I need to inverse a dictionary so that each old value will now be a key and the old keys will be the new values.
The trick is that there could be multiple values that are the same in the old dictionary so I need each value in the new dictionary to be a list, and if there were identical values in the old dictionary then they both will be in the list of the value of the new dictionary.
for example:
the dictionary {"python" : 1, "is" : 1, "cool" : 2}
would end up as: {1 : ["python", "is"], 2 : ["cool"]}
this is what I tried:
def inverse_dict(my_dict):
new_dict = {}
values_list = list(my_dict.values())
new_dict = new_dict.fromkeys(values_list)
for key in new_dict:
new_dict[key] = []
for old_key in my_dict:
new_dict[my_dict[old_key]] = list(new_dict[my_dict[old_key]]).append(old_key)
return new_dict
Would greatly appreciate any help with my approach (and better approaches to the problem) as I am very new to Python, thanks!
You can use dict.setdefault check if a key exists in the dictionary and if not, create new value (in this case empty list []):
d = {"python" : 1, "is" : 1, "cool" : 2}
reversed_d = {}
for k, v in d.items():
reversed_d.setdefault(v, []).append(k)
print(reversed_d)
Prints:
{1: ['python', 'is'], 2: ['cool']}
This can be more explicitly rewritten as:
d = {"python" : 1, "is" : 1, "cool" : 2}
reversed_d = {}
for k, v in d.items():
if v not in reversed_d:
reversed_d[v] = [k]
else:
reversed_d[v].append(k)
print(reversed_d)
You can use a defaultdict to avoid the pre-fill step
from collections import defaultdict
def inverse_dict(my_dict: dict):
new_dict = defaultdict(list)
for k, v in my_dict.items():
new_dict[v].append(k)
return new_dict
Though I prefer #azro's answer with the default dict, another solution is doing it with dictionary and list comprehensions.
It looks like this:
{value : [key for key in my_dict if my_dict[key] == value] for value in set(my_dict.values())}
What it does is runs over the values of the dictionary without duplicates - set(my_dict.values()).
It builds every value as a key (because it's on the left side of the ":").
And its value is a list of the keys that point to that value - [key for key in my_dict if my_dict[key] == value].
Related
I have a dictionary that looks like this:
d = {key1 : {(key2,key3) : value}, ...}
so it is a dictionary of dictionaries and in the inside dict the keys are tuples.
I would like to get a triple nested dict:
{key1 : {key2 : {key3 : value}, ...}
I know how to do it with 2 loops and a condition:
new_d = {}
for key1, inside_dict in d.items():
new_d[key1] = {}
for (key2,key3), value in inside_dict.items():
if key2 in new_d[key1].keys():
new_d[key1][key2][key3] = value
else:
new_d[key1][key2] = {key3 : value}
Edit: key2 values are not guaranteed to be unique. This is why I added the condition
It feels very unpythonic to me.
Is there a faster and/or shorter way to do this?
You could use the common trick for nesting dicts arbitrarily, using collections.defaultdict:
from collections import defaultdict
tree = lambda: defaultdict(tree)
new_d = tree()
for k1, dct in d.items():
for (k2, k3), val in dct.items():
new_d[k1][k2][k3] = val
If I understand the problem correctly, for this case you can wrap all the looping up in a dict comprehension. This assumes that your data is unique:
data = {"key1": {("key2", "key3"): "val"}}
{k: {keys[0]: {keys[1]: val}} for k,v in data.items() for keys, val in v.items()}
Example:
last_names = ['Bakir','Jose','Jose','Pierce']
university = ['Davis', 'Stanford', 'George Town', 'Berkeley']
Desire the Following
resulting_dictionary = {'Bakir':'Davis', 'Jose': ['Stanford', 'George Town'], 'Pierce':'Berkeley'}
I've tried the following
dictionary = {key:value for key, value in zip(last_names, university)}
But obtained the following:
{'Bakir': 'Davis', 'Jose': 'George Town', 'Pierce': 'Berkeley'}
Due to duplicate key value in last name list.
Thoughts?
Use defaultdict
from collections import defaultdict
d = defaultdict(list)
for k, v in zip(last_names, university):
d[k].append(v)
You can use the dict.setdefault method to initialize new keys with sub-lists:
dictionary = {}
for k, v in zip(last_names, university):
dictionary.setdefault(k, []).append(v)
You need to set the value to a list, and check whether the key already exists and append to it.
dictionary = {}
for key, value in zip(last_names, university):
if key in dictionary:
dictionary[key].append(value)
else:
dictionary[key] = [value]
This will make all the values lists. It's generally easier if you're consistent about your data format, not mixing strings and lists. So the result will be:
{'Bakir':['Davis'], 'Jose': ['Stanford', 'George Town']], 'Pierce':['Berkeley']}
Assuming, as the question seems to imply, that a list is not desired if only one value is present for a key, a simple flattening post-step can be used to achieve the specified result.
from collections import defaultdict
d = defaultdict(list)
for k, v in zip(last_names, university):
d[k].append(v)
# flatten single entry lists back to a string
d = { k: v[0] if len(v) == 1 else v for k, v in d.items() }
Try this:
my_dict = {}
for key, value in zip(last_names, university):
if key in my_dict:
old_value = my_dict[key]
if isinstance (old_value, str):
my_dict[key] = [old_value, value]
else:
old_value.add (value)
else:
my_dict[key] = value
I have two dictionaries as follows:
mydictionary_1 = {1:'apple',2:'banana'}
mydictionary_2 = {1:50,2:30}
The resultant dictionary should be such that it takes the key as the value of first dictionary.
Result_dictionary= {'apple':50, 'banana':30}
You can use a dictionary comprehension using the values of the first dictionary as the keys of the resulting dictionary. This assumes all keys of the first are present in the second dict
{v: dict2[k] for k, v in dict1.items()}
you can also add a check for the presence of the keys in the second dictionary
{v: dictionary_2[k] for k, v in dictionary_1.items() if k in dictionary_2}
Loop through one of the dictionaries and check if the value for a key in mydictionary_1 exists in mydictionary_2.
You can achieve this using python's dictionary comprehension -
Result_dictionary = { v:mydictionary_2[k] for k,v in mydictionary_1.iteritems() if k in mydictionary_2.keys()}
To see how this list comprehension is working you can even use general for loop to loop through each key, value pair in mydictionary_1
for key,value in mydictionary_1.iteritems():
if key in mydictionary_2.keys():
Result_dictionary[value]=mydictionary_2[key]
Dictionary comprehension is an ideal solution for this one, as previously mentioned. Here is a for loop example:
def combine_dictionaries(dict1, dict2):
result_dictionary = {}
for key in dict1.keys():
result_dictionary[dict1[key]] = dict2[key]
return result_dictionary
combine_dictionaries({1:'apple', 2:'banana'}, {1:50, 2:30})
>>>{'apple': 50, 'banana': 30}
This assumes all values of the dict1 are present in the dict2.
def dict_cross_match(dict1, dict2):
new_dict = {}
for item in dict1.keys():
if item in dict2.keys():
new_dict[dict1[item]] = dict2[item]
return new_dict
mydictionary_1 = {1:'apple',2:'banana'}
mydictionary_2 = {1:50,2:30}
print(dict_cross_match(mydictionary_1, mydictionary_2))
I've been struggling on something for the day,
I have a dictionnary under the format
dict = {a:[element1, element2, element3], b:[element4, element5, element6]...}
I want a new dictionnary under the form
newdict = {a:element1, b:element4...}
Meaning only keeping the first element of the lists contained for each value.
You can use a dictionary comprehension:
{k: v[0] for k, v in d.items()}
# {'a': 'element1', 'b': 'element4'}
Hopefully this helps.
I like to check if the dictionary has a key before overwriting a keys value.
dict = {a:[element1, element2, element3], b:[element4, element5, element6]}
Python 2
newDict = {}
for k, v in dict.iteritems():
if k not in newDict:
# add the first list value to the newDict's key
newDick[k] = v[0]
Python 3
newDict = {}
for k, v in dict.items():
if k not in newDict:
# add the first list value to the newDict's key
newDick[k] = v[0]
I need to do a not "natural" operation on a dictionary so i wondering what is the best pythonic way to do this.
I need to simplify a dictionary by removing all the keys on it with the same value (keys are differents, values are the same)
For example:
Input:
dict = {key1 : [1,2,3], key2: [1,2,6], key3: [1,2,3]}
expected output:
{key1 : [1,2,3], key2:[1,2,6]}
I dont care about which key is delete (on the example: key1 or key3)
Exchange keys and values; duplicated key-value pairs will be removed as a side effect (because dictionary does not allow duplicated keys). Exchange keys and values again.
>>> d = {'key1': [1,2,3], 'key2': [1,2,6], 'key3': [1,2,3]}
>>> d2 = {tuple(v): k for k, v in d.items()} # exchange keys, values
>>> d = {v: list(k) for k, v in d2.items()} # exchange again
>>> d
{'key2': [1, 2, 6], 'key1': [1, 2, 3]}
NOTE: tuple(v) was used because list is not hashable; cannot be used as key directly.
BTW, don't use dict as a variable name. It will shadow builtin function/type dict.
This solution deletes the keys with same values without creating a new dictionary.
seen = set()
for key in mydict.keys():
value = tuple(mydict[key])
if value in seen:
del mydict[key]
else:
seen.add(value)
I think you can do it this way also. But I don't say as there seems to be more efficient ways. It is in-line.
for i in dictionary.keys():
if dictionary.values().count(dictionary[i]) > 1:
del dictionary[i]
You can iterate over your dict items and use a set to check what we have seen so far, deleting a key if we have already seen the value:
d = {"key1" : [1,2,3], "key2": [1,2,6], "key3": [1,2,3]}
seen = set()
for k, v in d.items(): # list(items) for python3
temp = tuple(v)
if temp in seen:
del d[k]
seen.add(temp)
print(d)
{'key1': [1, 2, 3], 'key2': [1, 2, 6]}
This will be more efficient that using creating a dict and reversing the values as you only have to cast to tuple once not from a tuple back to a list.
this worked for me:
seen = set()
for key in mydict.copy():
value = tuple(mydict[key])
if value in seen:
del mydict[key]
else:
seen.add(value)