将数字格式化为字符串

Question

How do you format a number as a string so that it takes a number of spaces in front of it? I want the shorter number 5 to have enough spaces in front of it so that the spaces plus the 5 have the same length as 52500. The procedure below works, but is there a built in way to do this?

a = str(52500)
b = str(5)
lengthDiff = len(a) - len(b)
formatted = '%s/%s' % (' '*lengthDiff + b, a)
# formatted looks like:'     5/52500'

Answer 1

Format operator :

>>> "%10d" % 5
'         5'
>>> 

Using * spec, the field length can be an argument:

>>> "%*d" % (10,5)
'         5'
>>> 

Answer 2

You can just use the %*d formatter to give a width. int(math.ceil(math.log(x, 10))) will give you the number of digits. The * modifier consumes a number, that number is an integer that means how many spaces to space by. So by doing '%*d' % (width, num)` you can specify the width AND render the number without any further python string manipulation.

Here is a solution using math.log to ascertain the length of the 'outof' number.

import math
num = 5
outof = 52500
formatted = '%*d/%d' % (int(math.ceil(math.log(outof, 10))), num, outof)

Another solution involves casting the outof number as a string and using len(), you can do that if you prefer:

num = 5
outof = 52500
formatted = '%*d/%d' % (len(str(outof)), num, outof)

Answer 3

'%*s/%s' % (len(str(a)), b, a)

Answer 4

See String Formatting Operations :

s = '%5i' % (5,)

You still have to dynamically build your formatting string by including the maximum length:

fmt = '%%%ii' % (len('52500'),)
s = fmt % (5,)

Answer 5

Not sure exactly what you're after, but this looks close:

>>> n = 50
>>> print "%5d" % n
   50

If you want to be more dynamic, use something like rjust :

>>> big_number = 52500
>>> n = 50
>>> print ("%d" % n).rjust(len(str(52500)))
   50

Or even:

>>> n = 50
>>> width = str(len(str(52500)))
>>> ('%' + width + 'd') % n
'   50'