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.
410 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] = colAccording 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.
1UPDATE: 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] = valFor 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_key3pip install addictfrom addict import Dict
mapping = Dict()
mapping.a.b.c.d.e = 2
print(mapping) # {'a': {'b': {'c': {'d': {'e': 2}}}}}References:
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 dataExample:
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}
}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
