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I have a list of dictionaries. I want to loop through this list of dictionary and for each specific name (an attribute inside each dictionary), I want to create a dictionary where the key is the name and the value of this key is a list which dynamically appends to the list in accordance with a specific condition.
For example, I have
d = [{'Name': 'John', 'id': 10},
{'Name': 'Mark', 'id': 21},
{'Name': 'Matthew', 'id': 30},
{'Name': 'Luke', 'id': 11},
{'Name': 'John', 'id': 20}]
I then built a list with only the names using names=[i['Name'] for i in dic1] so I have a list of names. Notice John will appear twice in this list (at the beginning and end). Then, I want to create a for-loop (for name in names), which creates a dictionary 'ID' that for its value is a list which appends this id field as it goes along.
So in the end I'm looking for this ID dictionary to have:
John: [10,20]
Mark: [21]
Matthew: [30]
Luke: [11]
Notice that John has a list length of two because his name appears twice in the list of dictionaries.
But I can't figure out a way to dynamically append these values to a list inside the for-loop. I tried:
ID={[]} #I also tried with just {}
for name in names:
ID[names].append([i['id'] for i in dic1 if i['Name'] == name])
Please let me know how one can accomplish this. Thanks.
Don't loop over the list of names and go searching for every one in the list; that's very inefficient, since you're scanning the whole list all over again for every name. Just loop over the original list once and update the ID dict as you go. Also, if you build the ID dict first, then you can get the list of names from it and avoid another list traversal:
names = ID.keys()
The easiest solution for ID itself is a dictionary with a default value of the empty list; that way ID[name].append will work for names that aren't in the dict yet, instead of blowing up with a KeyError.
from collections import defaultdict
ID = defaultdict(list)
for item in d:
ID[item['Name']].append(item['id'])
You can treat a defaultdict like a normal dict for almost every purpose, but if you need to, you can turn it into a plain dict by calling dict on it:
plain_id = dict(ID)
The Thonnu has a solution using get and list concatenation which works without defaultdict. Here's another take on a no-import solution:
ID = {}
for item in d:
name, number = item['Name'], item['id']
if name in ID:
ID[name].append(number)
else:
ID[name] = [ number ]
Using collections.defaultdict:
from collections import defaultdict
out = defaultdict(list)
for item in dic1:
out[item['Name']].append(item['id'])
print(dict(out))
Or, without any imports:
out = {}
for item in dic1:
out[item['Name']] = out.get(item['Name'], []) + [item['id']]
print(out)
Or, with a list comprehension:
out = {}
[out.update({item['Name']: out.get(item['Name'], []) + [item['id']]}) for item in dic1]
print(out)
Output:
{'John': [10, 20], 'Mark': [21], 'Matthew': [30], 'Luke': [11]}
dic1 = [{'Name': 'John', 'id': 10}, {'Name': 'Mark', 'id': 21}, {'Name': 'Matthew', 'id': 30}, {'Name': 'Luke', 'id': 11}, {'Name': 'John', 'id': 20}]
id_dict = {}
for dic in dic1:
key = dic['Name']
if key in id_dict:
id_dict[key].append(dic['id'])
else:
id_dict[key] = [dic['id']]
print(id_dict) # {'John': [10, 20], 'Mark': [21], 'Matthew': [30], 'Luke': [11]}
You can use defaultdict for this to initiate a dictionary with a default value. In this case the default value will be empty list.
from collections import defaultdict
d=defaultdict(list)
for item in dic1:
d[item['Name']].append(item['id'])
Output
{'John': [10, 20], 'Mark': [21], 'Matthew': [30], 'Luke': [11]} # by converting (not required) into pure dict dict(d)
You can do in a easy version
dic1=[{'Name': 'John', 'id':10}, {'Name': 'Mark', 'id':21},{'Name': 'Matthew', 'id':30}, {'Name': 'Luke', 'id':11}, {'Name': 'John', 'id':20}]
names=[i['Name'] for i in dic1]
ID = {}
for i, name in enumerate(names):
if name in ID:
ID[name].append(dic1[i]['id'])
else:
ID[name] = [dic1[i]['id']]
print(ID)
I have a big list of dictionaries. And the dictionaries don´t have the same structure. But, I don´t know all the structures (my list has 1 million elements).
For example:
mylist = [{'name': 'Juan Carlos','age':38},{'name':'David','country':'Brazil'},
{'name':'Agustina', 'country': 'Argentina'},{'name': 'Renzo','age':24}]
I want to separate mylist into different lists, and each list has dictionaries with the same keys.
For example:
list1 = [{'name': 'Juan Carlos','age':38},{'name': 'Renzo','age':24}]
list2 = [{'name':'David','country':'Brazil'},{'name':'Agustina', 'country': 'Argentina'}]
The problem is I don´t know how many sublists I'm going to have.
To do this, you need to create a dictionary where the key is the set of keys that are present:
mylist = [{'name': 'Juan Carlos','age':38},{'name':'David','country':'Brazil'},{'name':'Agustina', 'country': 'Argentina'},{'name': 'Renzo','age':24}]
sublists = {}
for sl in mylist:
keys = tuple(set(sl.keys()))
if keys not in sublists:
sublists[keys] = []
sublists[keys].append( sl )
print( sublists )
Output:
{('name', 'age'): [{'name': 'Juan Carlos', 'age': 38}, {'name': 'Renzo', 'age': 24}], ('name', 'country'): [{'name': 'David', 'country': 'Brazil'}, {'name': 'Agustina', 'country': 'Argentina'}]}
You can use a frozenset of keys as a new key:
mylist = [{'name': 'Juan Carlos','age':38},{'name':'David','country':'Brazil'},
{'name':'Agustina', 'country': 'Argentina'},{'name': 'Renzo','age':24}]
sep = {}
for d in mylist:
sep.setdefault(frozenset(d), []).append(d)
and then you can get a list of lists:
print(sep.values())
I have the data
data = [{'name': 'Dave', 'role': 'Manager'}, {'name': 'David', 'role': 'Engineer'}]
I am intending to add the values based on same keys and merging it by semicolon.
Expected output is like this as follows
{'name': 'Dave;David', 'role': 'Manager;Engineer'}
What I am trying to do is to convert into matrix then adding all the elements based on the index but it is very tedious approach.
result = [[i[e] for e in sorted(i.keys())] for i in data]
print(result)
current output: [['Dave', 'Manager'], ['David', 'Engineer']]
The following does what you want, assuming that data always contains dictionaries of strings.
I'm checking if each dictionary key already exists in the final dictionary. If it doesn't I add it, if it does I append it to the current value using the semicolon as a separator.
data = [{'name': 'Dave', 'role': 'Manager'}, {'name': 'David', 'role': 'Engineer'}]
final = {}
for dict in data:
for k,v in dict.items():
if k in final:
final[k] += ';' + v
else:
final[k] = v
print(final)
I have a list of dictionary for ex:
names = [{'Mark':'Volvo'}, {'John':'BMW'}, {'Eliza':'Merci'}, {'Calen':'Audi'}]
I would like to set the explicit ordering by the key names of the dictionary.
For example if I give this order:
['John','Mark','Calen','Eliza']
The expected output would be:
[{'John':'BMW'},{'Mark':'Volvo'},{'Calen':'Audi'},{'Eliza':'Merci'}]
I want to add a custom logic order. To be displayed through the template by their names, based on how I defined the order of their names.
Similar to Is there a way to sort a list of string by a “predicate” list?: Since names is a list of dictionaries with just one key-value pair each, use the index of the person's name in the order list as the key for the sort:
>>> names = [{'Mark': 'Volvo'}, {'John': 'BMW'}, {'Eliza': 'Merci'}, {'Calen' :'Audi'}]
>>> order = ['John', 'Mark', 'Calen', 'Eliza']
>>>
>>> # with `sorted()`
>>> sorted(names, key=lambda d: order.index(list(d.keys())[0]))
[{'John': 'BMW'}, {'Mark': 'Volvo'}, {'Calen': 'Audi'}, {'Eliza': 'Merci'}]
>>>
>>> # or with `list.sort()`
>>> names.sort(key=lambda d: order.index(list(d.keys())[0]))
>>> names
[{'John': 'BMW'}, {'Mark': 'Volvo'}, {'Calen': 'Audi'}, {'Eliza': 'Merci'}]
dict.keys() is not not subscriptable, so dict.keys()[0] doesn't work. So first convert that to a list and then use its one-and-only key list(dict.keys())[0]. That would give 'Mark', 'John', etc. Then get the index of that person's name in the order list. Note: it will fail if a person is not listed in order.
Even if names is a list of dictionaries with more than one key-value pair each, as long as the person's name is the first key, it will still work as of Python 3.7/3.6. See the note below this item:
Changed in version 3.7: Dictionary order is guaranteed to be insertion order. This behavior was an implementation detail of CPython from 3.6.
>>> names = [{'Mark': 'Volvo', 'age': 30},
... {'John': 'BMW', 'age': 40},
... {'Eliza': 'Merci', 'age': 50},
... {'Calen': 'Audi', 'age': 60}]
>>> sorted(names, key=lambda d: order.index(list(d.keys())[0]))
[{'John': 'BMW', 'age': 40}, {'Mark': 'Volvo', 'age': 30}, {'Calen': 'Audi', 'age': 60}, {'Eliza': 'Merci', 'age': 50}]
>>>
First, if your dictionaries only have one entry, tuples seem to be a better choice for this data:
>>> names = [('Mark', 'Volvo'), ('John', 'BMW'), ('Eliza', 'Merci'), ('Calen', 'Audi')]
Now, given this order:
>>> order = ['John', 'Mark', 'Calen', 'Eliza']
you can create a dict that maps the names to the indices:
>>> order_map = { k: v for v, k in enumerate(order) }
>>> order_map
{'John': 0, 'Mark': 1, 'Calen': 2, 'Eliza': 3}
and use it in a key function for sort:
>>> names.sort(key=lambda v: order_map[v[0]])
>>> names
[('John', 'BMW'), ('Mark', 'Volvo'), ('Calen', 'Audi'), ('Eliza', 'Merci')]
names = [{'Mark':'Volvo'}, {'John':'BMW'}, {'Eliza':'Merci'}, {'Calen':'Audi'}]
ordered_keys = ['John','Mark','Calen','Eliza']
sorted_names = [name for key in ordered_keys for name in names if key in name]
It iterates over the ordered_keys in order and extracts any name in the list of dict that has that key.
My code is
index = 0
for key in dataList[index]:
print(dataList[index][key])
Seems to work fine for printing the values of dictionary keys for index = 0. However, I can't figure out how to iterate through an unknown number of dictionaries in dataList.
You could just iterate over the indices of the range of the len of your list:
dataList = [{'a': 1}, {'b': 3}, {'c': 5}]
for index in range(len(dataList)):
for key in dataList[index]:
print(dataList[index][key])
or you could use a while loop with an index counter:
dataList = [{'a': 1}, {'b': 3}, {'c': 5}]
index = 0
while index < len(dataList):
for key in dataList[index]:
print(dataList[index][key])
index += 1
you could even just iterate over the elements in the list directly:
dataList = [{'a': 1}, {'b': 3}, {'c': 5}]
for dic in dataList:
for key in dic:
print(dic[key])
It could be even without any lookups by just iterating over the values of the dictionaries:
dataList = [{'a': 1}, {'b': 3}, {'c': 5}]
for dic in dataList:
for val in dic.values():
print(val)
Or wrap the iterations inside a list-comprehension or a generator and unpack them later:
dataList = [{'a': 1}, {'b': 3}, {'c': 5}]
print(*[val for dic in dataList for val in dic.values()], sep='\n')
the possibilities are endless. It's a matter of choice what you prefer.
You can easily do this:
for dict_item in dataList:
for key in dict_item:
print(dict_item[key])
It will iterate over the list, and for each dictionary in the list, it will iterate over the keys and print its values.
use=[{'id': 29207858, 'isbn': '1632168146', 'isbn13': '9781632168146', 'ratings_count': 0}]
for dic in use:
for val,cal in dic.items():
print(f'{val} is {cal}')
def extract_fullnames_as_string(list_of_dictionaries):
return list(map(lambda e : "{} {}".format(e['first'],e['last']),list_of_dictionaries))
names = [{'first': 'Zhibekchach', 'last': 'Myrzaeva'}, {'first': 'Gulbara', 'last': 'Zholdoshova'}]
print(extract_fullnames_as_string(names))
#Well...the shortest way (1 line only) in Python to extract data from the list of dictionaries is using lambda form and map together.
"""The approach that offers the most flexibility and just seems more dynamically appropriate to me is as follows:"""
Loop thru list in a Function called.....
def extract_fullnames_as_string(list_of_dictionaries):
result = ([val for dic in list_of_dictionaries for val in
dic.values()])
return ('My Dictionary List is ='result)
dataList = [{'first': 3, 'last': 4}, {'first': 5, 'last': 7},{'first':
15, 'last': 9},{'first': 51, 'last': 71},{'first': 53, 'last': 79}]
print(extract_fullnames_as_string(dataList))
"""This way, the Datalist can be any format of a Dictionary you throw at it, otherwise you can end up dealing with format issues, I found. Try the following and it will still works......."""
dataList1 = [{'a': 1}, {'b': 3}, {'c': 5}]
dataList2 = [{'first': 'Zhibekchach', 'last': 'Myrzaeva'}, {'first':
'Gulbara', 'last': 'Zholdoshova'}]
print(extract_fullnames_as_string(dataList1))
print(extract_fullnames_as_string(dataList2))
Another pythonic solution is using collections module.
Here is an example where I want to generate a dict containing only 'Name' and 'Last Name' values:
from collections import defaultdict
test_dict = [{'Name': 'Maria', 'Last Name': 'Bezerra', 'Age': 31},
{'Name': 'Ana', 'Last Name': 'Mota', 'Age': 31},
{'Name': 'Gabi', 'Last Name': 'Santana', 'Age': 31}]
collect = defaultdict(dict)
# at this moment, 'key' becomes every dict of your list of dict
for key in test_dict:
collect[key['Name']] = key['Last Name']
print(dict(collect))
Output should be:
{'Name': 'Maria', 'Last Name': 'Bezerra'}, {'Name': 'Ana', 'Last Name': 'Mota'}, {'Name': 'Gabi', 'Last Name': 'Santana'}
There are multiple ways to iterate through a list of dictionaries. However, if you are into Pythonic code, consider the following ways, but first, let's use data_list instead of dataList because in Python snake_case is preferred over camelCase.
Way #1: Iterating over a dictionary's keys
# let's assume that data_list is the following dictionary
data_list = [{'Alice': 10}, {'Bob': 7}, {'Charlie': 5}]
for element in data_list:
for key in element:
print(key, element[key])
Output
Alice 10
Bob 7
Charlie 5
Explanation:
for element in data_list: -> element will be a dictionary in data_list at each iteration, i.e., {'Alice': 10} in the first iteration,
{'Bob': 7} in the second iteration, and {'Charlie': 5}, in the third iteration.
for key in element: -> key will be a key of element at each iteration, so when element is {'Alice': 10}, the values for key will be 'Alice'. Keep in mind that element could contain more keys, but in this particular example it has just one.
print(key, element[key]) -> it prints key and the value of element for key key, i.e., it access the value of key in `element.
Way #2: Iterating over a dictionary's keys and values
# let's assume that data_list is the following dictionary
data_list = [{'Alice': 10}, {'Bob': 7}, {'Charlie': 5}]
for element in data_list:
for key, value in element.items():
print(key, value)
The output for this code snippet is the same as the previous one.
Explanation:
for element in data_list: -> it has the same explanation as the one in the code before.
for key, value in element.items(): -> at each iteration, element.items() will return a tuple that contains two elements. The former element is the key, and the latter is the value associated with that key, so when element is {'Alice': 10}, the value for key will be 'Alice', and the value for value will be 10. Keep in mind that this dictionary has only one key-value pair.
print(key, value) -> it prints key and value.
As stated before, there are multiple ways to iterate through a list of dictionaries, but to keep your code more Pythonic, avoid using indices or while loops.
had a similar issue, fixed mine by using a single for loop to iterate over the list, see code snippet
de = {"file_name":"jon","creation_date":"12/05/2022","location":"phc","device":"s3","day":"1","time":"44692.5708703703","year":"1900","amount":"3000","entity":"male"}
se = {"file_name":"bone","creation_date":"13/05/2022","location":"gar","device":"iphone","day":"2","time":"44693.5708703703","year":"2022","amount":"3000","entity":"female"}
re = {"file_name":"cel","creation_date":"12/05/2022","location":"ben car","device":"galaxy","day":"1","time":"44695.5708703703","year":"2022","amount":"3000","entity":"male"}
te = {"file_name":"teiei","creation_date":"13/05/2022","location":"alcon","device":"BB","day":"2","time":"44697.5708703703","year":"2022","amount":"3000","entity":"female"}
ye = {"file_name":"js","creation_date":"12/05/2022","location":"woji","device":"Nokia","day":"1","time":"44699.5708703703","year":"2022","amount":"3000","entity":"male"}
ue = {"file_name":"jsdjd","creation_date":"13/05/2022","location":"town","device":"M4","day":"5","time":"44700.5708703703","year":"2022","amount":"3000","entity":"female"}
d_list = [de,se,re,te,ye,ue]
for dic in d_list:
print (dic['file_name'],dic['creation_date'])