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SO is already filled with quite a number of questions regarding to Python's locale settings for formatting a number with corresponding thousands and decimal separators, the best of which I believe is found here.

However, I have been trying to build upon this solution by extending its functionality to percentages as well, to no avail:

import locale

locale.setlocale(locale.LC_ALL, 'pt_BR.UTF-8')

def pbr(number, isfloat=True):
    if isfloat:
        return locale.format_string('%.2f', number, grouping=True)
    else:
        return locale.format_string('%.2%', number, grouping=True) # Not working

Obviously '%.2%' is not a valid formatter, but I don't know what should be done so that

In [1]: pbr(81.23421039401, isfloat=False)
Out[1]: '8.123,42%'

Also, is this implementation of the locale solution (including the format_string function syntax which seems to be some leftover of the Python 2 era) still optimal, considering Python 3.8?

manoelpqueiroz
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    As the documentation states, [`format_string()`](https://docs.python.org/3/library/locale.html#locale.format_string) "follows the conventions of the `%` operator" which means you can use `'%.2f%%'` as the format string. It's probably "left-over" for backwards compatibility. The `%` notation itself is a carry-over from the C `printf()` function. Regardless, note that the result will be `81,23%`, not what you have shown in your question. – martineau Sep 27 '20 at 19:26
  • Are you sure that '%.2%' is a valid specifier? Also I am confused by your function logic, in the example you passed a float into the function but specified that isfloat=False. If I am correct, you are trying to convert a float in American notation: 81.23421039401, to European convention: 81,23421039401 and round off the decimals with '%.2f' to: 81,23, and also make it a string: '81,23'? And then if it is not a float return a string with some other specifier '%.2%'? – Sebastian Peterlin Sep 27 '20 at 19:31
  • @martineau, thanks for the info, I know nothing about old notation, so it always bugs me when I have to deal with it in Python 3. As pointed out by Anab's answer, the lack of proper percentage notation can be easily circumvented by altering the implementation of the function. – manoelpqueiroz Sep 27 '20 at 19:46

1 Answers1

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Looking at the Python documentation and Python reference for string formatting, it seems that you are trying to use the % operator from the "new formatting style" in the "old formatting style", where it meant something else. This explains why your method fails.

One way to achieve what you want with the locale is to do the multiplication and add the % sign yourself:

import locale

locale.setlocale(locale.LC_ALL, 'pt_BR.UTF-8')

def pbr(number, isfloat=True):
    if isfloat:
        return locale.format_string('%.2f', number, grouping=True)
    else:
        return locale.format_string('%.2f%%', number*100, grouping=True)

I cannot say whether this is still optimal in Python 3.8, I feel there should be a way to simply use string.format with the new style but I haven't yet found the way to make it use the locale.

Anab
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