How do I pad a string with zeroes?

To pad strings:

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

To pad numbers:

>>> 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

String formatting documentation.

Just use the rjust method of the string object.

This example creates a 10-character length string, padding as necessary:

>>> s = 'test'
>>> s.rjust(10, '0')
>>> '000000test'

Besides zfill, you can use general string formatting:

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'))

Documentation for string formatting and f-strings.

What is the most pythonic way to pad a numeric string with zeroes to the left, i.e., so the numeric string has a specific length?

str.zfill is specifically intended to do this:

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

Note that it is specifically intended to handle numeric strings as requested, and moves a + or - to the beginning of the string:

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

Here's the help on 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.

Performance

This is also the most performant of alternative methods:

>>> 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

To best compare apples to apples for the % method (note it is actually slower), which will otherwise pre-calculate:

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

Implementation

With a little digging, I found the implementation of the zfill method in 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;
}

Let's walk through this C code.

It first parses the argument positionally, meaning it doesn't allow keyword arguments:

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

It then checks if it's the same length or longer, in which case it returns the string.

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

zfill calls pad (this pad function is also called by ljust, rjust, and center as well). This basically copies the contents into a new string and fills in the padding.

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;
}

After calling pad, zfill moves any originally preceding + or - to the beginning of the string.

Note that for the original string to actually be numeric is not required:

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

When using Python >= 3.6, the cleanest way is to use f-strings with string formatting:

>>> 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'

I would prefer formatting with an int, since only then the sign is handled correctly:

>>> f"{-1:08}"
'-0000001'

>>> f"{1:+08}"
'+0000001'

>>> f"{'-1':0>8}"
'000000-1'

For numbers:

i = 12
print(f"{i:05d}")

Output

00012

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

See the print documentation for all the exciting details!

Update for Python 3.x (7.5 years later)

That last line should now be:

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

I.e. print() is now a function, not a statement. Note that I still prefer the Old School printf() style because, IMNSHO, it reads better, and because, um, I've been using that notation since January, 1980. Something ... old dogs .. something something ... new tricks.

I am adding how to use a int from a length of a string within an f-string because it didn't appear to be covered:

>>> pad_number = len("this_string")
11
>>> s = f"{1:0{pad_number}}" }
>>> s
'00000000001'

Its ok too:

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

so output will be: "02:07:03"

If you're looking to pad an integer, and limit the significant figures at the same time (with f strings):

a = 4.432
>> 4.432
a = f'{a:04.1f}'
>> '04.4'

f'{a:04.1f}' this translates to 1 decimal/(float) point, left pad the digit until 4 characters total.

Another approach would be to use a list comprehension with a condition checking for lengths. Below is a demonstration:

# 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']

I made a function :

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, "#")

The output :

0099
b'00001011
#007BEF