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
这里有一个也适用于浮动:
def float2comma(f):
s = str(abs(f)) # Convert to a string
decimalposition = s.find(".") # Look for decimal point
if decimalposition == -1:
decimalposition = len(s) # If no decimal, then just work from the end
out = ""
for i in range(decimalposition+1, len(s)): # do the decimal
if not (i-decimalposition-1) % 3 and i-decimalposition-1: out = out+","
out = out+s[i]
if len(out):
out = "."+out # add the decimal point if necessary
for i in range(decimalposition-1,-1,-1): # working backwards from decimal point
if not (decimalposition-i-1) % 3 and decimalposition-i-1: out = ","+out
out = s[i]+out
if f < 0:
out = "-"+out
return out
使用的例子:
>>> float2comma(10000.1111)
'10,000.111,1'
>>> float2comma(656565.122)
'656,565.122'
>>> float2comma(-656565.122)
'-656,565.122'
其他回答
在浮点数中同时使用分隔符和小数: (在这个例子中,小数点后两位)
large_number = 4545454.26262666
print(f"Formatted: {large_number:,.2f}")
结果: 格式:4545454点
下面是一行正则表达式替换:
re.sub("(\d)(?=(\d{3})+(?!\d))", r"\1,", "%d" % val)
仅适用于积分输出:
import re
val = 1234567890
re.sub("(\d)(?=(\d{3})+(?!\d))", r"\1,", "%d" % val)
# Returns: '1,234,567,890'
val = 1234567890.1234567890
# Returns: '1,234,567,890'
或者对于小于4位的浮点数,将格式说明符更改为%.3f:
re.sub("(\d)(?=(\d{3})+(?!\d))", r"\1,", "%.3f" % val)
# Returns: '1,234,567,890.123'
注意:不能正确工作与超过三个十进制数字,因为它将尝试分组小数部分:
re.sub("(\d)(?=(\d{3})+(?!\d))", r"\1,", "%.5f" % val)
# Returns: '1,234,567,890.12,346'
它是如何工作的
让我们来分析一下:
re.sub(pattern, repl, string)
pattern = \
"(\d) # Find one digit...
(?= # that is followed by...
(\d{3})+ # one or more groups of three digits...
(?!\d) # which are not followed by any more digits.
)",
repl = \
r"\1,", # Replace that one digit by itself, followed by a comma,
# and continue looking for more matches later in the string.
# (re.sub() replaces all matches it finds in the input)
string = \
"%d" % val # Format the string as a decimal to begin with
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('-,','-')
当地unaware
'{:,}'.format(value) # For Python ≥2.7
f'{value:,}' # For Python ≥3.6
区域设置感知
import locale
locale.setlocale(locale.LC_ALL, '') # Use '' for auto, or force e.g. to 'en_US.UTF-8'
'{:n}'.format(value) # For Python ≥2.7
f'{value:n}' # For Python ≥3.6
参考
根据格式规范迷你语言,
','选项表示使用逗号作为千位分隔符。对于支持区域设置的分隔符,请使用'n'整数表示类型。
我相信这一定有一个标准的库函数,但尝试自己用递归写它很有趣,所以这是我想到的:
def intToStringWithCommas(x):
if type(x) is not int and type(x) is not long:
raise TypeError("Not an integer!")
if x < 0:
return '-' + intToStringWithCommas(-x)
elif x < 1000:
return str(x)
else:
return intToStringWithCommas(x / 1000) + ',' + '%03d' % (x % 1000)
话虽如此,如果其他人确实找到了一种标准的方法,那么您应该使用它。