String format printing with python3: print from unpacked array *some* of the time

Question

In my question a few minutes ago, I asked about how to print using python's str.format printing, when the strings are stored in an array.

Then answer was clearly unpack the list, like this:

# note that I had to play with the whitespace because the {} text is 2 characters, while its replacement is always one

hex_string = r'''
            _____
           /     \
          /       \
    ,----(    {}    )----.
   /      \       /      \
  /   {}    \_____/   {}    \
  \        /     \        /
   \      /       \      /
    )----(    {}    )----(
   /      \       /      \
  /        \_____/        \
  \   {}    /     \   {}    /
   \      /       \      /
    `----(    {}    )----'
          \       /
           \_____/
'''

letters = list('1234567')

print(hex_string.format(*letters))

But if I always want the center hexagon to have printed inside the first item in the array: letters[0], how can I mix unpacking the array with keeping the 4th printed string from the first array element?

I'm open to other printing types, like f-strings.

For example:

print(hex_string.format(letters[3], letters[1], letters[2], letters[0], letters[4], letters[5], letters[6]))

So that my output actually looks like this:

            _____
           /     \
          /       \
    ,----(    4    )----.
   /      \       /      \
  /   2    \_____/   3    \
  \        /     \        /
   \      /       \      /
    )----(    1    )----(
   /      \       /      \
  /        \_____/        \
  \   5    /     \   6    /
   \      /       \      /
    `----(    7    )----'
          \       /
           \_____/

Answer 1

You can try something like this with .format():

a = '123'
print('{2}, {0}, {1}'.format(*a))

which would print 3, 1, 2

With this approach, your initial hex_string will "self document" where the letters from your array will go exactly.

Answer 2

If you know the required order beforehand:

letters = list('1234567')
reordering = [4,2,3,1,5,4,7]

You can apply this to the letters:

new_letters = [letters[index-1] for index in reordering]

Output:

['4', '2', '3', '1', '5', '4', '7']

Now you can create your formatted string:

print(hex_string.format(*new_letters))

Answer 3

I think the simplest approach for it to change your letters to list('4231567'), which allows you to mix unpacking as well as avoids the need to move around elements in the array

hex_string = r'''
            _____
           /     \
          /       \
    ,----(    {}    )----.
   /      \       /      \
  /   {}    \_____/   {}    \
  \        /     \        /
   \      /       \      /
    )----(    {}    )----(
   /      \       /      \
  /        \_____/        \
  \   {}    /     \   {}    /
   \      /       \      /
    `----(    {}    )----'
          \       /
           \_____/
'''

letters = list('4231547')

print(hex_string.format(*letters))

The output will be

            _____
           /     \
          /       \
    ,----(    4    )----.
   /      \       /      \
  /   2    \_____/   3    \
  \        /     \        /
   \      /       \      /
    )----(    1    )----(
   /      \       /      \
  /        \_____/        \
  \   5    /     \   6    /
   \      /       \      /
    `----(    7    )----'
          \       /
           \_____/

Or expanding on the idea suggested below, use str.format with indexes defined to unpack the original string, but assigning different items at different positions

hex_string = r'''
            _____
           /     \
          /       \
    ,----(    {3}    )----.
   /      \       /      \
  /   {1}    \_____/   {2}    \
  \        /     \        /
   \      /       \      /
    )----(    {0}    )----(
   /      \       /      \
  /        \_____/        \
  \   {4}    /     \   {5}    /
   \      /       \      /
    `----(    {6}    )----'
          \       /
           \_____/
'''

letters = list('1234567')

print(hex_string.format(*letters))

This also give you a better reference point in the sense that you know that {0} will contain the first item of the list, which is 1

The output will be same as above