# Dictionaries {% file src="/files/LOhlJMHWtjOVX49d1B6h" %} ![](/files/bsN7FdX93CC6h2hMk2JF) ![](/files/GAtNHH552hJ2HTvkBclB) ![](/files/qh58H5JkUx39fwjm9AMQ) {% embed url="" %} ## Dictionaries A dictionary is a collection of unordered, modifiable(mutable) paired (key: value) data type. ### Creating a Dictionary To create a dictionary we use curly brackets, {} or the *dict()* built-in function. ```python # syntax empty_dict = {} # Dictionary with data values dct = {'key1':'value1', 'key2':'value2', 'key3':'value3', 'key4':'value4'} ``` **Example:** ```python person = { 'first_name':'Asabeneh', 'last_name':'Yetayeh', 'age':250, 'country':'Finland', 'is_marred':True, 'skills':['JavaScript', 'React', 'Node', 'MongoDB', 'Python'], 'address':{ 'street':'Space street', 'zipcode':'02210' } } ``` The dictionary above shows that a value could be any data types:string, boolean, list, tuple, set or a dictionary. ### Dictionary Length It checks the number of 'key: value' pairs in the dictionary. ```python # syntax dct = {'key1':'value1', 'key2':'value2', 'key3':'value3', 'key4':'value4'} print(len(dct)) # 4 ``` **Example:** ```python person = { 'first_name':'Asabeneh', 'last_name':'Yetayeh', 'age':250, 'country':'Finland', 'is_marred':True, 'skills':['JavaScript', 'React', 'Node', 'MongoDB', 'Python'], 'address':{ 'street':'Space street', 'zipcode':'02210' } } print(len(person)) # 7 ``` ### Accessing Dictionary Items We can access Dictionary items by referring to its key name. ```python # syntax dct = {'key1':'value1', 'key2':'value2', 'key3':'value3', 'key4':'value4'} print(dct['key1']) # value1 print(dct['key4']) # value4 ``` **Example:** ```python person = { 'first_name':'Asabeneh', 'last_name':'Yetayeh', 'age':250, 'country':'Finland', 'is_marred':True, 'skills':['JavaScript', 'React', 'Node', 'MongoDB', 'Python'], 'address':{ 'street':'Space street', 'zipcode':'02210' } } print(person['first_name']) # Asabeneh print(person['country']) # Finland print(person['skills']) # ['JavaScript', 'React', 'Node', 'MongoDB', 'Python'] print(person['skills'][0]) # JavaScript print(person['address']['street']) # Space street print(person['city']) # Error ``` Accessing an item by key name raises an error if the key does not exist. To avoid this error first we have to check if a key exist or we can use the *get* method. The get method returns None, which is a NoneType object data type, if the key does not exist. ```python person = { 'first_name':'Asabeneh', 'last_name':'Yetayeh', 'age':250, 'country':'Finland', 'is_marred':True, 'skills':['JavaScript', 'React', 'Node', 'MongoDB', 'Python'], 'address':{ 'street':'Space street', 'zipcode':'02210' } } print(person.get('first_name')) # Asabeneh print(person.get('country')) # Finland print(person.get('skills')) #['HTML','CSS','JavaScript', 'React', 'Node', 'MongoDB', 'Python'] print(person.get('city')) # None ``` ### Adding Items to a Dictionary We can add new key and value pairs to a dictionary ```python # syntax dct = {'key1':'value1', 'key2':'value2', 'key3':'value3', 'key4':'value4'} dct['key5'] = 'value5' ``` **Example:** ```python person = { 'first_name':'Asabeneh', 'last_name':'Yetayeh', 'age':250, 'country':'Finland', 'is_marred':True, 'skills':['JavaScript', 'React', 'Node', 'MongoDB', 'Python'], 'address':{ 'street':'Space street', 'zipcode':'02210' } } person['job_title'] = 'Instructor' person['skills'].append('HTML') print(person) ``` ### Modifying Items in a Dictionary We can modify items in a dictionary ```python # syntax dct = {'key1':'value1', 'key2':'value2', 'key3':'value3', 'key4':'value4'} dct['key1'] = 'value-one' ``` **Example:** ```python person = { 'first_name':'Asabeneh', 'last_name':'Yetayeh', 'age':250, 'country':'Finland', 'is_marred':True, 'skills':['JavaScript', 'React', 'Node', 'MongoDB', 'Python'], 'address':{ 'street':'Space street', 'zipcode':'02210' } } person['first_name'] = 'Eyob' person['age'] = 252 ``` ### Checking Keys in a Dictionary We use the *in* operator to check if a key exist in a dictionary ```python # syntax dct = {'key1':'value1', 'key2':'value2', 'key3':'value3', 'key4':'value4'} print('key2' in dct) # True print('key5' in dct) # False ``` ### Removing Key and Value Pairs from a Dictionary * *pop(key)*: removes the item with the specified key name: * *popitem()*: removes the last item * *del*: removes an item with specified key name ```python # syntax dct = {'key1':'value1', 'key2':'value2', 'key3':'value3', 'key4':'value4'} dct.pop('key1') # removes key1 item dct = {'key1':'value1', 'key2':'value2', 'key3':'value3', 'key4':'value4'} dct.popitem() # removes the last item del dct['key2'] # removes key2 item ``` **Example:** ```python person = { 'first_name':'Asabeneh', 'last_name':'Yetayeh', 'age':250, 'country':'Finland', 'is_marred':True, 'skills':['JavaScript', 'React', 'Node', 'MongoDB', 'Python'], 'address':{ 'street':'Space street', 'zipcode':'02210' } } person.pop('first_name') # Removes the firstname item person.popitem() # Removes the address item del person['is_married'] # Removes the is_married item ``` ### Changing Dictionary to a List of Items The *items()* method changes dictionary to a list of tuples. ```python # syntax dct = {'key1':'value1', 'key2':'value2', 'key3':'value3', 'key4':'value4'} print(dct.items()) # dict_items([('key1', 'value1'), ('key2', 'value2'), ('key3', 'value3'), ('key4', 'value4')]) ``` ### Clearing a Dictionary If we don't want the items in a dictionary we can clear them using *clear()* method ```python # syntax dct = {'key1':'value1', 'key2':'value2', 'key3':'value3', 'key4':'value4'} print(dct.clear()) # None ``` ### Deleting a Dictionary If we do not use the dictionary we can delete it completely ```python # syntax dct = {'key1':'value1', 'key2':'value2', 'key3':'value3', 'key4':'value4'} del dct ``` ### Copy a Dictionary We can copy a dictionary using a *copy()* method. Using copy we can avoid mutation of the original dictionary. ```python # syntax dct = {'key1':'value1', 'key2':'value2', 'key3':'value3', 'key4':'value4'} dct_copy = dct.copy() # {'key1':'value1', 'key2':'value2', 'key3':'value3', 'key4':'value4'} ``` ### Getting Dictionary Keys as a List The *keys()* method gives us all the keys of a a dictionary as a list. ```python # syntax dct = {'key1':'value1', 'key2':'value2', 'key3':'value3', 'key4':'value4'} keys = dct.keys() print(keys) # dict_keys(['key1', 'key2', 'key3', 'key4']) ``` ### Getting Dictionary Values as a List The *values* method gives us all the values of a a dictionary as a list. ```python # syntax dct = {'key1':'value1', 'key2':'value2', 'key3':'value3', 'key4':'value4'} values = dct.values() print(values) # dict_values(['value1', 'value2', 'value3', 'value4']) ``` ```python import itertools # Given an array of coins and an array of quantities for each coin with the # same index, determine how many distinct sums can be made from non-zero # sets of the coins # Note: This problem took a little more working-through, with a failed brute- # force attempt that consisted of finding every combination of coins and # adding them, which failed when I needed to consider >50k coins # the overall number of coins was guaranteed to be less than about 1 million, # so the solution appeared to be a form of divide-and-conquer where each # possible sum for each coin was put into a set at that coin's index in the # original coins array, and then the sums were repeatedly combined into an # aggregate set until every coin possible coin value (given by the coins # array) had been added into the set of sums # problem considered "hard," asked by Google def possibleSums(coins, quantity): # sum_map = set() # start with brute force # total_arr = [coins[i] for i, q in enumerate(quantity) for l in range(q)] # for i in range(1, len(total_arr)+1): # combos = itertools.combinations(total_arr, i) # print(combos) # for combo in combos: # sum_map.add(sum(combo)) # return len(sum_map) # faster? comb_indices = [i for i in range(len(coins))] possible_sums = [] for i, c in enumerate(coins): this_set = set() for q in range(1, 1 + quantity[i]): this_set.add(c * q) possible_sums.append(this_set) # print(possible_sums) while len(possible_sums) > 1: possible_sums[0] = combine_sets(possible_sums[0], possible_sums[1]) possible_sums.pop(1) return len(possible_sums[0]) def combine_sets(set1, set2): together_set = set() for item1 in set1: for item2 in set2: together_set.add(item1 + item2) together_set.add(item1) for item2 in set2: together_set.add(item2) return together_set ``` ```python # need strings[i] = strings[j] for all patterns[i] = patterns[j] to be true - # give false if strings[i] != strings[j] and patterns[i] = patterns[j] or # strings[i] = strings[j] and patterns[j] != patterns[j] - this last condition # threw me for a bit as an edge case! Need to ensure that each string is unique # to each key, not just that each key corresponds to the given string! # from a google interview set, apparently def areFollowingPatterns(strings, patterns): pattern_to_string = {} string_to_pattern = {} for i in range(len(patterns)): # first, check condition that strings are equal for patterns[i]=patterns[j] this_pattern = patterns[i] if patterns[i] in pattern_to_string: if strings[i] != pattern_to_string[this_pattern]: return False else: pattern_to_string[this_pattern] = strings[i] # now check condition that patterns are equal for strings[i]=strings[j] # if there are more keys than values, then there is not 1:1 correspondence if len(pattern_to_string.keys()) != len(set(pattern_to_string.values())): return False return True ``` ```python # gives True if two duplicate numbers in the nums array are within k distance # (inclusive) of one another, measuring by absolute difference in index # did relatively well on this one, made a greater-than/less-than flip error on # the conditional for the true case and needed to rewrite my code to remove # keys from the dictionary without editing it while looping over it, but # otherwise went well! # problem considered medium difficulty, from Palantir def containsCloseNums(nums, k): num_dict = {} # setup keys for each number seen, then list their indices for i, item in enumerate(nums): if item in num_dict: num_dict[item].append(i) else: num_dict[item] = [i] # remove all nums that are not repeated # first make a set of keys to remove to prevent editing the dictionary size while iterating over it removals = set() for key in num_dict.keys(): if len(num_dict[key]) < 2: removals.add(key) # now remove each key from the num_dict that has fewer than two values for key in removals: num_dict.pop(key) # now check remaining numbers to see if they fall within the desired range for key in num_dict.keys(): last_ind = num_dict[key][0] for next_ind in num_dict[key][1:]: if next_ind - last_ind <= k: return True last_ind = next_ind return False ``` --- # Agent Instructions: Querying This Documentation If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question. 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