我有一些表示特定货币格式的数字字符串,例如:
money="$6,150,593.22"
我想将这个字符串转换为数字
6150593.22
如何最好地实现这个目标?
我有一些表示特定货币格式的数字字符串,例如:
money="$6,150,593.22"
我想将这个字符串转换为数字
6150593.22
如何最好地实现这个目标?
试试这个:
from re import sub
from decimal import Decimal
money = '$6,150,593.22'
value = Decimal(sub(r'[^\d.]', '', money))
这种方法有其优点,因为它使用Decimal
而不是float
(在表示货币时更好),并且通过不硬编码特定的货币符号也避免了任何区域设置问题。
value = Decimal(sub(r'[^\d\-.]', '', money))
翻译为:value = Decimal(sub(r'[^\d\-.]', '', money))
。 - Dave如果您的本地设置正确,可以使用locale.atof
,但仍需要手动去掉“$”:
>>> import locale
>>> locale.setlocale(locale.LC_ALL, 'en_US.UTF8')
'en_US.UTF8'
>>> money = "$6,150,593.22"
>>> locale.atof(money.strip("$"))
6150593.2199999997
locale.atof
时要加1分,但对于金融应用程序来说,float
显然不是最佳选择。 - Fred Foo如果要实现一种不硬编码货币位置或符号的解决方案:
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)
conv['currency_symbol'].decode('utf-8')
对我来说失败了(" 'str' object has no attribute 'decode'), 但是这个不需要 decode 就可以运行。 - cphlewisbabel
包非常有用,可以解决以下问题:
这使得在本地化的场景下轻松解析数字。>>> babel.numbers.parse_decimal('1,024.64', locale='en')
Decimal('1024.64')
>>> babel.numbers.parse_decimal('1.024,64', locale='de')
Decimal('1024.64')
>>>
babel.numbers.get_currency_symbol('USD')
来获取货币符号,而不需要硬编码前缀/后缀。扩展包括带括号的负数:
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
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('', '', '$,)')
%timeit -n10000 [float( x.replace('(','-').translate(tbl3)) for x in n]
3.51 µs ± 84.8 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
在 @Andrew Clark 答案 的基础上进行拓展。
对于非 en_US 的其他语言环境:
>>> import re
>>> import locale
>>> locale.setlocale(locale.LC_NUMERIC, 'pt_BR.UTF8') # this is for atof()
'pt_BR.UTF8'
>>> locale.setlocale(locale.LC_MONETARY, 'pt_BR.UTF8') # this is for currency()
'pt_BR.UTF8'
>>> curr = locale.currency(6150593.22, grouping = True)
>>> curr
'R$ 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
必须提醒的是:在Python中,货币的适当类型是Decimal,而不是浮点数。
我将提供我的解决方案,希望它能帮助那些不仅遇到 ,
,而且还遇到 .
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
我发现最简单的方法是不需要硬编码或混淆货币检测,而且使用Decimal
类型可以避免float
类型的问题:
>>> from decimal import Decimal
>>> money="$6,150,593.22"
>>> amount = Decimal("".join(d for d in money if d.isdigit() or d == '.'))
>>> amount
Decimal('6150593.22')
credit: https://www.reddit.com/r/learnpython/comments/2248mp/how_to_format_currency_without_currency_sign/cgjd1o4?utm_source=share&utm_medium=web2x
我几年前编写了这个函数以解决同样的问题。
def money(number):
number = number.strip('$')
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
except
,否则会阻止使用CTRL-C等功能。 - Mark Lawrence这个函数将土耳其货币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