I am trying to develop a code and half of it is done, I am grouping my diction. I want to create a function to go back to the a_dict from b_dict
I want to print it as this;
Expected output;
a_dict: {'A': 1, 'B': 2, 'C': 3, 'D': 1, 'E': 2, 'F': 3} # Original Diction
Grouped dict: {1: ['A', 'D'], 2: ['B', 'E'], 3: ['C', 'F']} # Grouped Diction
Expected dict: {'A': 1, 'D': 1, 'B': 2, 'E': 2, 'C': 3, 'F': 3} # Expected second output with go_back function. Current output can not do this
Code:
a_dict = {'A': 1, 'B': 2, 'C': 3, 'D': 1, 'E': 2, 'F': 3}
print('a_dict: ', a_dict)
def fun_dict(a_dict):
b_dict = {}
for i, v in a_dict.items():
b_dict[v] = [i] if v not in b_dict.keys() else b_dict[v] + [i]
return b_dict
def go_back(b_dict):
#
# Need a function to convert b_dict to c_dict to go back as the expected output
#
b_dict = fun_dict(a_dict)
print('Grouped dict: ', b_dict)
c_dict = fun_dict(b_dict)
print('Went to the original dict: ', c_dict)
The go_back you want could be like this:
def go_back(b_dict):
r = {}
for k, vv in b_dict.items():
for v in vv:
r[v] = k
return r
Result:
Went to the original dict: {'A': 1, 'D': 1, 'B': 2, 'E': 2, 'C': 3, 'F': 3}
Here is a proposal:
def go_back(b_dict):
return {e: k for k, v in b_dict.items() for e in v}
Related
I want to change just Keys (not Values) in a Dictionary in Python. Is there any way to do that?
You can pop the value of the old key and reassign:
d = {'A': 1, 'B': 2, 'C': 3}
d['b'] = d.pop('B')
print(d)
# {'A': 1, 'C': 3, 'b': 2}
Note that this won't maintain the order of the keys (python 3.6+). The renamed key will be instead at the end.
maintaining order
If order is important you need to create a new dictionary
d = {'A': 1, 'B': 2, 'C': 3}
rename = {'B': 'b', 'A': 'a'}
d = {rename.get(k, k): v for k,v in d.items()}
print(d)
# {'a': 1, 'b': 2, 'C': 3}
in place modification while maintaining order
If you want to modify the dictionary in place (i.e. not creating a new object), you need to pop and reinsert all keys in order:
d = {'A': 1, 'B': 2, 'C': 3}
rename = {'B': 'b', 'A': 'a'}
keys = list(d)
for k in keys:
d[rename.get(k, k)] = d.pop(k)
print(d)
{'a': 1, 'b': 2, 'C': 3}
I want to merge list of dictionaries in python. The number of dictionaries contained inside the list is not fixed and the nested dictionaries are being merged on both same and different keys. The dictionaries within the list do not contain nested dictionary. The values from same keys can be stored in a list.
My code is:
list_of_dict = [{'a': 1, 'b': 2, 'c': 3}, {'a': 3, 'b': 5}, {'k': 5, 'j': 5}, {'a': 3, 'k': 5, 'd': 4}, {'a': 3} ...... ]
output = {}
for i in list_of_dict:
for k,v in i.items():
if k in output:
output[k].append(v)
else:
output[k] = [v]
Is there a shorter and faster way of implementing this?
I am actually trying to implement the most fast way of doing this because the list of dictionary is very large and then there are lots of rows with such data.
One way using collections.defaultdict:
from collections import defaultdict
res = defaultdict(list)
for d in list_of_dict:
for k, v in d.items():
res[k].append(v)
Output:
defaultdict(list,
{'a': [1, 3, 3, 3],
'b': [2, 5],
'c': [3],
'k': [5, 5],
'j': [5],
'd': [4]})
items() is a dictionary method, but list_of_dict is a list. You need a nested loop so you can loop over the dictionaries and then loop over the items of each dictionary.
ou = {}
for d in list_of_dict:
for key, value in d.items():
output.setdefault(key, []).append(value)
another shorten version can be,
list_of_dict = [{'a': 1, 'b': 2, 'c': 3}, {'a': 3, 'b': 5}, {'k': 5, 'j': 5}, {'a': 3, 'k': 5, 'd': 4}, {'a': 3}]
output = {
k: [d[k] for d in list_of_dict if k in d]
for k in set().union(*list_of_dict)
}
print(output)
{'d': [4], 'k': [5, 5], 'a': [1, 3, 3, 3], 'j': [5], 'c': [3], 'b': [2, 5]}
Python 3.9+ you can use the merge operator for this.
def merge_dicts(dicts):
result = dict()
for _dict in dicts:
result |= _dict
return result
One of the shortest way would be to
prepare a list/set of all the keys from all the dictionaries
and call that key on all the dictionary in the list.
list_of_dict = [{'a': 1, 'b': 2, 'c': 3}, {'a': 3, 'b': 5}, {'k': 5, 'j': 5}, {'a': 3, 'k': 5, 'd': 4}, {'a': 3}]
# prepare a list/set of all the keys from all the dictionaries
# method 1: use sum
all_keys = sum([[a for a in x.keys()] for x in list_of_dict], [])
# method 2: use itertools
import itertools
all_keys = list(itertools.chain.from_iterable(list_of_dict))
# method 3: use union of the set
all_keys = set().union(*list_of_dict)
print(all_keys)
# ['a', 'b', 'c', 'a', 'b', 'k', 'j', 'a', 'k', 'd', 'a']
# convert the list to set to remove duplicates
all_keys = set(all_keys)
print(all_keys)
# {'a', 'k', 'c', 'd', 'b', 'j'}
# now merge the dictionary
merged = {k: [d.get(k) for d in list_of_dict if k in d] for k in all_keys}
print(merged)
# {'a': [1, 3, 3, 3], 'k': [5, 5], 'c': [3], 'd': [4], 'b': [2, 5], 'j': [5]}
In short:
all_keys = set().union(*list_of_dict)
merged = {k: [d.get(k) for d in list_of_dict if k in d] for k in all_keys}
print(merged)
# {'a': [1, 3, 3, 3], 'k': [5, 5], 'c': [3], 'd': [4], 'b': [2, 5], 'j': [5]}
I want to merge list of dictionaries in python. The number of dictionaries contained inside the list is not fixed and the nested dictionaries are being merged on both same and different keys. The dictionaries within the list do not contain nested dictionary. The values from same keys can be stored in a list.
My code is:
list_of_dict = [{'a': 1, 'b': 2, 'c': 3}, {'a': 3, 'b': 5}, {'k': 5, 'j': 5}, {'a': 3, 'k': 5, 'd': 4}, {'a': 3} ...... ]
output = {}
for i in list_of_dict:
for k,v in i.items():
if k in output:
output[k].append(v)
else:
output[k] = [v]
Is there a shorter and faster way of implementing this?
I am actually trying to implement the most fast way of doing this because the list of dictionary is very large and then there are lots of rows with such data.
One way using collections.defaultdict:
from collections import defaultdict
res = defaultdict(list)
for d in list_of_dict:
for k, v in d.items():
res[k].append(v)
Output:
defaultdict(list,
{'a': [1, 3, 3, 3],
'b': [2, 5],
'c': [3],
'k': [5, 5],
'j': [5],
'd': [4]})
items() is a dictionary method, but list_of_dict is a list. You need a nested loop so you can loop over the dictionaries and then loop over the items of each dictionary.
ou = {}
for d in list_of_dict:
for key, value in d.items():
output.setdefault(key, []).append(value)
another shorten version can be,
list_of_dict = [{'a': 1, 'b': 2, 'c': 3}, {'a': 3, 'b': 5}, {'k': 5, 'j': 5}, {'a': 3, 'k': 5, 'd': 4}, {'a': 3}]
output = {
k: [d[k] for d in list_of_dict if k in d]
for k in set().union(*list_of_dict)
}
print(output)
{'d': [4], 'k': [5, 5], 'a': [1, 3, 3, 3], 'j': [5], 'c': [3], 'b': [2, 5]}
Python 3.9+ you can use the merge operator for this.
def merge_dicts(dicts):
result = dict()
for _dict in dicts:
result |= _dict
return result
One of the shortest way would be to
prepare a list/set of all the keys from all the dictionaries
and call that key on all the dictionary in the list.
list_of_dict = [{'a': 1, 'b': 2, 'c': 3}, {'a': 3, 'b': 5}, {'k': 5, 'j': 5}, {'a': 3, 'k': 5, 'd': 4}, {'a': 3}]
# prepare a list/set of all the keys from all the dictionaries
# method 1: use sum
all_keys = sum([[a for a in x.keys()] for x in list_of_dict], [])
# method 2: use itertools
import itertools
all_keys = list(itertools.chain.from_iterable(list_of_dict))
# method 3: use union of the set
all_keys = set().union(*list_of_dict)
print(all_keys)
# ['a', 'b', 'c', 'a', 'b', 'k', 'j', 'a', 'k', 'd', 'a']
# convert the list to set to remove duplicates
all_keys = set(all_keys)
print(all_keys)
# {'a', 'k', 'c', 'd', 'b', 'j'}
# now merge the dictionary
merged = {k: [d.get(k) for d in list_of_dict if k in d] for k in all_keys}
print(merged)
# {'a': [1, 3, 3, 3], 'k': [5, 5], 'c': [3], 'd': [4], 'b': [2, 5], 'j': [5]}
In short:
all_keys = set().union(*list_of_dict)
merged = {k: [d.get(k) for d in list_of_dict if k in d] for k in all_keys}
print(merged)
# {'a': [1, 3, 3, 3], 'k': [5, 5], 'c': [3], 'd': [4], 'b': [2, 5], 'j': [5]}
I am a beginner in python trying to create a function that filters through my nested dictionary through by asking multiple values in a dictionary like
filtered_options = {'a': 5, 'b': "Cloth'}
For my dictionary
my_dict = {1.0:{'a': 1, 'b': "Food', 'c': 500, 'd': 'Yams'},
2.0:{'a': 5, 'v': "Cloth', 'c': 210, 'd': 'Linen'}}
If I input my dictionary in the filter function with such options I should get something that looks like
filtered_dict(my_dict, filtered_options = {'a': 5, 'b': "Cloth'})
which outputs the 2nd key and other keys with the same filtered options in my dictionary.
This should do what you want.
def dict_matches(d, filters):
return all(k in d and d[k] == v for k, v in filters.items())
def filter_dict(d, filters=None):
filters = filters or {}
return {k: v for k, v in d.items() if dict_matches(v, filters)}
Here's what happens when you test it:
>>> filters = {'a': 5, 'b': 'Cloth'}
>>> my_dict = {
... 1.0: {'a': 1, 'b': 'Food', 'c': 500, 'd': 'Yams'},
... 2.0: {'a': 5, 'b': 'Cloth', 'c': 210, 'd': 'Linen'}
... }
>>> filter_dict(my_dict, filters)
{2.0: {'b': 'Cloth', 'a': 5, 'd': 'Linen', 'c': 210}}
You can do this :
import operator
from functools import reduce
def multi_level_indexing(nested_dict, key_list):
"""Multi level index a nested dictionary, nested_dict through a list of keys in dictionaries, key_list
"""
return reduce(operator.getitem, key_list, nested_dict)
def filtered_dict(my_dict, filtered_options):
return {k : v for k, v in my_dict.items() if all(multi_level_indexing(my_dict, [k,f_k]) == f_v for f_k, f_v in filtered_options.items())}
So that:
my_dict = {1.0:{'a': 1, 'b': 'Food', 'c': 500, 'd': 'Yams'},
2.0:{'a': 5, 'b': 'Cloth', 'c': 210, 'd': 'Linen'}}
will give you:
print(filtered_dict(my_dict, {'a': 5, 'b': 'Cloth'}))
# prints {2.0: {'a': 5, 'b': 'Cloth', 'c': 210, 'd': 'Linen'}}
Lets say I have two dictionaries:
a = {'a': 1, 'b': 2, 'c': 3}
b = {'b': 2, 'c': 3, 'd': 4, 'e': 5}
What's the most pythonic way to find the non mutual items between the two of them such that for a and b I would get:
{'a': 1, 'd': 4, 'e': 5}
I had thought:
{key: b[key] for key in b if not a.get(key)}
but that only goes one way (b items not in a) and
a_only = {key: a[key] for key in a if not b.get(key)}.items()
b_only = {key: b[key] for key in b if not a.get(key)}.items()
dict(a_only + b_only)
seams very messy. Any other solutions?
>>> dict(set(a.iteritems()) ^ set(b.iteritems()))
{'a': 1, 'e': 5, 'd': 4}
Try with the symetric difference of set() :
out = {}
for key in set(a.keys()) ^ set(b.keys()):
out[key] = a.get(key, b.get(key))
diff = {key: a[key] for key in a if key not in b}
diff.update((key,b[key]) for key in b if key not in a)
just a bit cheaper version of what you have.
>>> a = {'a': 1, 'b': 2, 'c': 3}
>>> b = {'b': 2, 'c': 3, 'd': 4, 'e': 5}
>>> keys = set(a.keys()).symmetric_difference(set(b.keys()))
>>> result = {}
>>> for k in keys: result[k] = a.get(k, b.get(k))
...
>>> result
{'a': 1, 'e': 5, 'd': 4}
Whether this is less messy than your version is debatable, but at least it doesn't re-implement symmetric_difference.