如何在 Python 中将货币字符串转换为浮点数？

Question

我有一些表示具有特定货币格式的数字的字符串，例如：

money="$6,150,593.22"

我想将此字符串转换为数字

6150593.22

实现此目的的最佳方法是什么？

Answer 1

试试这个：

from re import sub
from decimal import Decimal

money = '$6,150,593.22'
value = Decimal(sub(r'[^\d.]', '', money))

这有一些优点，因为它使用 Decimal 而不是 float （这更适合表示货币），并且它还通过不硬编码特定的货币符号来避免任何区域设置问题。

Answer 2

如果您的语言环境设置正确，您可以使用 locale.atof ，但您仍然需要手动去掉“$”：

>>> import locale
>>> locale.setlocale(locale.LC_ALL, 'en_US.UTF8')
'en_US.UTF8'
>>> money = "$6,150,593.22"
>>> locale.atof(money.strip("$"))
6150593.2199999997

Answer 3

我发现 babel 包对于解决问题非常有帮助

• <一href="https://stackoverflow.com/questions/8421922/how-do-i-convert-a-currency-string-to-a-floating-point-number-in-python#comment32554847_8422055">本地化 <一href="https://stackoverflow.com/questions/8421922/how-do-i-convert-a-currency-string-to-a-floating-point-number-in-python/8422012#comment10405860_8422012">解析< /a>
• 以及需要 更改locale 全局

它使得解析本地化版本中的数字变得容易：

>>> babel.numbers.parse_decimal('1,024.64', locale='en')                                                                                                                           
Decimal('1024.64')
>>> babel.numbers.parse_decimal('1.024,64', locale='de')
Decimal('1024.64')
>>>

您可以使用 babel.numbers.get_currency_symbol('USD')去除前/后缀而不对其进行硬编码。

哈，
dtk

Answer 4

对于无需对货币位置或符号进行硬编码的解决方案：

raw_price = "17,30 €"
import locale
locale.setlocale(locale.LC_ALL, 'fr_FR.UTF8')
conv = locale.localeconv()
raw_numbers = raw_price.strip(conv['currency_symbol'])
amount = locale.atof(raw_numbers)

Answer 5

扩展以在括号中包含负数：

In [1]: import locale, string

In [2]: from decimal import Decimal

In [3]: n = ['$1,234.56','-$1,234.56','($1,234.56)', '$ -1,234.56']

In [4]: tbl = string.maketrans('(','-')

In [5]: %timeit -n10000 [locale.atof( x.translate(tbl, '$)')) for x in n]
10000 loops, best of 3: 31.9 æs per loop

In [6]: %timeit -n10000 [Decimal( x.translate(tbl, '$,)')) for x in n]
10000 loops, best of 3: 21 æs per loop

In [7]: %timeit -n10000 [float( x.replace('(','-').translate(None, '$,)')) for x in n]
10000 loops, best of 3: 3.49 æs per loop

In [8]: %timeit -n10000 [float( x.translate(tbl, '$,)')) for x in n]
10000 loops, best of 3: 2.19 æs per loop

请注意，必须从 float()/Decimal() 中删除逗号。带翻译表的replace() 或translate() 可用于将开头( 转换为-，翻译稍快一些。float() 最快10-15 倍，但缺乏精度，可能会出现区域设置问题。Decimal( ）具有精度，并且比 locale.atof() 快 50%，但也存在区域设置问题。 locale.atof() 是最慢的，但最通用。

编辑：新的 str.translate API（字符）映射到 None 从 str.translate 函数移至翻译表）

In [1]: import locale, string
        from decimal import Decimal

        locale.setlocale(locale.LC_ALL, '')

        n = ['$1,234.56','-$1,234.56','($1,234.56)', '$ -1,234.56']

In [2]: tbl = str.maketrans('(', '-', '$)')
        %timeit -n10000 [locale.atof( x.translate(tbl)) for x in n]
18 µs ± 296 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [3]: tbl2 = str.maketrans('(', '-', '$,)')
        %timeit -n10000 [Decimal( x.translate(tbl2)) for x in n]
3.77 µs ± 50.8 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [4]: %timeit -n10000 [float( x.translate(tbl2)) for x in n]
3.13 µs ± 66.3 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [5]: tbl3 = str.maketrans('', '', '$,)')
        %timeit -n10000 [float( x.replace('(','-').translate(tbl3)) for x in n]
3.51 µs ± 84.8 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

Answer 6

扩展 @Andrew Clark answer

对于不同于 en_US 的其他语言环境：

>>> import re
>>> import locale
>>> locale.setlocale(locale.LC_NUMERIC, 'pt_BR.UTF8') # this is for atof()
'pt_BR.UTF8'
>>> locale.setlocale(locale.LC_MONETARY, 'pt_BR.UTF8') # this is for currency()
'pt_BR.UTF8'
>>> curr = locale.currency(6150593.22, grouping = True)
>>> curr
'R$ 6.150.593,22'
>>> re.sub('[^(\d,.)]', '', curr)
'15,00'
>>> locale.atof(re.sub('[^(\d,.)]', '', curr))
6150593.22
>>> 6150593.22 == locale.atof(re.sub('[^(\d,.)]', '', locale.currency(6150593.22, grouping = True)))
True

强制提醒：适当的Python 的货币类型是 Decimal，而不是浮点数。

Answer 7

几年前我做了这个函数来解决同样的问题。

def money(number):
    number = number.strip('
)
    try:
        [num,dec]=number.rsplit('.')
        dec = int(dec)
        aside = str(dec)
        x = int('1'+'0'*len(aside))
        price = float(dec)/x
        num = num.replace(',','')
        num = int(num)
        price = num + price
    except:
        price = int(number)
    return price

Answer 8

该函数已将土耳其价格格式转换为十进制数。

money = '1.234,75'
def make_decimal(string):
    result = 0
    if string:
        [num, dec] = string.rsplit(',')
        result += int(num.replace('.', ''))
        result += (int(dec) / 100)
    return result
print(make_decimal(money))
1234.75

Answer 9

我发现的最简单的方法，无需对货币检测进行硬编码，还可以使用 Decimal 类型，从而避免 float 类型的问题：

>>> from decimal import Decimal
>>> money="$6,150,593.22"
>>> amount = Decimal("".join(d for d in money if d.isdigit() or d == '.'))
>>> amount
Decimal('6150593.22')

credit: https://www.reddit.com/r/learnpython/comments/2248mp/how_to_format_currency_without_currency_sign/cgjd1o4?utm_source=share&utm_medium=web2x

Answer 10

我将提供我的解决方案，希望它能帮助那些不仅面临 , 问题，而且还面临 . 问题的人。

def process_currency_adaptive(currency_string: str, decimal_sep_char: str) -> float:
    """
    Converts the currency string to common float format:
        Format: 
            ######.### 
        Example: 
            6150593.22
    """
    # Get rid of currency symbol
    currency_symbols = ["$", "€", "£", "₺"]
    
    # Replace any occurrence of currency symbol with empty string
    for symbol in currency_symbols:
        currency_string = currency_string.replace(symbol, "")
    
    
    if decimal_sep_char == ",":
        triple_sep_char = "."
    elif decimal_sep_char == ".":
        triple_sep_char = ","
    else:
        raise ValueError("Invalid decimal separator character: {}".format(decimal_sep_char))

    # Get rid of the triple separator
    currency_string = currency_string.replace(triple_sep_char, "")
    
    # There should be only one decimal_sep_char.
    if currency_string.count(decimal_sep_char) != 1:
        print("Error: Invalid currency format with value: {}".format(currency_string))
        raise ValueError
    
    return float(currency_string.replace(decimal_sep_char, "."))

# test process_currency
print(process_currency_adaptive("942,695", decimal_sep_char=","))  # 942.695
print(process_currency_adaptive("$6,150,593.22", decimal_sep_char="."))  # 6150593.22