Creating nested dictionary dynamically in python

I am trying to create a nested dictionary dynamically using python.

for example I need to create a function that will take the nodes and construct a nested dictionary with these nodes.

For example:

inputs:
'customers.applicant.individual.first_name'

output:

customers : 
 applicant: 
 individual:
 firstname:

and for each node, i need to make sure if it exist already if it does than skip else create the node. Can anyone please provide any help on this.

Thanks

edited Nov 14 '18 at 4:12

khelwood

31.6k74364

asked Nov 14 '18 at 3:37

skidwa

What have you tried so far?

– b-fg
Nov 14 '18 at 5:52

add a comment |

I am trying to create a nested dictionary dynamically using python.

for example I need to create a function that will take the nodes and construct a nested dictionary with these nodes.

For example:

inputs:
'customers.applicant.individual.first_name'

output:

customers : 
 applicant: 
 individual:
 firstname:

and for each node, i need to make sure if it exist already if it does than skip else create the node. Can anyone please provide any help on this.

Thanks

edited Nov 14 '18 at 4:12

khelwood

31.6k74364

asked Nov 14 '18 at 3:37

skidwa

What have you tried so far?

– b-fg
Nov 14 '18 at 5:52

add a comment |

I am trying to create a nested dictionary dynamically using python.

for example I need to create a function that will take the nodes and construct a nested dictionary with these nodes.

For example:

inputs:
'customers.applicant.individual.first_name'

output:

customers : 
 applicant: 
 individual:
 firstname:

and for each node, i need to make sure if it exist already if it does than skip else create the node. Can anyone please provide any help on this.

Thanks

edited Nov 14 '18 at 4:12

khelwood

31.6k74364

asked Nov 14 '18 at 3:37

skidwa

I am trying to create a nested dictionary dynamically using python.

for example I need to create a function that will take the nodes and construct a nested dictionary with these nodes.

For example:

inputs:
'customers.applicant.individual.first_name'

output:

customers : 
 applicant: 
 individual:
 firstname:

and for each node, i need to make sure if it exist already if it does than skip else create the node. Can anyone please provide any help on this.

Thanks

python python-3.x

edited Nov 14 '18 at 4:12

khelwood

31.6k74364

asked Nov 14 '18 at 3:37

skidwa

edited Nov 14 '18 at 4:12

khelwood

31.6k74364

asked Nov 14 '18 at 3:37

skidwa

edited Nov 14 '18 at 4:12

khelwood

31.6k74364

edited Nov 14 '18 at 4:12

khelwood

31.6k74364

edited Nov 14 '18 at 4:12

khelwood

31.6k74364

asked Nov 14 '18 at 3:37

skidwa

asked Nov 14 '18 at 3:37

skidwa

asked Nov 14 '18 at 3:37

skidwa

What have you tried so far?

– b-fg
Nov 14 '18 at 5:52

add a comment |

What have you tried so far?

– b-fg
Nov 14 '18 at 5:52

What have you tried so far?

– b-fg
Nov 14 '18 at 5:52

add a comment |

1 Answer
1

active

oldest

votes

You can take advantage of the fact that python dictionaries are mutable and do something like the following:

input_item1 = 'customers.applicant.individual.first_name.Bob'
input_item2 = 'customers.applicant.individual.first_name.Jim'

inputs = [input_item1, input_item2]
output_dictionary = dict()


for input_item in inputs:
 current_dict = output_dictionary
 for item in input_item.split('.'): 
 if item in current_dict:
 current_dict = current_dict[item]
 else:
 current_dict[item] = dict()
 current_dict = current_dict[item]

print(output_dictionary)

Basically, because dictionaries are mutable, if you modify 'current_dict', the entry in the larger dictionary that you are referencing gets updated too.

This gives an output of:

'customers': 'applicant': 'individual': 'first_name': 'Bob': , 'Jim':

edited Nov 14 '18 at 9:30

answered Nov 14 '18 at 9:04

Andrew McDowell

1,9181416

add a comment |

Your Answer

StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);

);

draft saved

draft discarded

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53292843%2fcreating-nested-dictionary-dynamically-in-python%23new-answer', 'question_page');

);

Post as a guest

Name

Required, but never shown

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

You can take advantage of the fact that python dictionaries are mutable and do something like the following:

input_item1 = 'customers.applicant.individual.first_name.Bob'
input_item2 = 'customers.applicant.individual.first_name.Jim'

inputs = [input_item1, input_item2]
output_dictionary = dict()


for input_item in inputs:
 current_dict = output_dictionary
 for item in input_item.split('.'): 
 if item in current_dict:
 current_dict = current_dict[item]
 else:
 current_dict[item] = dict()
 current_dict = current_dict[item]

print(output_dictionary)

Basically, because dictionaries are mutable, if you modify 'current_dict', the entry in the larger dictionary that you are referencing gets updated too.

This gives an output of:

'customers': 'applicant': 'individual': 'first_name': 'Bob': , 'Jim':

edited Nov 14 '18 at 9:30

answered Nov 14 '18 at 9:04

Andrew McDowell

1,9181416

add a comment |

You can take advantage of the fact that python dictionaries are mutable and do something like the following:

input_item1 = 'customers.applicant.individual.first_name.Bob'
input_item2 = 'customers.applicant.individual.first_name.Jim'

inputs = [input_item1, input_item2]
output_dictionary = dict()


for input_item in inputs:
 current_dict = output_dictionary
 for item in input_item.split('.'): 
 if item in current_dict:
 current_dict = current_dict[item]
 else:
 current_dict[item] = dict()
 current_dict = current_dict[item]

print(output_dictionary)

Basically, because dictionaries are mutable, if you modify 'current_dict', the entry in the larger dictionary that you are referencing gets updated too.

This gives an output of:

'customers': 'applicant': 'individual': 'first_name': 'Bob': , 'Jim':

edited Nov 14 '18 at 9:30

answered Nov 14 '18 at 9:04

Andrew McDowell

1,9181416

add a comment |

You can take advantage of the fact that python dictionaries are mutable and do something like the following:

input_item1 = 'customers.applicant.individual.first_name.Bob'
input_item2 = 'customers.applicant.individual.first_name.Jim'

inputs = [input_item1, input_item2]
output_dictionary = dict()


for input_item in inputs:
 current_dict = output_dictionary
 for item in input_item.split('.'): 
 if item in current_dict:
 current_dict = current_dict[item]
 else:
 current_dict[item] = dict()
 current_dict = current_dict[item]

print(output_dictionary)

Basically, because dictionaries are mutable, if you modify 'current_dict', the entry in the larger dictionary that you are referencing gets updated too.

This gives an output of:

'customers': 'applicant': 'individual': 'first_name': 'Bob': , 'Jim':

edited Nov 14 '18 at 9:30

answered Nov 14 '18 at 9:04

Andrew McDowell

1,9181416

You can take advantage of the fact that python dictionaries are mutable and do something like the following:

input_item1 = 'customers.applicant.individual.first_name.Bob'
input_item2 = 'customers.applicant.individual.first_name.Jim'

inputs = [input_item1, input_item2]
output_dictionary = dict()


for input_item in inputs:
 current_dict = output_dictionary
 for item in input_item.split('.'): 
 if item in current_dict:
 current_dict = current_dict[item]
 else:
 current_dict[item] = dict()
 current_dict = current_dict[item]

print(output_dictionary)

Basically, because dictionaries are mutable, if you modify 'current_dict', the entry in the larger dictionary that you are referencing gets updated too.

This gives an output of:

'customers': 'applicant': 'individual': 'first_name': 'Bob': , 'Jim':

edited Nov 14 '18 at 9:30

answered Nov 14 '18 at 9:04

Andrew McDowell

1,9181416

edited Nov 14 '18 at 9:30

answered Nov 14 '18 at 9:04

Andrew McDowell

1,9181416

answered Nov 14 '18 at 9:04

Andrew McDowell

1,9181416

answered Nov 14 '18 at 9:04

Andrew McDowell

1,9181416

add a comment |

draft saved

draft discarded

Thanks for contributing an answer to Stack Overflow!

Please be sure to answer the question. Provide details and share your research!

But avoid …

Asking for help, clarification, or responding to other answers.

Making statements based on opinion; back them up with references or personal experience.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Pfthb