Given 2 dictionaries, assign values to value list elements mapping from dictionary 2.
Input : test_dict = {‘Gfg’ : [3, 6], ‘best’ :[9]}, look_dict = {3 : [1, 5], 6 : “Best”, 9 : 12} Output : {‘Gfg’: {3: [1, 5], 6: ‘Best’}, ‘best’: {9: 12}} Explanation : 3 is replaced by key 3 and value [1, 5] and so on. Input : test_dict = {‘Gfg’ : [3, 6]}, look_dict = {3 : [1, 5], 6 : “Best”} Output : {‘Gfg’: {3: [1, 5], 6: ‘Best’}} Explanation : 3 is replaced by key 3 and value [1, 5] and so on.
Method #1 : Using nested dictionary comprehension
In this, we use inner dictionary comprehension to map values elements to dict 2, and outer dict is used to extract all keys from dictionary 1.
Python3
# Python3 code to demonstrate working of# Assign values to Values List# Using nested dictionary comprehension# initializing dictionarytest_dict = {'Gfg' : [3, 6], 'is' : [4, 2], 'best' :[9]}# printing original dictionaryprint("The original dictionary is : " + str(test_dict))# initializing lookup dictlook_dict = {3 : [1, 5], 6 : "Best", 4 : 10, 9 : 12, 2 : "CS"}# nested dictionaries to sought solutionres = {idx: {ikey: look_dict[ikey] for ikey in test_dict[idx]} for idx in test_dict}# printing resultprint("The mapped dictionary : " + str(res)) |
The original dictionary is : {'Gfg': [3, 6], 'is': [4, 2], 'best': [9]}
The mapped dictionary : {'Gfg': {3: [1, 5], 6: 'Best'}, 'is': {4: 10, 2: 'CS'}, 'best': {9: 12}}
Time Complexity: O(N*M), where N is the number of key-value pairs in the test_dict and M is number of values in any value list.
Auxiliary space: O(N + K), where N is the number of key-value pairs in the test_dict and K is the number of keys in the look_dict.
Method #2 : Using items() + dictionary comprehension
Similar to above method, another one-liner, difference being that items() is used for element access.
Python3
# Python3 code to demonstrate working of# Assign values to Values List# Using items() + dictionary comprehension# initializing dictionarytest_dict = {'Gfg': [3, 6], 'is': [4, 2], 'best': [9]}# printing original dictionaryprint("The original dictionary is : " + str(test_dict))# initializing lookup dictlook_dict = {3: [1, 5], 6: "Best", 4: 10, 9: 12, 2: "CS"}# nested dictionaries to sought solution# items() used to access key-val pairsres = {key: {ikey: ival for (ikey, ival) in look_dict.items() if ikey in val} for (key, val) in test_dict.items()}# printing resultprint("The mapped dictionary : " + str(res)) |
The original dictionary is : {'Gfg': [3, 6], 'is': [4, 2], 'best': [9]}
The mapped dictionary : {'Gfg': {3: [1, 5], 6: 'Best'}, 'is': {4: 10, 2: 'CS'}, 'best': {9: 12}}
Method#3: Using for loop
Approach
this approach uses dictionary comprehension to create a new dictionary with values assigned based on the look_dict.
Algorithm
1. Create a set of all the values in the look_dict.
2. Iterate over each key-value pair in the test_dict.
3. Create a new dictionary with the same keys as the test_dict and an empty dictionary as the value.
Iterate over each value in the value list of the current key.
4. If the value is present in the set of values from the look_dict, then add a key-value pair to the value dictionary with the key as the value and the value as the corresponding value from the look_dict.
5. Return the new dictionary.
Python3
def assign_values(test_dict, look_dict): values_set = set(look_dict.keys()) new_dict = {key: {} for key in test_dict} for key, values in test_dict.items(): for value in values: if value in values_set: new_dict[key][value] = look_dict[value] return new_dicttest_dict = {'Gfg' : [3, 6], 'is' : [4, 2], 'best' :[9]}look_dict = {3 : [1, 5], 6 : "Best", 4 : 10, 9 : 12, 2 : "CS"}print(assign_values(test_dict, look_dict)) |
{'Gfg': {3: [1, 5], 6: 'Best'}, 'is': {4: 10, 2: 'CS'}, 'best': {9: 12}}
Time Complexity: O(NM), where N is the number of key-value pairs in the test_dict and M is the maximum number of values in any value list.
Space Complexity: O(N + K), where N is the number of key-value pairs in the test_dict and K is the number of keys in the look_dict.
Method #4: Using defaultdict and for loop
- Import the defaultdict module from the collections library.
- Create a dictionary called test_dict that contains the original dictionary data.
- Create a dictionary called look_dict that contains the lookup data.
- Create an empty defaultdict called res to store the result.
- Iterate through each key-value pair in test_dict using the items() method.
- For each key-value pair, iterate through the values using a for loop.
- For each value, use it as a key to look up the corresponding value in look_dict.
- Add a new key-value pair to the corresponding dictionary in res, where the key is the current value and the value is the value from look_dict that was looked up in the previous step.
- Repeat steps 6-8 for all values in the current key-value pair.
- Repeat steps 5-9 for all key-value pairs in test_dict.
- Print the resulting res
Python3
from collections import defaultdict# initializing dictionarytest_dict = {'Gfg' : [3, 6], 'is' : [4, 2], 'best' :[9]}# initializing lookup dictlook_dict = {3 : [1, 5], 6 : "Best", 4 : 10, 9 : 12, 2 : "CS"}# creating a defaultdict to store the resultres = defaultdict(dict)# iterating through the test_dict and its valuesfor key, values in test_dict.items(): # iterating through the values for value in values: # using the value to lookup the corresponding values in the look_dict res[key][value] = look_dict[value]# printing resultprint("The mapped dictionary : " + str(res)) |
The mapped dictionary : defaultdict(<class 'dict'>, {'Gfg': {3: [1, 5], 6: 'Best'}, 'is': {4: 10, 2: 'CS'}, 'best': {9: 12}})
Time complexity: O(NM), where N is the number of keys in the test_dict and M is the average number of values per key.
Auxiliary space: O(NM), since we are storing the result in a dictionary.
