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
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。
其他回答
我做了一个函数:
def PadNumber(number, n_pad, add_prefix=None):
number_str = str(number)
paded_number = number_str.zfill(n_pad)
if add_prefix:
paded_number = add_prefix+paded_number
print(paded_number)
PadNumber(99, 4)
PadNumber(1011, 8, "b'")
PadNumber('7BEF', 6, "#")
输出:
0099
b'00001011
#007BEF
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。
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]。
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'
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”
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