如何在数字字符串的左边加上零,使字符串具有特定的长度?
当前回答
如果您希望填充一个整数,同时限制有效数字(使用f字符串):
a = 4.432
>> 4.432
a = f'{a:04.1f}'
>> '04.4'
f“{a:04.1f}”这转换为1个十进制/(浮点)点,将数字向左填充,直到总共4个字符。
其他回答
对于使用f-string的Python 3.6+:
>>> i = 1
>>> f"{i:0>2}" # Works for both numbers and strings.
'01'
>>> f"{i:02}" # Works only for numbers.
'01'
对于Python 2.6到Python 3.5:
>>> "{:0>2}".format("1") # Works for both numbers and strings.
'01'
>>> "{:02}".format(1) # Works only for numbers.
'01'
这些标准格式说明符是[[fill]align][minimumwid]和[0][minimupwid]。
除了zfill,您还可以使用常规字符串格式:
print(f'{number:05d}') # (since Python 3.6), or
print('{:05d}'.format(number)) # or
print('{0:05d}'.format(number)) # or (explicit 0th positional arg. selection)
print('{n:05d}'.format(n=number)) # or (explicit `n` keyword arg. selection)
print(format(number, '05d'))
字符串格式和f-string的文档。
对于数字:
i = 12
print(f"{i:05d}")
输出
00012
快速定时比较:
setup = '''
from random import randint
def test_1():
num = randint(0,1000000)
return str(num).zfill(7)
def test_2():
num = randint(0,1000000)
return format(num, '07')
def test_3():
num = randint(0,1000000)
return '{0:07d}'.format(num)
def test_4():
num = randint(0,1000000)
return format(num, '07d')
def test_5():
num = randint(0,1000000)
return '{:07d}'.format(num)
def test_6():
num = randint(0,1000000)
return '{x:07d}'.format(x=num)
def test_7():
num = randint(0,1000000)
return str(num).rjust(7, '0')
'''
import timeit
print timeit.Timer("test_1()", setup=setup).repeat(3, 900000)
print timeit.Timer("test_2()", setup=setup).repeat(3, 900000)
print timeit.Timer("test_3()", setup=setup).repeat(3, 900000)
print timeit.Timer("test_4()", setup=setup).repeat(3, 900000)
print timeit.Timer("test_5()", setup=setup).repeat(3, 900000)
print timeit.Timer("test_6()", setup=setup).repeat(3, 900000)
print timeit.Timer("test_7()", setup=setup).repeat(3, 900000)
> [2.281613943830961, 2.2719342631547077, 2.261691106209631]
> [2.311480238815406, 2.318420542148333, 2.3552384305184493]
> [2.3824197456864304, 2.3457239951596485, 2.3353268829498646]
> [2.312442972404032, 2.318053102249902, 2.3054072168069872]
> [2.3482314132374853, 2.3403386400002475, 2.330108825844775]
> [2.424549090688892, 2.4346475296851438, 2.429691196530058]
> [2.3259756401716487, 2.333549212826732, 2.32049893822186]
我做了不同重复的不同测试。差异不大,但在所有测试中,zfill解决方案都是最快的。
使用Python>=3.6时,最干净的方法是将f-string与字符串格式一起使用:
>>> s = f"{1:08}" # inline with int
>>> s
'00000001'
>>> s = f"{'1':0>8}" # inline with str
>>> s
'00000001'
>>> n = 1
>>> s = f"{n:08}" # int variable
>>> s
'00000001'
>>> c = "1"
>>> s = f"{c:0>8}" # str variable
>>> s
'00000001'
我更喜欢使用int格式,因为只有这样符号才能正确处理:
>>> f"{-1:08}"
'-0000001'
>>> f"{1:+08}"
'+0000001'
>>> f"{'-1':0>8}"
'000000-1'