将零填充到字符串的最好方法

使用零填充数字字符串的最Pythonic方法是什么,即数字字符串是否具有特定长度？

字符串:

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

对于数字:

>>> 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.

只需使用字符串对象的rjust方法.

此示例将生成一个长度为10个字符的字符串,根据需要填充.

>>> t = 'test'
>>> t.rjust(10, '0')
>>> '000000test'

对于数字:

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'))

另请参见:Python:字符串格式.

编辑:这是值得注意的是,截至昨日 2008年12月3日,格式化的这种方法有利于被弃用zfill字符串的方法:

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'))

有关详细信息,请参阅PEP 3101.

适用于Python 2和Python 3:

>>> "{:0>2}".format("1")  # Works for both numbers and strings.
'01'
>>> "{:02}".format(1)  # Works only for numbers.
'01'

>>> '99'.zfill(5)
'00099'
>>> '99'.rjust(5,'0')
'00099'

如果你想要相反:

>>> '99'.ljust(5,'0')
'99000'

str(n).zfill(width)将使用strings,ints,floats ...并且是Python 2. x和3. x兼容:

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

对于那些来这里了解而不仅仅是快速回答的人.我特别为时间字符串做这些:

hour = 4
minute = 3
"{:0>2}:{:0>2}".format(hour,minute)
# prints 04:03

"{:0>3}:{:0>5}".format(hour,minute)
# prints '004:00003'

"{:0<3}:{:0<5}".format(hour,minute)
# prints '400:30000'

"{:$<3}:{:#<5}".format(hour,minute)
# prints '4$$:3####'

"0"符号用"2"填充字符替换什么,默认为空格

">"符号将所有2"0"字符对齐到字符串的左侧

":"符号format_spec

使用零填充数字字符串的最pythonic方法是什么,即数字字符串是否具有特定长度？

str.zfill 特别打算这样做:

>>> '1'.zfill(4)
'0001'

请注意,它专门用于处理请求的数字字符串,并将a +或移动-到字符串的开头:

>>> '+1'.zfill(4)
'+001'
>>> '-1'.zfill(4)
'-001'

这是帮助str.zfill:

>>> help(str.zfill)
Help on method_descriptor:

zfill(...)
    S.zfill(width) -> str

    Pad a numeric string S with zeros on the left, to fill a field
    of the specified width. The string S is never truncated.

性能

这也是替代方法中性能最高的:

>>> min(timeit.repeat(lambda: '1'.zfill(4)))
0.18824880896136165
>>> min(timeit.repeat(lambda: '1'.rjust(4, '0')))
0.2104538488201797
>>> min(timeit.repeat(lambda: f'{1:04}'))
0.32585487607866526
>>> min(timeit.repeat(lambda: '{:04}'.format(1)))
0.34988890308886766

为了最好地比较该%方法的苹果与苹果(注意它实际上较慢),否则将预先计算:

>>> min(timeit.repeat(lambda: '1'.zfill(0 or 4)))
0.19728074967861176
>>> min(timeit.repeat(lambda: '%04d' % (0 or 1)))
0.2347015216946602

履行

通过一点挖掘,我发现了该zfill方法的实现Objects/stringlib/transmogrify.h:

static PyObject *
stringlib_zfill(PyObject *self, PyObject *args)
{
    Py_ssize_t fill;
    PyObject *s;
    char *p;
    Py_ssize_t width;

    if (!PyArg_ParseTuple(args, "n:zfill", &width))
        return NULL;

    if (STRINGLIB_LEN(self) >= width) {
        return return_self(self);
    }

    fill = width - STRINGLIB_LEN(self);

    s = pad(self, fill, 0, '0');

    if (s == NULL)
        return NULL;

    p = STRINGLIB_STR(s);
    if (p[fill] == '+' || p[fill] == '-') {
        /* move sign to beginning of string */
        p[0] = p[fill];
        p[fill] = '0';
    }

    return s;
}

>>> '1'.zfill(width=4)
Traceback (most recent call last):
  File "", line 1, in 
TypeError: zfill() takes no keyword arguments

>>> '1'.zfill(0)
'1'

static inline PyObject *
pad(PyObject *self, Py_ssize_t left, Py_ssize_t right, char fill)
{
    PyObject *u;

    if (left < 0)
        left = 0;
    if (right < 0)
        right = 0;

    if (left == 0 && right == 0) {
        return return_self(self);
    }

    u = STRINGLIB_NEW(NULL, left + STRINGLIB_LEN(self) + right);
    if (u) {
        if (left)
            memset(STRINGLIB_STR(u), fill, left);
        memcpy(STRINGLIB_STR(u) + left,
               STRINGLIB_STR(self),
               STRINGLIB_LEN(self));
        if (right)
            memset(STRINGLIB_STR(u) + left + STRINGLIB_LEN(self),
                   fill, right);
    }

    return u;
}

>>> '+foo'.zfill(10)
'+000000foo'
>>> '-foo'.zfill(10)
'-000000foo'

width = 10
x = 5
print "%0*d" % (width, x)
> 0000000005

print("%0*d" % (width, x))