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_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 itsGDN
,Device_Type
, andDevice_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
1,77319 silver badges18 bronze badges
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
1512 silver badges4 bronze badges
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:
- easydict GitHub
- addict GitHub
answered Apr 20, 2021 at 10:06
XerCisXerCis
6095 silver badges4 bronze badges
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
9928 silver badges13 bronze badges
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
4484 silver badges9 bronze badges
#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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