How to build a 2-level dictionary?

Content Index :

How to build a 2-level dictionary?
Is there any elegant way to build a multi-level dictionary in python?
Android studio top level build.gradle file - how to set property of lower level build file
Build a dictionary in the form of a multi-level tree
Flattening a multi-level dictionary into single level dictionary
Jenkins Build Error: Build step 'Invoke top-level Maven targets' marked build as failure

How to build a 2-level dictionary?

Tag : python , By : user152319

Date : November 24 2020, 03:01 PM

wish helps you Instead of using the literal construc, you should use the assign operator:

base_dict = {} # same that base_dict = dict()

for i in range(10):
    base_dict[i] = {'file_name': 'test' + str(i+1) + '.csv', 'file_type': i+1}

def check_files(directory=os.path.dirname(os.path.realpath(__file__))):
    files = {}
    for file in os.listdir(directory):
        if os.path.isfile(file):
            i = len(files)
            file_name = os.fsdecode(file)
            files[i] = {'file_name': file_name}
            with open(file_name,'r', encoding='utf-8', errors='ignore') as f:
                line = f.readline()
                if line == firstline['one']:
                    files[i]['file_type'] = 'one'
                elif line == firstline['two']:
                    files[i]['file_type'] = 'one'
                else:
                    files[i]['file_type'] = 'unknown'
    return files

def check_files(directory=os.path.dirname(os.path.realpath(__file__))):
    files = []
    for file in os.listdir(directory):
        if os.path.isfile(file):
            file_name = os.fsdecode(file)
            files.append({'file_name': file_name})
            with open(file_name,'r', encoding='utf-8', errors='ignore') as f:
                line = f.readline()
                if line == firstline['one']:
                    files[-1]['file_type'] = 'one'
                elif line == firstline['two']:
                    files[-1]['file_type'] = 'one'
                else:
                    files[-1]['file_type'] = 'unknown'
    return files

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How to build a 2-level dictionary , Is there any elegant way to build a multi-level dictionary in python , Build a dictionary in the form of a multi-level tree , Flattening a multi-level dictionary into single level dictionary , high level vs low level programming ,

Is there any elegant way to build a multi-level dictionary in python?

Tag : python , By : redha

Date : March 29 2020, 07:55 AM

I hope this helps . This uses the copy function to allow you specify a different leaf node. Otherwise all the leaves will point to the same dictionary.

from copy import copy

def multidict(*args):
    if len(args) == 1:
        return copy(args[0])
    out = {}
    for x in args[0]:
        out[x] = multidict(*args[1:])
    return out

print multidict(['a', 'b'], ['A', 'B'], ['1', '2'], {})

Android studio top level build.gradle file - how to set property of lower level build file

Tag : android-studio , By : Bharath

Date : March 29 2020, 07:55 AM

Any of those help It is not the same. As I know you can't define the gradle file with a setProperty method.
You can define a custom name in the settings.gradle file. Something like:

include ':app'
project(':app').buildFileName = 'mybuild.gradle'

Build a dictionary in the form of a multi-level tree

Tag : python , By : Simon Capewell

Date : March 29 2020, 07:55 AM

wish of those help you can keep track of a mapping of ID to where it is in the tree and another for the actual tree.

db_tree = [
{"id":2, "parent_id":1, "level":1,  "name":"parent 1"},
{"id":5, "parent_id":2, "level":2,  "name":"child 1 - 1"},
{"id":6, "parent_id":2, "level":2,  "name":"child 1 - 2"},
{"id":9, "parent_id":2, "level":2,  "name":"child 1- 3"},
{"id":7, "parent_id":5, "level":3,  "name":"child 1 - 1 - 1"},
{"id":11, "parent_id":6, "level":3, "name":"children 2- 1"},
{"id":10, "parent_id":7, "level":4, "name":"child 4 levl parent 1"},
{"id":3, "parent_id":1, "level":1,  "name":"parent 2"},
{"id":13, "parent_id":3, "level":2, "name":"parent 2- 1 - chil"},
{"id":4, "parent_id":1, "level":1,  "name":"parent 3"},
{"id":8, "parent_id":1, "level":1,  "name":"parent 4"}
]

tree = {}
id_to_children = {1: tree}


for entry in db_tree:
    id = entry["id"]
    name = entry["name"]
    parent_id = entry["parent_id"]
    # assume all elements are new, could check if it already exists
    my_children = {}
    id_to_children[id] = my_children
    #this is where we should be added in the tree.
    parent_children = id_to_children[parent_id]
    # add our node as a new child of our parent.
    parent_children[name] = my_children


import pprint

pprint.pprint(tree)

Flattening a multi-level dictionary into single level dictionary

Tag : python , By : sam

Date : March 29 2020, 07:55 AM

Hope this helps I want to flatten a multi-level dictionary into a single level dictionary. The new single levels keys should be concatenated with a separator "_". It's not a duplicate of the other flatten dict question since this has arrays that are concatenated into a string properly separated. , The following should work:
Code:

d_new = dict()

for k, v in d.items():
    for k2 in set(k for d in v for k in d.keys()):
        d_new[f"{k}_{k2}"] = '; '.join(d2.get(k2, '') for d2 in v)

>>> d_new
{'dates_unparsed_date': '; 1987; ',
 'dates_date': '1970-08-01; ; 2000-00-00',
 'dates_type': 'Date of birth; Year of birth; Date of birth2'}

>>> {f"{k}_{k2}" : '; '.join(d2.get(k2, '') for d2 in v) for k, v in d.items() for k2 in set(k for d in v for k in d.keys())}
{'dates_unparsed_date': '; 1987; ',
 'dates_date': '1970-08-01; ; 2000-00-00',
 'dates_type': 'Date of birth; Year of birth; Date of birth2'}