k, v in d.items() if k != field)
if sort:
items = sorted(items, key=_key)
res = []
for k, g in groupby(items, key=_key):
d = dict(k)
d[field] = reduce(op, (el[field] for el in g))
res.append(d)
return res
reduce_field(arr, 'Language')
You can try it manually :
new_dict={}
#
#
#
d = [{'FirstName': 'David', 'LastName': 'Smith', 'Language': {'en'}},
{'FirstName': 'David', 'LastName': 'Smith', 'Language': {'fr'}},
{'FirstName': 'Bob', 'LastName': 'Jones', 'Language': {'en'}}]
for i in d:
if (i['FirstName'],i['LastName']) not in new_dict:
new_dict[(i['FirstName'],i['LastName'])]=i
else:
new_dict[(i['FirstName'],i['LastName'])]['Language']=set(list(new_dict[(i['FirstName'],i['LastName'])]['Language'])+list(i['Language']))
print(new_dict.values())
output:
# dict_values([{'FirstName': 'Bob',
# 'LastName': 'Jones',
# 'Language': {'en'}},
# {'FirstName': 'David',
# 'LastName': 'Smith',
# 'Language': {'fr', 'en'}}])