这个问题我来晚了，我在寻找解决办法的时候发现了这个问题。在

将,符号与format()一起使用可以很好地工作，但是会带来一些问题，因为不幸的是，,符号不能应用于字符串。因此，如果您从数字的文本表示开始，那么在调用format()之前，必须将它们转换为整数或浮点。如果您需要同时处理需要保留的不同精度级别的整数和浮点数，format()代码会迅速变得相当复杂。为了处理这种情况，我最终编写了自己的代码，而不是使用format()。它使用了最广泛使用的千位分隔符(,)和十进制标记(.)，但它可以很快地修改为与其他区域设置的符号一起使用，或者用于创建一个适用于所有区域设置的解决方案。在def separate_thousands_with_delimiter(num_str):

"""

Returns a modified version of "num_str" with thousand separators added.

e.g. "1000000" > "1,000,000", "1234567.1234567" > "1,234,567.1234567".

Numbers which require no thousand separators will be returned unchanged.

e.g. "123" > "123", "0.12345" > "0.12345", ".12345" > ".12345".

Signed numbers (a + or - prefix) will be returned with the sign intact.

e.g. "-12345" > "-12,345", "+123" > "+123", "-0.1234" > "-0.1234".

"""

decimal_mark = "."

thousands_delimiter = ","

sign = ""

fraction = ""

# If num_str is signed, store the sign and remove it.

if num_str[0] == "+" or num_str[0] == "-":

sign = num_str[0]

num_str = num_str[1:]

# If num_str has a decimal mark, store the fraction and remove it.

# Note that find() will return -1 if the substring is not found.

dec_mark_pos = num_str.find(decimal_mark)

if dec_mark_pos >= 0:

fraction = num_str[dec_mark_pos:]

num_str = num_str[:dec_mark_pos]

# Work backwards through num_str inserting a separator after every 3rd digit.

i = len(num_str) - 3

while i > 0:

num_str = num_str[:i] + thousands_delimiter + num_str[i:]

i -= 3

# Build and return the final string.

return sign + num_str + fraction

# Test with:

test_nums = ["1", "10", "100", "1000", "10000", "100000", "1000000",

"-1", "+10", "-100", "+1000", "-10000", "+100000", "-1000000",

"1.0", "10.0", "100.0", "1000.0", "10000.0", "100000.0",

"1000000.0", "1.123456", "10.123456", "100.123456", "1000.123456",

"10000.123456", "100000.123456", "1000000.123456", "+1.123456",

"-10.123456", "+100.123456", "-1000.123456", "+10000.123456",

"-100000.123456", "+1000000.123456", "1234567890123456789",

"1234567890123456789.1", "-1234567890123456789.1",

"1234567890123456789.123456789", "0.1", "0.12", "0.123", "0.1234",

"-0.1", "+0.12", "-0.123", "+0.1234", ".1", ".12", ".123",

".1234", "-.1", "+.12", "-.123", "+.1234"]

for num in test_nums:

print("%s > %s" % (num, separate_thousands_with_delimiter(num)))

# Beginners should note that an integer or float can be converted to a string

# very easily by simply using: str(int_or_float)

test_int = 1000000

test_int_str = str(test_int)

print("%d > %s" % (test_int, separate_thousands_with_delimiter(test_int_str)))

test_float = 1000000.1234567

test_float_str = str(test_float)

print("%f > %s" % (test_float, separate_thousands_with_delimiter(test_float_str)))

希望这有帮助。：)

我知道理论上大整数的数字可以按数千个分组,以提高可读性：

Python 3.5.2 (default, Nov 17 2016, 17:05:23)

[GCC 5.4.0 20160609] on linux

Type "help", "copyright", "credits" or "license" for more information.

>>> import locale

>>> locale.setlocale(locale.LC_ALL, 'en_US.UTF-8')

'en_US.UTF-8'

>>> locale.format('%d', 1234567890, grouping=True)

'1,234,567,890'

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

'1,234,567,890'

但是,令人惊讶的是,这不适用于每个语言环境：

>>> locale.setlocale(locale.LC_ALL, 'pl_PL.UTF-8')

'pl_PL.UTF-8'

>>> locale.format('%d', 1234567890, grouping=True)

'1234567890'

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

'1234567890'

为什么数字没有格式化？我觉得这很奇怪.我希望打印出类似于1 234 567 890的东西.

根据Format Specification Mini-Language,我们可以明确地强制执行两个可能的分隔符：逗号和下划线_.可悲的是,逗号不适合波兰语,因为它在那里被用作小数点分隔符,而像1_234_567_890这样的数字对于大多数人来说看起来很奇怪.

我们能否以某种方式强制使用不间断的空间作为千位分隔符？

解决方法:

pl_PL语言环境千位分隔符似乎是empty.我不知道这是否准确地表示波兰的常见用法,但Python正确地根据pl_PL语言环境的规则格式化您的数字.这可能是区域设置文件中的错误.

据我所知,没有选项可以手动指定千位分隔符和十进制标记字符.

标签：python,integer,formatting,locale

来源： https://codeday.me/bug/20190701/1350697.html