# How to pad a string with zeroes to the left?

This is a very frequently asked question, how do I pad a numeric string with zeroes to the left, so that the string has a specific length? In this article, we’ll answer this question.

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

``````>>> 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``````

## How to pad a string with zeroes to the left in Python?

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'))``````

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 to Python 3.5:

``````>>> "{: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'
``````

if you want the opposite:

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

`str(n).zfill(width)` will work with `string`s, `int`s, `float`s… 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'``````

## 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 = 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'``````

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

## How to pad a string with zeroes to the left in Python 3.6+?

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'``````

I am adding how to use an 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