Question:
1. Find the Binary numbers from the given list of Numbers .
2. sort on ascending order based on the no of 1's present
3. if no of 1's are same , then give preference to its decimal nos.
4. After sorting, convert to Decimal numbers and Display the output .
Enter total no of number : 3
Enter numbers:[ 0 ]: 9
Enter numbers:[ 1 ]: 6
Enter numbers:[ 2 ]: 7
Numbers in list are : [9, 6, 7]
9 1001 count ones are : 2
6 110 count ones are : 2
7 111 count ones are : 3
I am facing challenges how to sort binary nos based on counting no of 1's .
Did you read my link? We need to see your code
I can't read your link...
codes are :
#Binary conversion
cnt_num = int(input("Enter total no of number : "))
num = []
for i in range(cnt_num):
print("Enter numbers:[",i,"]: ",end = '')
num1 = int(input())
num.append(num1)
print("Numbers in list are : ",num)
for j in range(len(num)):
bin_num.append(bin(num[j])[2:])
bits = bin_num[j].count('1')
print(num[j],bin_num[j],' count ones are : ',bits)#Need to write logic to sort the binary numbers based on counting no of 1's
Read
https://docs.python.org/3/howto/sorting.html
and especially the part about key argument.
you need to specify a key in such a way that it will compare first the number of 1s and if equal - teh decimal representation of the number.