Align numbers at decimal point

[Alex Martelli]

Enno Middelberg <enno.middelbergREMOVE_ME at csiro.au> wrote:

> I'm sure someone asked this question before, but I can't find the 
> solution on the web or in the groups.
> 
> I want to print out columns of numbers which are aligned at the decimal
> point, eg:
> 
> 123.45
>    3.0
>   65.765486
> 
> I can't find how to do that. I use the following command as a 
> workaround, but it only works when the number of digits after the 
> decimal point is constant:
> 
> print ("%.2f" % reduced_time).rjust(9)
> 
> 
> Is there an easy solution?

Well, first of all you must have all the numbers at hand before you can
print even just one of them, of course -- you can't know how much space
you need on the left until you know what number out of them all needs
most such space.  So, say the numbers are coming from some iterable
all_numbers, just start by accumulating their formatted forms (without
yet any padding), as well as the space each occupies before the decimal
point (since we do know we'll need that information later):

formatted_numbers = []
space_on_the_left = []
for number in all_numbers:
    formatted_string = your_favourite_format % number
    formatted_numbers.append(formatted_string)
    on_the_left = formatted_string.find('.')
    if on_the_left < 0: on_the_left = len(formatted_string)
    space_on_the_left.append(on_the_left)

You might make this part a bit more compact with a couple of clever list
comprehensions, but I think it's more readable when written out this
way.  If you know every formatted string WILL include a period, you
should change the if statement into an assert (or change the call to the
find method into a call to the index method, and remove the following if
statement -- index will raise if it can't find the needle in the
haystack, while find returns -1 in that case).

After this loop you can easily find out how much space you need on the
left for the number that is longest there...:

left_total = max(space_on_the_left)

and now you do have all the info you need to do the printing:

for s, l in zip(formatted_numbers, space_on_the_left):
    padding = ' '*(left_total - l)
    print '%s%s' % (padding, s)

Again, you can of course make this a bit more compact, for example by
computing the padding string contextually to the print statement, but I
think it's clearer and more readable to give 'padding' its own name.

In many cases you can simplify things if you know left_total in advance,
for example you know that the '.' (if any) must always fall exactly on
the 21st column of each row you're printing (meaning you know that no
number will have more than 20 digits before the . -- and even if they're
all shorter you still want to leave that amount of space).  For such a
case, you need only a single loop and no extra memory...:

for number in all_numbers:
    formatted_string = your_favourite_format % number
    on_the_left = formatted_string.find('.')
    if on_the_left < 0: on_the_left = len(formatted_string)
    padding = ' '*(left_total - on_the_left)
    print '%s%s' % (padding, formatted_string)

(once again, you can trade off legibility for compactness, if you wish).


Alex