Question

在我的场景（python 2.7）中，我有：

str(var)

其中var是一个变量，有时需要数千个分隔符，例如1,500，但正如您所看到的那样，已转换为字符串（用于连接目的）。

我希望能够将该变量打印为具有数千个分隔符的字符串。

我已经阅读了为数字添加格式的解决方案，例如：

>>> '{:20,.2}'.format(f)
'18,446,744,073,709,551,616.00'

来自https://stackoverflow.com/a/1823189/1063287

但这似乎仅适用于数字，而不适用于已转换为字符串的数字。

谢谢你。

对于更具体的上下文，

编辑，这是我的实施方案：

print 'the cost = $' + str(var1) + '\n'
print 'the cost = $' + str(var2) + '\n'

我有几个'vars'。

Answer 1

请勿使用str(var)和连接，即{/ 1}} 。取决于.format()的类型选择以下之一：

var

取决于您拥有的号码类型。

'{:,}'.format(var) # format an integer '{:,.2f}'.format(var) # format a decimal or float

Answer 2

我在寻找同一事物的解决方案时发现这个问题的时间很晚。

对,使用format()表示法可以很好地工作但会带来一些问题，因为遗憾的是,表示法不能应用于字符串。因此，如果您从数字的文本表示开始，则必须在调用format()之前将它们转换为整数或浮点数。如果您需要处理需要保留的不同精度级别的整数和浮点数，format()代码将变得非常复杂。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()。它使用最广泛使用的千位分隔符（,）和十进制标记（.）但显然可以非常快速地修改它以使用其他语言环境的符号或用于创建适用于所有语言环境。

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)))

希望这会有所帮助。：）

如何将数千个分隔符添加到已在python中转换为字符串的数字？

2 个答案: