How to pad zeroes to a string?
PythonStringZero PaddingPython Problem Overview
What is a Pythonic way to pad a numeric string with zeroes to the left, i.e. so the numeric string has a specific length?
Python Solutions
Solution 1 - Python
Strings:
>>> n = '4'
>>> print(n.zfill(3))
004
And for 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
Solution 2 - Python
Just use the rjust method of the string object.
This example will make a string of 10 characters long, padding as necessary.
>>> t = 'test'
>>> t.rjust(10, '0')
>>> '000000test'
Solution 3 - Python
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.
Solution 4 - Python
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'
Solution 5 - Python
>>> '99'.zfill(5)
'00099'
>>> '99'.rjust(5,'0')
'00099'
if you want the opposite:
>>> '99'.ljust(5,'0')
'99000'
Solution 6 - Python
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'
Solution 7 - Python
>## 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'
Solution 8 - Python
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
Solution 9 - Python
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'
Solution 10 - Python
For numbers:
i = 12
print(f"{i:05d}")
Output
00012
Solution 11 - Python
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.
Solution 12 - Python
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'
Solution 13 - Python
For zip codes saved as integers:
>>> a = 6340
>>> b = 90210
>>> print '%05d' % a
06340
>>> print '%05d' % b
90210
Solution 14 - Python
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.
Solution 15 - Python
Its ok too:
h = 2
m = 7
s = 3
print("%02d:%02d:%02d" % (h, m, s))
so output will be: "02:07:03"
Solution 16 - Python
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']
Solution 17 - Python
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
Solution 18 - Python
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.
Solution 19 - Python
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:]