我想在Python中实现一个HashMap。我想让用户输入。根据他的输入，我正在从HashMap中检索一些信息。如果用户输入HashMap的一个键，我想检索相应的值。
我如何在Python中实现这个功能?
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")
当前回答
Python Counter在这种情况下也是一个很好的选择:
from collections import Counter
counter = Counter(["Sachin Tendulkar", "Sachin Tendulkar", "other things"])
print(counter)
这将返回一个包含列表中每个元素计数的dict:
Counter({'Sachin Tendulkar': 2, 'other things': 1})
其他回答
Python中的字典是实现这一点的最佳方式。我们可以使用给定的<key,value>对创建以下字典:
d = {"1": "Sachin Tendulkar", "2": "Dravid", "3": "Sehwag", "4": "Laxman", "5": "Kohli"}
要提取特定键的值，我们可以直接使用d[key]:
name = d["1"] # The value of name would be "Sachin Tendulkar" here
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()
下面是使用python实现哈希映射 为了简单起见，哈希映射的大小是固定的16。 这很容易改变。 重散列超出了这段代码的范围。
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"))
输出:
sachin
sehwag
ganguly
srinath
kumble
dhoni
kohli
pandya
rohit
dhawan
shastri
manjarekar
gupta
agarkar
nehra
gawaskar
vengsarkar
你所需要的(在最初问问题的时候)只是一个暗示。这里有一个提示:在Python中，您可以使用字典。
Python Counter在这种情况下也是一个很好的选择:
from collections import Counter
counter = Counter(["Sachin Tendulkar", "Sachin Tendulkar", "other things"])
print(counter)
这将返回一个包含列表中每个元素计数的dict:
Counter({'Sachin Tendulkar': 2, 'other things': 1})
