I'm writing a program in python 3 to convert input string to integers. There is just one problem with the code. That whenever a space comes it prints -64. I've tried editing a code but it prints -64 along with the space. Any advice?
n = input("please enter the text:").lower() print(n) a = [] for i in n: a.append(ord(i)-96) if (ord(i)-96) == -64: a.append(" ") print(a) Thanks
Input: "BatMan is Awesome" Output: [2, 1, 20, 13, 1, 14, -64, ' ', 9, 19, -64, ' ', 1, 23, 5, 19, 15, 13, 5] If I understand you correctly you want to convert "abc def" to [1, 2, 3, " ", 4, 5, 6]. Currently you are first adding ord(i) - 96 to your list and then, if the character is a space, you add an additional space. You want only to add ord(i) - 96 if it is not a space.
n = input("please enter the text:").lower() print(n) a = [] for i in n: if (ord(i)-96) == -64: a.append(" ") else: a.append(ord(i)-96) print(a) 
You can check if the character is a space with str.isspace() adding ord(i)-96 if it not a space else just add the char:
n = "BatMan is Awesome".lower() print([ord(i)-96 if not i.isspace() else i for i in n]) [2, 1, 20, 13, 1, 14, ' ', 9, 19, ' ', 1, 23, 5, 19, 15, 13, 5] The equivalent code in your loop would be:
a = [] for i in n: if not i.isspace(): a.append(ord(i)-96) else: a.append(i) 
Actually you were appending ord(i)-96 to a before checking the condition if (ord(i)-96) == -64, So the correct way is to first check the condition and if it matches then append " " else simple append ord(i)-96, You can simply do the same with only one if condition and ignoring the else cause by reverting the conditional as :
n = input("please enter the text:").lower() print(n) a = [] for i in n: if (ord(i)-96) != -64: a.append(ord(i)-96) print(a) You could also do this as a one(ish)-liner:
import string n = input("please enter the text:").lower() a = [ord(c) - 96 if c not in string.whitespace else c for c in n] print(a) Using the string.whitespace list also means that other types of whitespace will be kept, which might be useful to you?
