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I want to create function that keep appending new data to nested dictionary

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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;

0

I want to store new data(Key, value) to the nested dictionary.
But I don't have any clue how to fix it.

def add(a,b,c,d,container)
 container = # as database
 data =
 data[a] = ["first": b, "second": c, "third": d]

 for e in data:
 if date[date] not in calendar:
 container[date[a]] = 
 container[date[a]].update(["first": b, "second": c, "third": d)


add(name, 1, 2, hi, container)
add(name1, 2, 1, hi, container)

I see the following output:

name: ["first": 1, "second": 2, "third":hi ]

name1: ["first": 2, "second": 1, "third":hi ]

I expect the output as:

name: ["first": 1, "second": 2, "third":hi ], name1: ["first": 2, "second": 1, "third":hi ]

Please help me out!

python

share|improve this question

edited Mar 25 at 4:18

Neal Lee

asked Mar 25 at 3:51

Neal LeeNeal Lee

32

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What's your function? Did you mean add?
  
  – knh190
  Mar 25 at 4:01
  

  
  
  
  

add a comment | 

0

I want to store new data(Key, value) to the nested dictionary.
But I don't have any clue how to fix it.

def add(a,b,c,d,container)
 container = # as database
 data =
 data[a] = ["first": b, "second": c, "third": d]

 for e in data:
 if date[date] not in calendar:
 container[date[a]] = 
 container[date[a]].update(["first": b, "second": c, "third": d)


add(name, 1, 2, hi, container)
add(name1, 2, 1, hi, container)

I see the following output:

name: ["first": 1, "second": 2, "third":hi ]

name1: ["first": 2, "second": 1, "third":hi ]

I expect the output as:

name: ["first": 1, "second": 2, "third":hi ], name1: ["first": 2, "second": 1, "third":hi ]

Please help me out!

python

share|improve this question

edited Mar 25 at 4:18

Neal Lee

asked Mar 25 at 3:51

Neal LeeNeal Lee

32

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What's your function? Did you mean add?
  
  – knh190
  Mar 25 at 4:01
  

  
  
  
  

add a comment | 

I want to store new data(Key, value) to the nested dictionary.
But I don't have any clue how to fix it.

def add(a,b,c,d,container)
 container = # as database
 data =
 data[a] = ["first": b, "second": c, "third": d]

 for e in data:
 if date[date] not in calendar:
 container[date[a]] = 
 container[date[a]].update(["first": b, "second": c, "third": d)


add(name, 1, 2, hi, container)
add(name1, 2, 1, hi, container)

I see the following output:

name: ["first": 1, "second": 2, "third":hi ]

name1: ["first": 2, "second": 1, "third":hi ]

I expect the output as:

name: ["first": 1, "second": 2, "third":hi ], name1: ["first": 2, "second": 1, "third":hi ]

Please help me out!

python

share|improve this question

edited Mar 25 at 4:18

Neal Lee

asked Mar 25 at 3:51

Neal LeeNeal Lee

32

def add(a,b,c,d,container)
 container = # as database
 data =
 data[a] = ["first": b, "second": c, "third": d]

 for e in data:
 if date[date] not in calendar:
 container[date[a]] = 
 container[date[a]].update(["first": b, "second": c, "third": d)


add(name, 1, 2, hi, container)
add(name1, 2, 1, hi, container)

name: ["first": 1, "second": 2, "third":hi ]

name1: ["first": 2, "second": 1, "third":hi ]

name: ["first": 1, "second": 2, "third":hi ], name1: ["first": 2, "second": 1, "third":hi ]

share|improve this question

edited Mar 25 at 4:18

Neal Lee

asked Mar 25 at 3:51

Neal LeeNeal Lee

32

share|improve this question

edited Mar 25 at 4:18

Neal Lee

asked Mar 25 at 3:51

Neal LeeNeal Lee

32

asked Mar 25 at 3:51

Neal LeeNeal Lee

32

asked Mar 25 at 3:51

Neal LeeNeal Lee

32

3 Answers
3

active

oldest

votes

0

i am creating a global_dict which is main dictionary where storing of all the
dict element is taken place.
for your given problem, if you want to add the data in a dict continously and that dict have same structure, then better:
is to create a default dict , and in that dict update the values and finally add that new small dict to the main dict using update method.

def funcion(a,b,c,d, container):
 new_dic=a:['first':b,"Second":c,"third":d]
 container.update(new_dic)
funcion('name', 1, 2, 'hi')
funcion('name1', 2, 1,'hi')
print(container)

"""
output 

'name': ['first': 1, 'Second': 2, 'third': 'hi'], 
'name1': ['first': 2, 'Second': 1, 'third': 'hi']

"

share|improve this answer

edited Mar 25 at 4:09

answered Mar 25 at 3:59

prashant ranaprashant rana

2,20611123

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Add some explanation is more informative.
  
  – knh190
  Mar 25 at 3:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @knh190 done added explanation
  
  – prashant rana
  Mar 25 at 4:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you for your comment. Do you have another way expect global method?
  
  – Neal Lee
  Mar 25 at 4:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @NealLee pass the main dictionary in the function and return the global_dict dictionary
  
  – prashant rana
  Mar 25 at 4:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @NealLee updated the solution, check
  
  – prashant rana
  Mar 25 at 4:09
  

  
  
  
  

add a comment | 

1

You are creating a local dictionary in your add function. Did you see that?

def add(a,b,c,d,container)
 container = # it's a local new dict
 # ...

Instead you should create the dictionary outside of the function, otherwise you'll always get a new dictionary containing only one key. For example:

container = 

def add(a, b, c, d):
 container[a] = b
 container[c] = d

share|improve this answer

answered Mar 25 at 4:04

knh190knh190

1,708822

add a comment | 

-1

I recently made a module for this and posted it on github, it doesnt use recursion so therefore it has absolutely no limits to it. It allows you to edit, add, and remove from a nested dictionary using a keypath. Here it is on stackoverflow(this will answer your question):

How can you add entries, and retrieve, alter, or remove values from specific keys in any nested dictionary without recursion?

and here it is on github:

https://github.com/kthewhispers/Nested-Dictionary-Tools-Python/tree/master/src

share|improve this answer

answered Mar 25 at 4:07

Keith CroninKeith Cronin

1368

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Better to add solution here, your link might get delete in future, which will create problem for the future readers
  
  – prashant rana
  Mar 25 at 4:10
  

  
  
  
  

add a comment | 

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0

i am creating a global_dict which is main dictionary where storing of all the
dict element is taken place.
for your given problem, if you want to add the data in a dict continously and that dict have same structure, then better:
is to create a default dict , and in that dict update the values and finally add that new small dict to the main dict using update method.

def funcion(a,b,c,d, container):
 new_dic=a:['first':b,"Second":c,"third":d]
 container.update(new_dic)
funcion('name', 1, 2, 'hi')
funcion('name1', 2, 1,'hi')
print(container)

"""
output 

'name': ['first': 1, 'Second': 2, 'third': 'hi'], 
'name1': ['first': 2, 'Second': 1, 'third': 'hi']

"

share|improve this answer

edited Mar 25 at 4:09

answered Mar 25 at 3:59

prashant ranaprashant rana

2,20611123

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Add some explanation is more informative.
  
  – knh190
  Mar 25 at 3:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @knh190 done added explanation
  
  – prashant rana
  Mar 25 at 4:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you for your comment. Do you have another way expect global method?
  
  – Neal Lee
  Mar 25 at 4:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @NealLee pass the main dictionary in the function and return the global_dict dictionary
  
  – prashant rana
  Mar 25 at 4:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @NealLee updated the solution, check
  
  – prashant rana
  Mar 25 at 4:09
  

  
  
  
  

add a comment | 

0

i am creating a global_dict which is main dictionary where storing of all the
dict element is taken place.
for your given problem, if you want to add the data in a dict continously and that dict have same structure, then better:
is to create a default dict , and in that dict update the values and finally add that new small dict to the main dict using update method.

def funcion(a,b,c,d, container):
 new_dic=a:['first':b,"Second":c,"third":d]
 container.update(new_dic)
funcion('name', 1, 2, 'hi')
funcion('name1', 2, 1,'hi')
print(container)

"""
output 

'name': ['first': 1, 'Second': 2, 'third': 'hi'], 
'name1': ['first': 2, 'Second': 1, 'third': 'hi']

"

share|improve this answer

edited Mar 25 at 4:09

answered Mar 25 at 3:59

prashant ranaprashant rana

2,20611123

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Add some explanation is more informative.
  
  – knh190
  Mar 25 at 3:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @knh190 done added explanation
  
  – prashant rana
  Mar 25 at 4:03
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  thank you for your comment. Do you have another way expect global method?
  
  – Neal Lee
  Mar 25 at 4:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @NealLee pass the main dictionary in the function and return the global_dict dictionary
  
  – prashant rana
  Mar 25 at 4:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @NealLee updated the solution, check
  
  – prashant rana
  Mar 25 at 4:09
  

  
  
  
  

add a comment | 

i am creating a global_dict which is main dictionary where storing of all the
dict element is taken place.
for your given problem, if you want to add the data in a dict continously and that dict have same structure, then better:
is to create a default dict , and in that dict update the values and finally add that new small dict to the main dict using update method.

def funcion(a,b,c,d, container):
 new_dic=a:['first':b,"Second":c,"third":d]
 container.update(new_dic)
funcion('name', 1, 2, 'hi')
funcion('name1', 2, 1,'hi')
print(container)

"""
output 

'name': ['first': 1, 'Second': 2, 'third': 'hi'], 
'name1': ['first': 2, 'Second': 1, 'third': 'hi']

"

share|improve this answer

edited Mar 25 at 4:09

answered Mar 25 at 3:59

prashant ranaprashant rana

2,20611123

def funcion(a,b,c,d, container):
 new_dic=a:['first':b,"Second":c,"third":d]
 container.update(new_dic)
funcion('name', 1, 2, 'hi')
funcion('name1', 2, 1,'hi')
print(container)

"""
output 

'name': ['first': 1, 'Second': 2, 'third': 'hi'], 
'name1': ['first': 2, 'Second': 1, 'third': 'hi']

"

share|improve this answer

edited Mar 25 at 4:09

answered Mar 25 at 3:59

prashant ranaprashant rana

2,20611123

answered Mar 25 at 3:59

prashant ranaprashant rana

2,20611123

answered Mar 25 at 3:59

prashant ranaprashant rana

2,20611123

1

You are creating a local dictionary in your add function. Did you see that?

def add(a,b,c,d,container)
 container = # it's a local new dict
 # ...

Instead you should create the dictionary outside of the function, otherwise you'll always get a new dictionary containing only one key. For example:

container = 

def add(a, b, c, d):
 container[a] = b
 container[c] = d

share|improve this answer

answered Mar 25 at 4:04

knh190knh190

1,708822

add a comment | 

1

You are creating a local dictionary in your add function. Did you see that?

def add(a,b,c,d,container)
 container = # it's a local new dict
 # ...

Instead you should create the dictionary outside of the function, otherwise you'll always get a new dictionary containing only one key. For example:

container = 

def add(a, b, c, d):
 container[a] = b
 container[c] = d

share|improve this answer

answered Mar 25 at 4:04

knh190knh190

1,708822

add a comment | 

You are creating a local dictionary in your add function. Did you see that?

def add(a,b,c,d,container)
 container = # it's a local new dict
 # ...

Instead you should create the dictionary outside of the function, otherwise you'll always get a new dictionary containing only one key. For example:

container = 

def add(a, b, c, d):
 container[a] = b
 container[c] = d

share|improve this answer

answered Mar 25 at 4:04

knh190knh190

1,708822

def add(a,b,c,d,container)
 container = # it's a local new dict
 # ...

container = 

def add(a, b, c, d):
 container[a] = b
 container[c] = d

share|improve this answer

answered Mar 25 at 4:04

knh190knh190

1,708822

answered Mar 25 at 4:04

knh190knh190

1,708822

answered Mar 25 at 4:04

knh190knh190

1,708822

-1

I recently made a module for this and posted it on github, it doesnt use recursion so therefore it has absolutely no limits to it. It allows you to edit, add, and remove from a nested dictionary using a keypath. Here it is on stackoverflow(this will answer your question):

How can you add entries, and retrieve, alter, or remove values from specific keys in any nested dictionary without recursion?

and here it is on github:

https://github.com/kthewhispers/Nested-Dictionary-Tools-Python/tree/master/src

share|improve this answer

answered Mar 25 at 4:07

Keith CroninKeith Cronin

1368

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Better to add solution here, your link might get delete in future, which will create problem for the future readers
  
  – prashant rana
  Mar 25 at 4:10
  

  
  
  
  

add a comment | 

-1

I recently made a module for this and posted it on github, it doesnt use recursion so therefore it has absolutely no limits to it. It allows you to edit, add, and remove from a nested dictionary using a keypath. Here it is on stackoverflow(this will answer your question):

How can you add entries, and retrieve, alter, or remove values from specific keys in any nested dictionary without recursion?

and here it is on github:

https://github.com/kthewhispers/Nested-Dictionary-Tools-Python/tree/master/src

share|improve this answer

answered Mar 25 at 4:07

Keith CroninKeith Cronin

1368

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Better to add solution here, your link might get delete in future, which will create problem for the future readers
  
  – prashant rana
  Mar 25 at 4:10
  

  
  
  
  

add a comment | 

I recently made a module for this and posted it on github, it doesnt use recursion so therefore it has absolutely no limits to it. It allows you to edit, add, and remove from a nested dictionary using a keypath. Here it is on stackoverflow(this will answer your question):

How can you add entries, and retrieve, alter, or remove values from specific keys in any nested dictionary without recursion?

and here it is on github:

https://github.com/kthewhispers/Nested-Dictionary-Tools-Python/tree/master/src

share|improve this answer

answered Mar 25 at 4:07

Keith CroninKeith Cronin

1368

share|improve this answer

answered Mar 25 at 4:07

Keith CroninKeith Cronin

1368

answered Mar 25 at 4:07

Keith CroninKeith Cronin

1368

answered Mar 25 at 4:07

Keith CroninKeith Cronin

1368