“切片”和“替换”作业的字符串

Question

I am stuck on a problem for my class. The pic of the problem is down below. I'm trying to convert this string "abcd efgh ijkl" to this "AbCd eFgH IjKl" using ordinal function/values to capitalize even index numbers.

This is what I got so far:

def mock(string):
    index = 0
    for i in string:
        if string.find(i, index) % 2 == 0:
            string.replace(i,(chr(ord(i) - 32)))
        index = string.find(i) + 1
    return string

print(mock("abcd efgh ijkl"))

Any help is appreciated.

Pic of the problem

Answer 1

str.replace returns string, but you aren't assigning the return value to anything. Also you need to handle spaces correctly (thanks @JacobStrauss for further simplification!)

def mock(string):
    for i in string:
        if i != ' ' and string.find(i) % 2 == 0:
            string = string.replace(i, (chr(ord(i) - 32)))
    return string

print(mock("abcd efgh ijkl"))

Prints:

AbCd eFgH IjKl

Answer 2

All credit to Andrej Kesely for providing a solution, but I just wanted to improve on it slightly. Since you are already iterating over a string, and the.find() method returns an index position, you do not need to keep track of the index. I've tested the code below and it works just as well.

def mock(string):
    for i in string:
        if i != ' ' and string.find(i) % 2 == 0:
            string = string.replace(i, (chr(ord(i) - 32)))
    return string

The above returns "AbCd eFgH IjKl"

Answer 3

use a for loop and a upper the string by index character

def capitalize(string):
    result=""
    for i in range(len(string)):
        if i%2==0:
            result+=string[i].upper()
        else:
            result+=string[i]
    return result

string="abcd efgaah ijklaaa"
print(capitalize(string))

output:

AbCd eFgAaH IjKlAaA