How do you create an empty nested dictionary in python?

I have 2 CSV files: 'Data' and 'Mapping':

• 'Mapping' file has 4 columns: Device_Name, GDN, Device_Type, and Device_OS. All four columns are populated.
• 'Data' file has these same columns, with Device_Name column populated and the other three columns blank.
• I want my Python code to open both files and for each Device_Name in the Data file, map its GDN, Device_Type, and Device_OS value from the Mapping file.

I know how to use dict when only 2 columns are present [1 is needed to be mapped] but I don't know how to accomplish this when 3 columns need to be mapped.

Following is the code using which I tried to accomplish mapping of Device_Type:

x = dict[[]]
with open["Pricing Mapping_2013-04-22.csv", "rb"] as in_file1:
    file_map = csv.reader[in_file1, delimiter=',']
    for row in file_map:
       typemap = [row[0],row[2]]
       x.append[typemap]

with open["Pricing_Updated_Cleaned.csv", "rb"] as in_file2, open["Data Scraper_GDN.csv", "wb"] as out_file:
    writer = csv.writer[out_file, delimiter=',']
    for row in csv.reader[in_file2, delimiter=',']:
         try:
              row[27] = x[row[11]]
         except KeyError:
              row[27] = ""
         writer.writerow[row]

It returns Attribute Error.

After some researching, I think I need to create a nested dict, but I don't have any idea how to do this.

smci

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asked May 2, 2013 at 8:14

4

A nested dict is a dictionary within a dictionary. A very simple thing.

>>> d = {}
>>> d['dict1'] = {}
>>> d['dict1']['innerkey'] = 'value'
>>> d['dict1']['innerkey2'] = 'value2'
>>> d
{'dict1': {'innerkey': 'value', 'innerkey2': 'value2'}}

You can also use a defaultdict from the collections package to facilitate creating nested dictionaries.

>>> import collections
>>> d = collections.defaultdict[dict]
>>> d['dict1']['innerkey'] = 'value'
>>> d  # currently a defaultdict type
defaultdict[, {'dict1': {'innerkey': 'value'}}]
>>> dict[d]  # but is exactly like a normal dictionary.
{'dict1': {'innerkey': 'value'}}

You can populate that however you want.

I would recommend in your code something like the following:

d = {}  # can use defaultdict[dict] instead

for row in file_map:
    # derive row key from something 
    # when using defaultdict, we can skip the next step creating a dictionary on row_key
    d[row_key] = {} 
    for idx, col in enumerate[row]:
        d[row_key][idx] = col

According to your comment:

may be above code is confusing the question. My problem in nutshell: I have 2 files a.csv b.csv, a.csv has 4 columns i j k l, b.csv also has these columns. i is kind of key columns for these csvs'. j k l column is empty in a.csv but populated in b.csv. I want to map values of j k l columns using 'i` as key column from b.csv to a.csv file

My suggestion would be something like this [without using defaultdict]:

a_file = "path/to/a.csv"
b_file = "path/to/b.csv"

# read from file a.csv
with open[a_file] as f:
    # skip headers
    f.next[]
    # get first colum as keys
    keys = [line.split[','][0] for line in f] 

# create empty dictionary:
d = {}

# read from file b.csv
with open[b_file] as f:
    # gather headers except first key header
    headers = f.next[].split[','][1:]
    # iterate lines
    for line in f:
        # gather the colums
        cols = line.strip[].split[',']
        # check to make sure this key should be mapped.
        if cols[0] not in keys:
            continue
        # add key to dict
        d[cols[0]] = dict[
            # inner keys are the header names, values are columns
            [headers[idx], v] for idx, v in enumerate[cols[1:]]]

Please note though, that for parsing csv files there is a csv module.

answered May 2, 2013 at 8:24

Inbar RoseInbar Rose

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1

UPDATE: For an arbitrary length of a nested dictionary, go to this answer.

Use the defaultdict function from the collections.

High performance: "if key not in dict" is very expensive when the data set is large.

Low maintenance: make the code more readable and can be easily extended.

from collections import defaultdict

target_dict = defaultdict[dict]
target_dict[key1][key2] = val

answered Dec 7, 2015 at 20:22

JunchenJunchen

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3

For arbitrary levels of nestedness:

In [2]: def nested_dict[]:
   ...:     return collections.defaultdict[nested_dict]
   ...:

In [3]: a = nested_dict[]

In [4]: a
Out[4]: defaultdict[, {}]

In [5]: a['a']['b']['c'] = 1

In [6]: a
Out[6]:
defaultdict[,
            {'a': defaultdict[,
                         {'b': defaultdict[,
                                      {'c': 1}]}]}]

answered Mar 30, 2016 at 4:18

andrewandrew

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1

It is important to remember when using defaultdict and similar nested dict modules such as nested_dict, that looking up a nonexistent key may inadvertently create a new key entry in the dict and cause a lot of havoc.

Here is a Python3 example with nested_dict module:

import nested_dict as nd
nest = nd.nested_dict[]
nest['outer1']['inner1'] = 'v11'
nest['outer1']['inner2'] = 'v12'
print['original nested dict: \n', nest]
try:
    nest['outer1']['wrong_key1']
except KeyError as e:
    print['exception missing key', e]
print['nested dict after lookup with missing key.  no exception raised:\n', nest]

# Instead, convert back to normal dict...
nest_d = nest.to_dict[nest]
try:
    print['converted to normal dict. Trying to lookup Wrong_key2']
    nest_d['outer1']['wrong_key2']
except KeyError as e:
    print['exception missing key', e]
else:
    print[' no exception raised:\n']

# ...or use dict.keys to check if key in nested dict
print['checking with dict.keys']
print[list[nest['outer1'].keys[]]]
if 'wrong_key3' in list[nest.keys[]]:

    print['found wrong_key3']
else:
    print[' did not find wrong_key3']

Output is:

original nested dict:   {"outer1": {"inner2": "v12", "inner1": "v11"}}

nested dict after lookup with missing key.  no exception raised:  
{"outer1": {"wrong_key1": {}, "inner2": "v12", "inner1": "v11"}} 

converted to normal dict. 
Trying to lookup Wrong_key2 

exception missing key 'wrong_key2' 

checking with dict.keys 

['wrong_key1', 'inner2', 'inner1']  
did not find wrong_key3

smci

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answered Mar 3, 2017 at 20:26

Gerard GGerard G

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pip install addict

from addict import Dict

mapping = Dict[]
mapping.a.b.c.d.e = 2
print[mapping]  # {'a': {'b': {'c': {'d': {'e': 2}}}}}

References:

1. easydict GitHub
2. addict GitHub

answered Apr 20, 2021 at 10:06

XerCisXerCis

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If you want to create a nested dictionary given a list [arbitrary length] for a path and perform a function on an item that may exist at the end of the path, this handy little recursive function is quite helpful:

def ensure_path[data, path, default=None, default_func=lambda x: x]:
    """
    Function:

    - Ensures a path exists within a nested dictionary

    Requires:

    - `data`:
        - Type: dict
        - What: A dictionary to check if the path exists
    - `path`:
        - Type: list of strs
        - What: The path to check

    Optional:

    - `default`:
        - Type: any
        - What: The default item to add to a path that does not yet exist
        - Default: None

    - `default_func`:
        - Type: function
        - What: A single input function that takes in the current path item [or default] and adjusts it
        - Default: `lambda x: x` # Returns the value in the dict or the default value if none was present
    """
    if len[path]>1:
        if path[0] not in data:
            data[path[0]]={}
        data[path[0]]=ensure_path[data=data[path[0]], path=path[1:], default=default, default_func=default_func]
    else:
        if path[0] not in data:
            data[path[0]]=default
        data[path[0]]=default_func[data[path[0]]]
    return data

Example:

data={'a':{'b':1}}
ensure_path[data=data, path=['a','c'], default=[1]]
print[data] #=> {'a':{'b':1, 'c':[1]}}
ensure_path[data=data, path=['a','c'], default=[1], default_func=lambda x:x+[2]]
print[data] #=> {'a': {'b': 1, 'c': [1, 2]}}

answered Oct 7, 2020 at 12:10

conmakconmak

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This thing is empty nested list from which ne will append data to empty dict

ls = [['a','a1','a2','a3'],['b','b1','b2','b3'],['c','c1','c2','c3'], 
['d','d1','d2','d3']]

this means to create four empty dict inside data_dict

data_dict = {f'dict{i}':{} for i in range[4]}
for i in range[4]:
    upd_dict = {'val' : ls[i][0], 'val1' : ls[i][1],'val2' : ls[i][2],'val3' : ls[i][3]}

    data_dict[f'dict{i}'].update[upd_dict]

print[data_dict]

The output

{'dict0': {'val': 'a', 'val1': 'a1', 'val2': 'a2', 'val3': 'a3'}, 'dict1': {'val': 'b', 'val1': 'b1', 'val2': 'b2', 'val3': 'b3'},'dict2': {'val': 'c', 'val1': 'c1', 'val2': 'c2', 'val3': 'c3'}, 'dict3': {'val': 'd', 'val1': 'd1', 'val2': 'd2', 'val3': 'd3'}}

answered Jan 11, 2021 at 8:38

Shah VipulShah Vipul

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#in jupyter
import sys
!conda install -c conda-forge --yes --prefix {sys.prefix} nested_dict 
import nested_dict as nd
d = nd.nested_dict[]

'd' can be used now to store the nested key value pairs.

answered Mar 10, 2021 at 14:44

travel_log = {
    "France" : {"cities_visited" : ["paris", "lille", "dijon"], "total_visits" : 10},
    "india" : {"cities_visited" : ["Mumbai", "delhi", "surat",], "total_visits" : 12}
}

4b0

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answered Mar 26, 2021 at 8:47

1

You can initialize an empty NestedDict and then assign values to new keys.

from ndicts.ndicts import NestedDict

nd = NestedDict[]
nd["level1", "level2", "level3"] = 0

>>> nd
NestedDict[{'level1': {'level2': {'level3': 0}}}]

ndicts is on Pypi

pip install ndicts

answered Mar 8 at 17:12

edd313edd313

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