The Question :
336 people think this question is useful
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?
The Question Comments :
The Answer 1
674 people think this answer is useful
You can use the zfill()
method to pad a string with zeros:
In [3]: str(1).zfill(2)
Out[3]: '01'
The Answer 2
182 people think this answer is useful
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:
i = random.randint(0, 99999)
print(f'{i:05d}')
… Python 2.6 and above:
print '{0:05d}'.format(i)
… before Python 2.6:
print "%05d" % i
See: https://docs.python.org/3/library/string.html
The Answer 3
91 people think this answer is useful
Python 3.6 f-strings allows us to add leading zeros easily:
number = 5
print(f' now we have leading zeros in {number:02d}')
Have a look at this good post about this feature.
The Answer 4
70 people think this answer is useful
You most likely just need to format your integer:
'%0*d' % (fill, your_int)
For example,
>>> '%0*d' % (3, 4)
'004'
The Answer 5
23 people think this answer is useful
Python 2.6 allows this:
add_nulls = lambda number, zero_count : "{0:0{1}d}".format(number, zero_count)
>>>add_nulls(2,3)
'002'
The Answer 6
17 people think this answer is useful
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
The Answer 7
6 people think this answer is useful
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.
The Answer 8
2 people think this answer is useful
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 Answer 9
1 people think this answer is useful
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
The Answer 10
-2 people think this answer is useful
This is my Python function:
def add_nulls(num, cnt=2):
cnt = cnt - len(str(num))
nulls = '0' * cnt
return '%s%s' % (nulls, num)