How do I pad a string with zeroes?

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.

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.

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'

For zip codes saved as integers:

>>> a = 6340
>>> b = 90210
>>> print '%05d' % a
06340
>>> print '%05d' % b
90210

str(n).zfill(width) will work with strings, ints, floats… and is Python 2.x and 3.x compatible:

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

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

if you want the opposite:

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

You could also repeat “0”, prepend it to str(n) and get the rightmost width slice. Quick and dirty little expression.

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

For Python 3.6+ using f-strings:

>>> i = 1
>>> f"{i:0>2}"  # Works for both numbers and strings.
'01'
>>> f"{i:02}"  # Works only for numbers.
'01'

For Python 2.6 to Python 3.5:

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

Those standard format specifiers are [[fill]align][minimumwidth] and [0][minimumwidth].

For the ones who came here to understand and not just a quick answer.
I do these especially for time strings:

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” symbols what to replace with the “2” padding characters, the default is an empty space

“>” symbols allign all the 2 “0” character to the left of the string

“:” symbols the format_spec

Quick timing comparison:

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]

I’ve made different tests of different repetitions. The differences are not huge, but in all tests, the zfill solution was fastest.

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'

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']

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'

Its ok too:

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

so output will be: “02:07:03”

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

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'

For numbers:

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

Output

00012

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.