Python 格式的带空格的字符串千分隔符

You'll have to use the 'n' locale-aware number format instead, and set your locale to one that uses a space as a thousands separator. Yes, this is painful.

'n' - Number. This is the same as 'd', except that it uses the current locale setting to insert the appropriate number separator characters.

Here is bad but simple solution if you don't want to mess with locale:

'{:,}'.format(1234567890.001).replace(',', ' ')

This will set the locale environment to your default OS location:

import locale
locale.setlocale(locale.LC_ALL, '')

Then you can use:

"{0:n}".format(1234567890)

Answer of @user136036 is quite good, but unfortunately it does not take into account reality of Python bugs. Full answer could be following:

Variant A

If locale of your platform is working right, then just use locale:

import locale
locale.setlocale(locale.LC_ALL, '')
print("{:,d}".format(7123001))

Result is dependent on your locale and Python implementation working right.

But what if Python formatting according to locale is broken, e.g. Python 3.5 on Linux?

Variant B

If Python does not respect grouping=True parameter, you can use locale and a workaround (use monetary format):

locale.setlocale(locale.LC_ALL, '')
locale._override_localeconv = {'mon_thousands_sep': '.'}
print(locale.format('%.2f', 12345.678, grouping=True, monetary=True))

Above gives 12.345,68 on my platform. Setting monetary to False or omitting it - Python does not group thousands. Specifying locale._override_localeconv = {'thousands_sep': '.'} do nothing.

Variant C

If you don't have time to check what is working OK and what is broken with Python on your platform, you can just use regular string replace function (if you want to swap commas and dot to dots and comma):

print("{:,.2f}".format(7123001.345).replace(",", "X").replace(".", ",").replace("X", "."))

Replacing comma for space is trivial (point is assumed decimal separator):

print("{:,.2f}".format(7123001.345).replace(",", " ")

If you don't want to use the local, a solution that works with every size of string :

    def format_number(string):
j = len(string) % 3
substrings = [string[3 * i + j:3 * (i + 1) + j] for i in                                         range(len(string)//3)]
if j != 0:
substrings.insert(0, string[:j])
return " ".join(substrings)

Python community has it defined under PEP 378, where you can see that the main proposal includes format() function. So, by following that proposal, I suggest using the following:

format(1234567890, ',d').replace(',',' ')

Lets say you code:

print(f'{name} | Yearly salary: ${salary}.')

PRINTS -> Kate | Yearly salary: $60000.

First put a , separator:

print(f'{name} | Yearly salary: ${salary:,}.')

PRINTS -> Kate | Yearly salary: $60,000.

Then replace the comma with a space:

print(f'{name} | Yearly salary: {salary:,} USD.'.replace(',', ' '))

PRINTS -> Kate | Yearly salary: 60 000 USD.

NB: Remember USD, $ or £, kroner etc. notation should match accordingly universal standards

You can use an f-string:

number = 1234567890
f"{number:,.2f}"