How do you create nested dict in Python?
Question:
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.
Answers:
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(<type 'dict'>, {'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.
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
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(<function __main__.nested_dict>, {})
In [5]: a['a']['b']['c'] = 1
In [6]: a
Out[6]:
defaultdict(<function __main__.nested_dict>,
{'a': defaultdict(<function __main__.nested_dict>,
{'b': defaultdict(<function __main__.nested_dict>,
{'c': 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
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]}}
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’}}
#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.
travel_log = {
"France" : {"cities_visited" : ["paris", "lille", "dijon"], "total_visits" : 10},
"india" : {"cities_visited" : ["Mumbai", "delhi", "surat",], "total_visits" : 12}
}
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:
You can initialize an empty NestedDict and then assign values to new keys.
from ndicts import NestedDict
nd = NestedDict()
nd["level1", "level2", "level3"] = 0
>>> nd
NestedDict({'level1': {'level2': {'level3': 0}}})
ndicts is on Pypi
pip install ndicts
You can create simple class inherited from dict and implement only __missing__ method:
class NestedDict(dict):
def __missing__(self, x):
self[x] = NestedDict()
return self[x]
d = NestedDict()
d[1][2] = 3
print(d)
# {1: {2: 3}}
I have 2 CSV files: ‘Data’ and ‘Mapping’:
- ‘Mapping’ file has 4 columns:
Device_Name,GDN,Device_Type, andDevice_OS. All four columns are populated. - ‘Data’ file has these same columns, with
Device_Namecolumn populated and the other three columns blank. - I want my Python code to open both files and for each
Device_Namein the Data file, map itsGDN,Device_Type, andDevice_OSvalue 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.
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(<type 'dict'>, {'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.
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
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(<function __main__.nested_dict>, {})
In [5]: a['a']['b']['c'] = 1
In [6]: a
Out[6]:
defaultdict(<function __main__.nested_dict>,
{'a': defaultdict(<function __main__.nested_dict>,
{'b': defaultdict(<function __main__.nested_dict>,
{'c': 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
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]}}
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’}}
#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.
travel_log = {
"France" : {"cities_visited" : ["paris", "lille", "dijon"], "total_visits" : 10},
"india" : {"cities_visited" : ["Mumbai", "delhi", "surat",], "total_visits" : 12}
}
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:
You can initialize an empty NestedDict and then assign values to new keys.
from ndicts import NestedDict
nd = NestedDict()
nd["level1", "level2", "level3"] = 0
>>> nd
NestedDict({'level1': {'level2': {'level3': 0}}})
ndicts is on Pypi
pip install ndicts
You can create simple class inherited from dict and implement only __missing__ method:
class NestedDict(dict):
def __missing__(self, x):
self[x] = NestedDict()
return self[x]
d = NestedDict()
d[1][2] = 3
print(d)
# {1: {2: 3}}