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。

str（n）.zfill（width）将处理字符串、int、浮点。。。并且兼容Python 2.x和3.x：

>>> n = 3
>>> str(n).zfill(5)
'00003'
>>> n = '3'
>>> str(n).zfill(5)
'00003'
>>> n = '3.0'
>>> str(n).zfill(5)
'003.0'

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'

另一种方法是使用列表理解和长度检查条件。下面是一个演示：

# input list of strings that we want to prepend zeros
In [71]: list_of_str = ["101010", "10101010", "11110", "0000"]

# prepend zeros to make each string to length 8, if length of string is less than 8
In [83]: ["0"*(8-len(s)) + s if len(s) < desired_len else s for s in list_of_str]
Out[83]: ['00101010', '10101010', '00011110', '00000000']

要填充字符串：

>>> n = '4'
>>> print(n.zfill(3))
004

要填充数字：

>>> n = 4
>>> print(f'{n:03}') # Preferred method, python >= 3.6
004
>>> print('%03d' % n)
004
>>> print(format(n, '03')) # python >= 2.6
004
>>> print('{0:03d}'.format(n))  # python >= 2.6 + python 3
004
>>> print('{foo:03d}'.format(foo=n))  # python >= 2.6 + python 3
004
>>> print('{:03d}'.format(n))  # python >= 2.7 + python3
004

字符串格式文档。

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。