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']

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。

您也可以重复“0”，将其前置到str（n）并获得最右侧的宽度切片。又快又脏的小表情。

def pad_left(n, width, pad="0"):
    return ((pad * width) + str(n))[-width:]

也可以：

 h = 2
 m = 7
 s = 3
 print("%02d:%02d:%02d" % (h, m, s))

因此输出将为：“02:07:03”

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。

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。