我如何打印一个整数与逗号作为千分隔符?
1234567 ⟶ 1,234,567
在句点和逗号之间决定不需要特定于区域设置。
我如何打印一个整数与逗号作为千分隔符?
1234567 ⟶ 1,234,567
在句点和逗号之间决定不需要特定于区域设置。
当前回答
意大利:
>>> import locale
>>> locale.setlocale(locale.LC_ALL,"")
'Italian_Italy.1252'
>>> f"{1000:n}"
'1.000'
其他回答
我使用的是python 2.5,所以我无法访问内置格式。
我查看了Django代码intcomma(下面代码中的intcomma_recurs),意识到它的效率很低,因为它是递归的,而且每次运行时都编译正则表达式也不是一件好事。这并不是一个必要的“问题”,因为django并没有真正关注这种低级性能。此外,我原本预计性能会有10倍的差异,但实际上只慢了3倍。
出于好奇,我实现了几个版本的intcomma,看看使用regex时的性能优势是什么。我的测试数据显示,这项任务有一点优势,但令人惊讶的是,优势并不大。
我也很高兴地看到了我的怀疑:在没有正则表达式的情况下,使用反向xrange方法是不必要的,但它确实使代码看起来更好,代价是性能下降了10%左右。
另外,我假设你传入的是一个字符串,看起来有点像一个数字。其他结果未定。
from __future__ import with_statement
from contextlib import contextmanager
import re,time
re_first_num = re.compile(r"\d")
def intcomma_noregex(value):
end_offset, start_digit, period = len(value),re_first_num.search(value).start(),value.rfind('.')
if period == -1:
period=end_offset
segments,_from_index,leftover = [],0,(period-start_digit) % 3
for _index in xrange(start_digit+3 if not leftover else start_digit+leftover,period,3):
segments.append(value[_from_index:_index])
_from_index=_index
if not segments:
return value
segments.append(value[_from_index:])
return ','.join(segments)
def intcomma_noregex_reversed(value):
end_offset, start_digit, period = len(value),re_first_num.search(value).start(),value.rfind('.')
if period == -1:
period=end_offset
_from_index,segments = end_offset,[]
for _index in xrange(period-3,start_digit,-3):
segments.append(value[_index:_from_index])
_from_index=_index
if not segments:
return value
segments.append(value[:_from_index])
return ','.join(reversed(segments))
re_3digits = re.compile(r'(?<=\d)\d{3}(?!\d)')
def intcomma(value):
segments,last_endoffset=[],len(value)
while last_endoffset > 3:
digit_group = re_3digits.search(value,0,last_endoffset)
if not digit_group:
break
segments.append(value[digit_group.start():last_endoffset])
last_endoffset=digit_group.start()
if not segments:
return value
if last_endoffset:
segments.append(value[:last_endoffset])
return ','.join(reversed(segments))
def intcomma_recurs(value):
"""
Converts an integer to a string containing commas every three digits.
For example, 3000 becomes '3,000' and 45000 becomes '45,000'.
"""
new = re.sub("^(-?\d+)(\d{3})", '\g<1>,\g<2>', str(value))
if value == new:
return new
else:
return intcomma(new)
@contextmanager
def timed(save_time_func):
begin=time.time()
try:
yield
finally:
save_time_func(time.time()-begin)
def testset_xsimple(func):
func('5')
def testset_simple(func):
func('567')
def testset_onecomma(func):
func('567890')
def testset_complex(func):
func('-1234567.024')
def testset_average(func):
func('-1234567.024')
func('567')
func('5674')
if __name__ == '__main__':
print 'Test results:'
for test_data in ('5','567','1234','1234.56','-253892.045'):
for func in (intcomma,intcomma_noregex,intcomma_noregex_reversed,intcomma_recurs):
print func.__name__,test_data,func(test_data)
times=[]
def overhead(x):
pass
for test_run in xrange(1,4):
for func in (intcomma,intcomma_noregex,intcomma_noregex_reversed,intcomma_recurs,overhead):
for testset in (testset_xsimple,testset_simple,testset_onecomma,testset_complex,testset_average):
for x in xrange(1000): # prime the test
testset(func)
with timed(lambda x:times.append(((test_run,func,testset),x))):
for x in xrange(50000):
testset(func)
for (test_run,func,testset),_delta in times:
print test_run,func.__name__,testset.__name__,_delta
下面是测试结果:
intcomma 5 5
intcomma_noregex 5 5
intcomma_noregex_reversed 5 5
intcomma_recurs 5 5
intcomma 567 567
intcomma_noregex 567 567
intcomma_noregex_reversed 567 567
intcomma_recurs 567 567
intcomma 1234 1,234
intcomma_noregex 1234 1,234
intcomma_noregex_reversed 1234 1,234
intcomma_recurs 1234 1,234
intcomma 1234.56 1,234.56
intcomma_noregex 1234.56 1,234.56
intcomma_noregex_reversed 1234.56 1,234.56
intcomma_recurs 1234.56 1,234.56
intcomma -253892.045 -253,892.045
intcomma_noregex -253892.045 -253,892.045
intcomma_noregex_reversed -253892.045 -253,892.045
intcomma_recurs -253892.045 -253,892.045
1 intcomma testset_xsimple 0.0410001277924
1 intcomma testset_simple 0.0369999408722
1 intcomma testset_onecomma 0.213000059128
1 intcomma testset_complex 0.296000003815
1 intcomma testset_average 0.503000020981
1 intcomma_noregex testset_xsimple 0.134000062943
1 intcomma_noregex testset_simple 0.134999990463
1 intcomma_noregex testset_onecomma 0.190999984741
1 intcomma_noregex testset_complex 0.209000110626
1 intcomma_noregex testset_average 0.513000011444
1 intcomma_noregex_reversed testset_xsimple 0.124000072479
1 intcomma_noregex_reversed testset_simple 0.12700009346
1 intcomma_noregex_reversed testset_onecomma 0.230000019073
1 intcomma_noregex_reversed testset_complex 0.236999988556
1 intcomma_noregex_reversed testset_average 0.56299996376
1 intcomma_recurs testset_xsimple 0.348000049591
1 intcomma_recurs testset_simple 0.34600019455
1 intcomma_recurs testset_onecomma 0.625
1 intcomma_recurs testset_complex 0.773999929428
1 intcomma_recurs testset_average 1.6890001297
1 overhead testset_xsimple 0.0179998874664
1 overhead testset_simple 0.0190000534058
1 overhead testset_onecomma 0.0190000534058
1 overhead testset_complex 0.0190000534058
1 overhead testset_average 0.0309998989105
2 intcomma testset_xsimple 0.0360000133514
2 intcomma testset_simple 0.0369999408722
2 intcomma testset_onecomma 0.207999944687
2 intcomma testset_complex 0.302000045776
2 intcomma testset_average 0.523000001907
2 intcomma_noregex testset_xsimple 0.139999866486
2 intcomma_noregex testset_simple 0.141000032425
2 intcomma_noregex testset_onecomma 0.203999996185
2 intcomma_noregex testset_complex 0.200999975204
2 intcomma_noregex testset_average 0.523000001907
2 intcomma_noregex_reversed testset_xsimple 0.130000114441
2 intcomma_noregex_reversed testset_simple 0.129999876022
2 intcomma_noregex_reversed testset_onecomma 0.236000061035
2 intcomma_noregex_reversed testset_complex 0.241999864578
2 intcomma_noregex_reversed testset_average 0.582999944687
2 intcomma_recurs testset_xsimple 0.351000070572
2 intcomma_recurs testset_simple 0.352999925613
2 intcomma_recurs testset_onecomma 0.648999929428
2 intcomma_recurs testset_complex 0.808000087738
2 intcomma_recurs testset_average 1.81900000572
2 overhead testset_xsimple 0.0189998149872
2 overhead testset_simple 0.0189998149872
2 overhead testset_onecomma 0.0190000534058
2 overhead testset_complex 0.0179998874664
2 overhead testset_average 0.0299999713898
3 intcomma testset_xsimple 0.0360000133514
3 intcomma testset_simple 0.0360000133514
3 intcomma testset_onecomma 0.210000038147
3 intcomma testset_complex 0.305999994278
3 intcomma testset_average 0.493000030518
3 intcomma_noregex testset_xsimple 0.131999969482
3 intcomma_noregex testset_simple 0.136000156403
3 intcomma_noregex testset_onecomma 0.192999839783
3 intcomma_noregex testset_complex 0.202000141144
3 intcomma_noregex testset_average 0.509999990463
3 intcomma_noregex_reversed testset_xsimple 0.125999927521
3 intcomma_noregex_reversed testset_simple 0.126999855042
3 intcomma_noregex_reversed testset_onecomma 0.235999822617
3 intcomma_noregex_reversed testset_complex 0.243000030518
3 intcomma_noregex_reversed testset_average 0.56200003624
3 intcomma_recurs testset_xsimple 0.337000131607
3 intcomma_recurs testset_simple 0.342000007629
3 intcomma_recurs testset_onecomma 0.609999895096
3 intcomma_recurs testset_complex 0.75
3 intcomma_recurs testset_average 1.68300008774
3 overhead testset_xsimple 0.0189998149872
3 overhead testset_simple 0.018000125885
3 overhead testset_onecomma 0.018000125885
3 overhead testset_complex 0.0179998874664
3 overhead testset_average 0.0299999713898
稍微扩展一下Ian Schneider的回答:
如果你想使用自定义千位分隔符,最简单的解决方案是:
'{:,}'.format(value).replace(',', your_custom_thousands_separator)
例子
'{:,.2f}'.format(123456789.012345).replace(',', ' ')
如果你想要像这样的德语表示,它会变得有点复杂:
('{:,.2f}'.format(123456789.012345)
.replace(',', ' ') # 'save' the thousands separators
.replace('.', ',') # dot to comma
.replace(' ', '.')) # thousand separators to dot
在低效率和不可读性方面,它是很难被击败的:
>>> import itertools
>>> s = '-1234567'
>>> ','.join(["%s%s%s" % (x[0], x[1] or '', x[2] or '') for x in itertools.izip_longest(s[::-1][::3], s[::-1][1::3], s[::-1][2::3])])[::-1].replace('-,','-')
我有这个代码的python 2和python 3版本。我知道这个问题是关于python2的,但是现在(8年过去了,哈哈)人们可能会使用python3。Python 3代码:
import random
number = str(random.randint(1, 10000000))
comma_placement = 4
print('The original number is: {}. '.format(number))
while True:
if len(number) % 3 == 0:
for i in range(0, len(number) // 3 - 1):
number = number[0:len(number) - comma_placement + 1] + ',' + number[len(number) - comma_placement + 1:]
comma_placement = comma_placement + 4
else:
for i in range(0, len(number) // 3):
number = number[0:len(number) - comma_placement + 1] + ',' + number[len(number) - comma_placement + 1:]
break
print('The new and improved number is: {}'.format(number))
Python 2代码:(编辑。python代码不能工作。我认为语法是不同的)。
import random
number = str(random.randint(1, 10000000))
comma_placement = 4
print 'The original number is: %s.' % (number)
while True:
if len(number) % 3 == 0:
for i in range(0, len(number) // 3 - 1):
number = number[0:len(number) - comma_placement + 1] + ',' + number[len(number) - comma_placement + 1:]
comma_placement = comma_placement + 4
else:
for i in range(0, len(number) // 3):
number = number[0:len(number) - comma_placement + 1] + ',' + number[len(number) - comma_placement + 1:]
break
print 'The new and improved number is: %s.' % (number)
从Python版本2.6,你可以这样做:
def format_builtin(n):
return format(n, ',')
对于< 2.6的Python版本,仅供参考,这里有两个手动解决方案,它们将浮点数转换为整数,但负数可以正常工作:
def format_number_using_lists(number):
string = '%d' % number
result_list = list(string)
indexes = range(len(string))
for index in indexes[::-3][1:]:
if result_list[index] != '-':
result_list.insert(index+1, ',')
return ''.join(result_list)
这里有几点需要注意:
string = '%d' % number漂亮地将数字转换为字符串,它支持负号,并从浮点数中删除分数,使它们成为整数; 这个切片索引[::-3]返回从开始的每第三个项 所以我使用了另一个切片[1:]来删除最后一项 因为我不需要在最后一个数字后面加逗号; 此条件如果l[index] != '-'被用于支持负数,则不要在减号后插入逗号。
还有一个更硬核的版本:
def format_number_using_generators_and_list_comprehensions(number):
string = '%d' % number
generator = reversed(
[
value+',' if (index!=0 and value!='-' and index%3==0) else value
for index,value in enumerate(reversed(string))
]
)
return ''.join(generator)