如何在Python 3中不对字符串本身进行计数来对字符串进行对齐？

Question

Just started learning python. And it looks to me that rjust() method is counting string that I'm trying to justify in justifying length.

Let me clarify a little bit, the book I'm using to study python had this type of small program:

order = "First\nSecond\nThird"
print("The order is: ", order)

Whose output is:

The order is: First  
Second  
Third

And I thought what could I change to make output as ( desired output ),

The order is: First   
              Second   
              Third  

Well there may be very easy way to solve this but bear in mind I just started learning python. I tried formatting to no avail. After learning about list I came up with this solution,

order = "First\nSecond\nThird"

listed_order = order.split("\n")
length_of_list = len(listed_order)

print_line = "The order is: "
length = len(print_line)

print(print_line + listed_order[0])
for i in range(1, length_of_list):
    print(listed_order[i].rjust(length))

But the output is,

The order is: First
        Second
         Third

That means rjust() method is counting Second and Third to make justified length of 14, while I wanted them to start from 14. Second has 8 spaces in front of it to make justified length of 14, while Third has 9 spaces.

So, what can I do to get my desired output? If there is another way to solve this let me know.

Answer 1

What you describe is left -justified, but with an additional offset (also known as indentation ).

Instead of

print(listed_order[i].rjust(length))

simply do

print(' ' * length + listed_order[i])

If there is nothing to the right of listed_order[i] , you don't need to use ljust either.

You might want to take a look at textwrap.indent , which lets you indent whole blocks of text without splitting the lines manually.

Answer 2

You don't have to do such things: length_of_list = len(listed_order) . You can run your loop without checking length of the list:

for i in listed_order:
    print(i.rjust(length))

EDIT: As pointed in the comments, this would also print first element which we want to skip. You can do it by telling python to process from second element by adding [1:] , which means from element of index 1 (which is second after index 0) untill end) , so:

for i in listed_order[1:]:
    print(i.ljust(length))

Now to the core.

What you need is the number of whitespaces. You can get it by running this:

number_of_whitespaces = len("The order is: ")

And then print it out before each line:

print(' ' * number_of_whitespaces, end='')

Full solution here:

#!/usr/bin/env python3

order = "First\nSecond\nThird"
print("The order is: ", order)

orders = order.split()
whitespaces = len('The order is: ')

print('The order is: {}'.format(orders[0]))
for o in orders[1:]:
    print(' ' * whitespaces, end='')
    print(o)

EDIT2: Added missing colon, Changed rjust to ljust EDIT3: Added full solution