I want to implement a HashMap in Python. I want to ask a user for an input. depending on his input I am retrieving some information from the HashMap. If the user enters a key of the HashMap, I would like to retrieve the corresponding value.
How do I implement this functionality in Python?
HashMap<String,String> streetno=new HashMap<String,String>();
streetno.put("1", "Sachin Tendulkar");
streetno.put("2", "Dravid");
streetno.put("3","Sehwag");
streetno.put("4","Laxman");
streetno.put("5","Kohli")
Answers:
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Method 1
Python dictionary is a built-in type that supports key-value pairs.
streetno = {"1": "Sachin Tendulkar", "2": "Dravid", "3": "Sehwag", "4": "Laxman", "5": "Kohli"}
as well as using the dict keyword:
streetno = dict({"1": "Sachin Tendulkar", "2": "Dravid"})
or:
streetno = {}
streetno["1"] = "Sachin Tendulkar"
Method 2
All you wanted (at the time the question was originally asked) was a hint. Here’s a hint: In Python, you can use dictionaries.
Method 3
It’s built-in for Python. See dictionaries.
Based on your example:
streetno = {"1": "Sachine Tendulkar",
"2": "Dravid",
"3": "Sehwag",
"4": "Laxman",
"5": "Kohli" }
You could then access it like so:
sachine = streetno["1"]
Also worth mentioning: it can use any non-mutable data type as a key. That is, it can use a tuple, boolean, or string as a key.
Method 4
streetno = { 1 : "Sachin Tendulkar",
2 : "Dravid",
3 : "Sehwag",
4 : "Laxman",
5 : "Kohli" }
And to retrieve values:
name = streetno.get(3, "default value")
Or
name = streetno[3]
That’s using number as keys, put quotes around the numbers to use strings as keys.
Method 5
Hash maps are built-in in Python, they’re called dictionaries:
streetno = {} #create a dictionary called streetno
streetno["1"] = "Sachin Tendulkar" #assign value to key "1"
Usage:
"1" in streetno #check if key "1" is in streetno streetno["1"] #get the value from key "1"
See the documentation for more information, e.g. built-in methods and so on. They’re great, and very common in Python programs (unsurprisingly).
Method 6
Here is the implementation of the Hash Map using python
For the simplicity hash map is of a fixed size 16.
This can be changed easily.
Rehashing is out of scope of this code.
class Node:
def __init__(self, key, value):
self.key = key
self.value = value
self.next = None
class HashMap:
def __init__(self):
self.store = [None for _ in range(16)]
def get(self, key):
index = hash(key) & 15
if self.store[index] is None:
return None
n = self.store[index]
while True:
if n.key == key:
return n.value
else:
if n.next:
n = n.next
else:
return None
def put(self, key, value):
nd = Node(key, value)
index = hash(key) & 15
n = self.store[index]
if n is None:
self.store[index] = nd
else:
if n.key == key:
n.value = value
else:
while n.next:
if n.key == key:
n.value = value
return
else:
n = n.next
n.next = nd
hm = HashMap()
hm.put("1", "sachin")
hm.put("2", "sehwag")
hm.put("3", "ganguly")
hm.put("4", "srinath")
hm.put("5", "kumble")
hm.put("6", "dhoni")
hm.put("7", "kohli")
hm.put("8", "pandya")
hm.put("9", "rohit")
hm.put("10", "dhawan")
hm.put("11", "shastri")
hm.put("12", "manjarekar")
hm.put("13", "gupta")
hm.put("14", "agarkar")
hm.put("15", "nehra")
hm.put("16", "gawaskar")
hm.put("17", "vengsarkar")
print(hm.get("1"))
print(hm.get("2"))
print(hm.get("3"))
print(hm.get("4"))
print(hm.get("5"))
print(hm.get("6"))
print(hm.get("7"))
print(hm.get("8"))
print(hm.get("9"))
print(hm.get("10"))
print(hm.get("11"))
print(hm.get("12"))
print(hm.get("13"))
print(hm.get("14"))
print(hm.get("15"))
print(hm.get("16"))
print(hm.get("17"))
Output:
sachin sehwag ganguly srinath kumble dhoni kohli pandya rohit dhawan shastri manjarekar gupta agarkar nehra gawaskar vengsarkar
Method 7
class HashMap:
def __init__(self):
self.size = 64
self.map = [None] * self.size
def _get_hash(self, key):
hash = 0
for char in str(key):
hash += ord(char)
return hash % self.size
def add(self, key, value):
key_hash = self._get_hash(key)
key_value = [key, value]
if self.map[key_hash] is None:
self.map[key_hash] = list([key_value])
return True
else:
for pair in self.map[key_hash]:
if pair[0] == key:
pair[1] = value
return True
else:
self.map[key_hash].append(list([key_value]))
return True
def get(self, key):
key_hash = self._get_hash(key)
if self.map[key_hash] is not None:
for pair in self.map[key_hash]:
if pair[0] == key:
return pair[1]
return None
def delete(self, key):
key_hash = self._get_hash(key)
if self.map[key_hash] is None :
return False
for i in range(0, len(self.map[key_hash])):
if self.map[key_hash][i][0] == key:
self.map[key_hash].pop(i)
return True
def print(self):
print('---Phonebook---')
for item in self.map:
if item is not None:
print(str(item))
h = HashMap()
Method 8
Python Counter is also a good option in this case:
from collections import Counter counter = Counter(["Sachin Tendulkar", "Sachin Tendulkar", "other things"]) print(counter)
This returns a dict with the count of each element in the list:
Counter({'Sachin Tendulkar': 2, 'other things': 1})
Method 9
In python you would use a dictionary.
It is a very important type in python and often used.
You can create one easily by
name = {}
Dictionaries have many methods:
# add entries:
>>> name['first'] = 'John'
>>> name['second'] = 'Doe'
>>> name
{'first': 'John', 'second': 'Doe'}
# you can store all objects and datatypes as value in a dictionary
# as key you can use all objects and datatypes that are hashable
>>> name['list'] = ['list', 'inside', 'dict']
>>> name[1] = 1
>>> name
{'first': 'John', 'second': 'Doe', 1: 1, 'list': ['list', 'inside', 'dict']}
You can not influence the order of a dict.
Method 10
A dictionary in Python is the best way to implement this. We can create the following dictionary using the given <key,value> pairs:
d = {"1": "Sachin Tendulkar", "2": "Dravid", "3": "Sehwag", "4": "Laxman", "5": "Kohli"}
To extract the value of a particular key, we can directly use d[key]:
name = d["1"] # The value of name would be "Sachin Tendulkar" here
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