Use:

'00'[len(str(i)):] + str(i)

或者使用数学模块:

import math
'00'[math.ceil(math.log(i, 10)):] + str(i)

所有这些都创建了字符串“01”:

>python -m timeit "'{:02d}'.format(1)"
1000000 loops, best of 5: 357 nsec per loop

>python -m timeit "'{0:0{1}d}'.format(1,2)"
500000 loops, best of 5: 607 nsec per loop

>python -m timeit "f'{1:02d}'"
1000000 loops, best of 5: 281 nsec per loop

>python -m timeit "f'{1:0{2}d}'"
500000 loops, best of 5: 423 nsec per loop

>python -m timeit "str(1).zfill(2)"
1000000 loops, best of 5: 271 nsec per loop

>python
Python 3.8.1 (tags/v3.8.1:1b293b6, Dec 18 2019, 23:11:46) [MSC v.1916 64 bit (AMD64)] on win32

如果处理的数字是一位数或两位数:

‘0’+ str (number)黑帽子-2:铝还是‘0那0’.format (number)黑-2:铝

python的方法是:

str(number).rjust(string_width, fill_char)

这样，如果原始字符串的长度大于string_width，则原始字符串将原封不动地返回。例子:

a = [1, 10, 100]
for num in a:
    print str(num).rjust(2, '0')

结果:

01
10
100

你可以使用str.zfill:

print(str(1).zfill(2))
print(str(10).zfill(2))
print(str(100).zfill(2))

打印:

01
10
100

print('{:02}'.format(1))
print('{:02}'.format(10))
print('{:02}'.format(100))

打印:

01
10
100