# How do I pad a string with zeroes?

## Question:

How do I pad a numeric string with zeroes to the left, so that the string has a specific length?

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.

``````>>> 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
``````
``````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 `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'
``````
``````>>> '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”

``````def PadNumber(number, n_pad, add_prefix=None):
number_str = str(number)

``````

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
'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.

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