5

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，但如您所见，它已转换为字符串（用于连接目的）。

我希望能够将该变量打印为带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，不适用于已转换为字符串的数字。

谢谢你。

编辑： 对于更具体的上下文，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'。

4

2 回答 2

8

不要使用和连接，str(var)这.format()就是. 根据选择的类型var之一：

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

取决于您拥有的号码类型。

>>> var = 12345678.123
>>> '{:,}'.format(int(var))  # ignore the `.123` part
'12,345,678'
>>> '{:,.2f}'.format(var)
'12,345,678.12'
于 2013-03-11T18:07:24.083 回答
0

我在为同一件事寻找解决方案时发现的这个问题来得太晚了。

使用,with 表示法format()可以很好地工作，但会带来一些问题，因为不幸的是，该,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，那么您必须在调用之前将它们转换为整数或浮点数format()。format()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，那么代码很快就会变得非常复杂。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,（.

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)))

希望这可以帮助。:)

于 2016-03-12T12:01:27.137 回答