# Best way to format integer as string with leading zeros?

## Question:

I need to add leading zeros to integer to make a string with defined quantity of digits (\$cnt).
What the best way to translate this simple function from PHP to Python:

``````function add_nulls(\$int, \$cnt=2) {
\$int = intval(\$int);
for(\$i=0; \$i<(\$cnt-strlen(\$int)); \$i++)
\$nulls .= '0';
return \$nulls.\$int;
}
``````

Is there a function that can do this?

You most likely just need to format your integer:

``````'%0*d' % (fill, your_int)
``````

For example,

``````>>> '%0*d' % (3, 4)
'004'
``````

You can use the `zfill()` method to pad a string with zeros:

``````In [3]: str(1).zfill(2)
Out[3]: '01'
``````

This is my Python function:

``````def add_nulls(num, cnt=2):
cnt = cnt - len(str(num))
nulls = '0' * cnt
return '%s%s' % (nulls, num)
``````

A straightforward conversion would be (again with a function):

``````def add_nulls2(int, cnt):
nulls = str(int)
for i in range(cnt - len(str(int))):
nulls = '0' + nulls
return nulls
``````

Python 2.6 allows this:

``````add_nulls = lambda number, zero_count : "{0:0{1}d}".format(number, zero_count)

'002'
``````

You have at least two options:

• str.zfill: `lambda n, cnt=2: str(n).zfill(cnt)`
• `%` formatting: `lambda n, cnt=2: "%0*d" % (cnt, n)`

If on Python >2.5, see a third option in clorz’s answer.

### One-liner alternative to the built-in `zfill`.

This function takes `x` and converts it to a string, and adds zeros in the beginning only and only if the length is too short:

``````def zfill_alternative(x,len=4): return ( (('0'*len)+str(x))[-l:] if len(str(x))<len else str(x) )
``````

To sum it up – build-in: `zfill` is good enough, but if someone is curious on how to implement this by hand, here is one more example.

The standard way is to use format string modifiers. These format string methods are available in most programming languages (via the sprintf function in c for example) and are a handy tool to know about.

To output a string of length 5:

… in Python 3.5 and above: f-strings.

``````i = random.randint(0, 99999)
print(f'{i:05d}')
``````

Search for f-strings here for more details.

… Python 2.6 and above:

``````print '{0:05d}'.format(i)
``````

… before Python 2.6:

``````print "%05d" % i
``````

For Python 3 and beyond:
str.zfill() is still the most readable option

But it is a good idea to look into the new and powerful str.format(), what if you want to pad something that is not 0?

``````    # if we want to pad 22 with zeros in front, to be 5 digits in length:
str_output = '{:0>5}'.format(22)
print(str_output)
# >>> 00022
# {:0>5} meaning: ":0" means: pad with 0, ">" means move 22 to right most, "5" means the total length is 5

# another example for comparision
str_output = '{:#<4}'.format(11)
print(str_output)
# >>> 11##

# to put it in a less hard-coded format:
int_inputArg = 22
int_desiredLength = 5
str_output = '{str_0:0>{str_1}}'.format(str_0=int_inputArg, str_1=int_desiredLength)
print(str_output)
# >>> 00022
``````

``````number = 5