+ 3
I have question that is we will ask user to input number like 2355597 and it will give answer like this 2466618(odd to even)
In this problem we have to convert odd digits to even except 9 to 1 and even remains same.I want answer in python
8 odpowiedzi
+ 2
Shelly Kapoor
Say the input is "9"
inp.index(9) #means position 9 in the list (index) which is 0 since 9 is the first and only element in the list.
So now inp.index(9) = 0
inp[inp.index(9)] = 1 #means inp[0] = 1
So this means get the first element in the list and make it 1 "only works is this is 9"
+ 12
Hello !!
This place is for asking programming related doubts. No one will solve the questions that you'll post here. If you want help then you have to show your attempt first or tell exactly where you are struct in the problem.
+ 10
This may work :
a=input("enter the value") # take normal input
list1 = [] # declare a blank list
res = [int(x) for x in str(a)] #store individual integer as a list
for i in res: # run for loop for each element
if i%2!=0: # check if the specific element is odd or not
if i==9: #if it is odd and equal to 9 store 1 at that point
i=1
else: # if it is odd and equal not to 9 make that number to even
i=i+1
list1.append(i) # store each element in a separate list with each iteration
print(list1) # print the list
+ 5
print(int(''.join(['1'if c=='9'else str(int(c)+1)if int(c)%2 else c for c in str(2355597)])))
+ 4
@keval
a=int(input("enter the value"))
x=list(str(a))
for i in range x:
if i%2!=0:
i=i+1
elif i==9:
i=i-1
else :
i
print(x)
+ 4
thnx both Mo hani and keval
+ 3
@Mo hani can u please explain the inp[inp.index(i)] method
+ 1
Remember to show us your own attempt first next time.
This is one way to do it:
inp = list(input())
for i in inp:
if int(i)%2!=0:
if int(i) == 9:
inp[inp.index(i)] = 1
else:
inp[inp.index(i)] = int(i)+1
This is what the code does:
#gets input and makes it list
#loops for elements in list
#checks if the element is odd
#checks for 9, replaces it with 1
#other than 9, adds one to it