you have an ipv4 network that has two IPv6-only hosts that need to be able to communicate with each other. You want to implement
1 answer:
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Answer:
Explanation:
The python code for the question is attached in the image below.
The objective here is to write a code in python with a function called has_duplicates that takes a string parameter and returns True if the string has any repeated characters. Otherwise, it should return False.
SEE BELOW FOR THE CODE.
alphabet = "abcdefghijklmnopqrstuvwxyz"
test_dups = ["zzz","dog","bookkeeper","subdermatoglyphic","subdermatoglyphics"]
test_miss = ["zzz","subdermatoglyphic","the quick brown fox jumps over the lazy dog"]
def histogram(s):
d = dict()
for c in s:
if c not in d:
d[c]=1
else:
d[c] += 1
return d
def has_duplicates():
for string in test_dups:
dictionary=histogram(string)
duplicates=False
for ch in dictionary.keys():
if dictionary[ch] > 1:
duplicates=True
break
if duplicates==True:
print(string,"has duplicates")
else:
print(string,"has no duplicates")
def missing_letters(string):
answer=""
for ch in alphabet:
if ch not in string:
answer=answer+ch
return answer
print("__________________________________________")
print(" Calling has_duplicates() function")
print("__________________________________________")
has_duplicates()
print("\n______________________________________________")
print("Calling missing_letters() function in for loop")
print("______________________________________________")
for i in test_miss:
answer=missing_letters(i)
if len(answer)>=1:
print(i,"is missing letters",answer)
else:
print(i,"uses all the letters")
Answer:
Here is the Python function:
def group_by(s, fn): #method definition
group = {} #dictionary
for e in s: #for each element from s
key = fn(e) #set key to fn(e)
if key in group: #if key is in dictionary group
group[key].append(e) #append e to the group[key]
else: #if key is not in group
group[key] = [e] #set group[key] to [e]
return group #returns dictionary group
Explanation:
To check the working of the above function call the function by passing a sequence and a function and use print function to display the results produced by the function:
print(group_by([12, 23, 14, 45], lambda p: p // 10))
print(group_by(range(-3, 4), lambda x: x * x))
function group_by takes in a sequence s and a function fn as parameters. It then creates an empty dictionary group. The values of the dictionary are lists of elements from s. Each element e in a list is constructed such that fn(e) is the same for all elements in that list. Finally, the key for each value is fn(e). The function returns a dictionary. The screenshot of the program along with its output is attached.
