Appending a Dictionary to a List in a Loop

Appending a dictionary to a list in a loop

You need to append a copy, otherwise you are just adding references to the same dictionary over and over again:

yourlist.append(yourdict.copy())

I used yourdict and yourlist instead of dict and list; you don't want to mask the built-in types.

Adding item to Dictionary within loop

In your current code, what Dictionary.update() does is that it updates (update means the value is overwritten from the value for same key in passed in dictionary) the keys in current dictionary with the values from the dictionary passed in as the parameter to it (adding any new key:value pairs if existing) . A single flat dictionary does not satisfy your requirement , you either need a list of dictionaries or a dictionary with nested dictionaries.

If you want a list of dictionaries (where each element in the list would be a diciotnary of a entry) then you can make case_list as a list and then append case to it (instead of update) .

Example -

case_list = []
for entry in entries_list:
    case = {'key1': entry[0], 'key2': entry[1], 'key3':entry[2] }
    case_list.append(case)

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Or you can also have a dictionary of dictionaries with the key of each element in the dictionary being entry1 or entry2 , etc and the value being the corresponding dictionary for that entry.

case_list = {}
for entry in entries_list:
    case = {'key1': value, 'key2': value, 'key3':value }
    case_list[entryname] = case  #you will need to come up with the logic to get the entryname.

How to append a multiple dictionaries to a list from a loop, python?

def func(*args):
    list_of_dicts = []

    for X, y in zip(X, y):
        a_dict = make_your_a_dict() # what you do in your example 
        list_of_dicts.append(a_dict)

    return list_of_dicts # check indentation level

maybe it works.

How to append dictionary to a list in loop

We're going to mix three important concepts to make this code really small and beautiful. First, a list comprehension, then, the zip method, and finally, the dict method, to build a dictionary out of a list of tuples:

my_list = [('a1', 'b1', 'c1', 'd1', 'e1'), ('a2', 'b2', 'c2', 'd2', 'e2')]
keys = ('key1', 'key2', 'key3', 'key4', 'key5')
final = [dict(zip(keys, elems)) for elems in my_list]

After that, the value of the final variable is:

>>> final
[{'key3': 'c1', 'key2': 'b1', 'key1': 'a1', 'key5': 'e1', 'key4': 'd1'},
{'key3': 'c2', 'key2': 'b2', 'key1': 'a2', 'key5': 'e2', 'key4': 'd2'}]

Also, you can get elements of a certain dictionary using the position of the dictionary in the list and the key you're looking for, i.e.:

>>> final[0]['key1']
'a1'

how to iterate over list and append dictionary to another list

How about this simpler version. Instead, the replace you can simply add 'd' to the string[1:] -

[{'id':i,'new_id':'d'+i[1:]} for i in lst]

[{'id': '12233223', 'new_id': 'd2233223'},
 {'id': '13232423', 'new_id': 'd3232423'},
 {'id': '23453443', 'new_id': 'd3453443'}]

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EDIT: just saw your if condition. Update with that -

[{'id':i,'new_id':'d'+i[1:]} for i in lst if i[0] if i[0]=='1' and len(i)==8]

[{'id': '12233223', 'new_id': 'd2233223'},
 {'id': '13232423', 'new_id': 'd3232423'}]

Python: Way to build a dictionary with a variable key and append to a list as the value inside a loop

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]}

Can't append all values to dictionary using for loop

try this

l = [(1,2),(3,4)]
a = {}
a['r'] = []
for i in l:
        a['r'].append(i)
print(a)

or simply you can do

l = [(1,2),(3,4)]
a = {}
a['r'] = l

print(a)

this is the output

{'r': [(1, 2), (3, 4)]}

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