str.index() 和 list.index() 有什么区别？

Question

Is it possible to use str.index() method instead of list.index() to solve this problem and what is the difference between them both after 2 times of the execution of while-loop in both cases (as shown below)?

During the code, I was working on a task that looks like this:

string = 'ABCDEFG'

1. Assign 0 to the variable var .
2. Add 1 to var .
3. Swap the letter at position var with the letter at position var + 2 .
4. if var < 4 go to Step 2.
5. End.

The main question of the task: What will be the string at Step 5?

The right output should be CDEFABG

After some time of thinking about it I came with the right solution using slist.index() :

string = 'ABCDEFG'
slist = list(string)
var = 0

while var < 4:
    #assigned a letter to a variable
    letter_1, letter_2 = slist[var], slist[var+2]
    #assigned a letter index to a variable
    l_1, l_2 = slist.index(letter_1), slist.index(letter_2)
    #swaped two letters with each other
    slist[l_1], slist[l_2] = letter_2, letter_1
    var = var+1

string = ''.join(slist)

print(slist)
print(string)

Output:

['C', 'D', 'E', 'F', 'A', 'B', 'G']
CDEFABG

But before I came out to this I tried first to implement string.index() instead of slist.index() :

#assigned a letter to a variable
letter_1, letter_2 = string[var], string[var+2]
#assigned a letter index to a variable
l_1, l_2 = string.index(letter_1), string.index(letter_2)

And after 2 times of looping while var < 4 it started to execute different outputs in contrast to the right solution with list.index() and gave the outputs like this:

['C', 'B', 'A', 'D', 'E', 'F', 'G']
CBADEFG

['C', 'D', 'A', 'B', 'E', 'F', 'G']
CDABEFG

['C', 'D', 'E', 'B', 'C', 'F', 'G']
CDEBCFG #the right one is CDEBAFG

['C', 'D', 'E', 'F', 'C', 'D', 'G']
CDEFCDG #the right one is CDEFABG

Answer 1

Is it possible to use str.index() method instead of list.index() to solve this problem and what is the difference between them both after 2 times of the execution of while-loop in both cases (as shown below)?

During the code, I was working on a task that looks like this:

string = 'ABCDEFG'

1. Assign 0 to the variable var .
2. Add 1 to var .
3. Swap the letter at position var with the letter at position var + 2 .
4. if var < 4 go to Step 2.
5. End.

The main question of the task: What will be the string at Step 5?

The right output should be CDEFABG

After some time of thinking about it I came with the right solution using slist.index() :

string = 'ABCDEFG'
slist = list(string)
var = 0

while var < 4:
    #assigned a letter to a variable
    letter_1, letter_2 = slist[var], slist[var+2]
    #assigned a letter index to a variable
    l_1, l_2 = slist.index(letter_1), slist.index(letter_2)
    #swaped two letters with each other
    slist[l_1], slist[l_2] = letter_2, letter_1
    var = var+1

string = ''.join(slist)

print(slist)
print(string)

Output:

['C', 'D', 'E', 'F', 'A', 'B', 'G']
CDEFABG

But before I came out to this I tried first to implement string.index() instead of slist.index() :

#assigned a letter to a variable
letter_1, letter_2 = string[var], string[var+2]
#assigned a letter index to a variable
l_1, l_2 = string.index(letter_1), string.index(letter_2)

And after 2 times of looping while var < 4 it started to execute different outputs in contrast to the right solution with list.index() and gave the outputs like this:

['C', 'B', 'A', 'D', 'E', 'F', 'G']
CBADEFG

['C', 'D', 'A', 'B', 'E', 'F', 'G']
CDABEFG

['C', 'D', 'E', 'B', 'C', 'F', 'G']
CDEBCFG #the right one is CDEBAFG

['C', 'D', 'E', 'F', 'C', 'D', 'G']
CDEFCDG #the right one is CDEFABG

Answer 2

Is it possible to use str.index() method instead of list.index() to solve this problem and what is the difference between them both after 2 times of the execution of while-loop in both cases (as shown below)?

During the code, I was working on a task that looks like this:

string = 'ABCDEFG'

1. Assign 0 to the variable var .
2. Add 1 to var .
3. Swap the letter at position var with the letter at position var + 2 .
4. if var < 4 go to Step 2.
5. End.

The main question of the task: What will be the string at Step 5?

The right output should be CDEFABG

After some time of thinking about it I came with the right solution using slist.index() :

string = 'ABCDEFG'
slist = list(string)
var = 0

while var < 4:
    #assigned a letter to a variable
    letter_1, letter_2 = slist[var], slist[var+2]
    #assigned a letter index to a variable
    l_1, l_2 = slist.index(letter_1), slist.index(letter_2)
    #swaped two letters with each other
    slist[l_1], slist[l_2] = letter_2, letter_1
    var = var+1

string = ''.join(slist)

print(slist)
print(string)

Output:

['C', 'D', 'E', 'F', 'A', 'B', 'G']
CDEFABG

But before I came out to this I tried first to implement string.index() instead of slist.index() :

#assigned a letter to a variable
letter_1, letter_2 = string[var], string[var+2]
#assigned a letter index to a variable
l_1, l_2 = string.index(letter_1), string.index(letter_2)

And after 2 times of looping while var < 4 it started to execute different outputs in contrast to the right solution with list.index() and gave the outputs like this:

['C', 'B', 'A', 'D', 'E', 'F', 'G']
CBADEFG

['C', 'D', 'A', 'B', 'E', 'F', 'G']
CDABEFG

['C', 'D', 'E', 'B', 'C', 'F', 'G']
CDEBCFG #the right one is CDEBAFG

['C', 'D', 'E', 'F', 'C', 'D', 'G']
CDEFCDG #the right one is CDEFABG