in python how do I convert a single digit number into a double digits string?

Question

So say i have

a = 5

i want to print it as a string '05'

Answer 1

In python 3.6, the fstring or "formatted string literal" mechanism was introduced.

f"{a:02}"

is the equivalent of the .format format below, but a little bit more terse.

---

python 3 before 3.6 prefers a somewhat more verbose formatting system:

"{0:0=2d}".format(a)

You can take shortcuts here, the above is probably the most verbose variant. The full documentation is available here: http://docs.python.org/3/library/string.html#string-formatting

---

print "%02d"%a is the python 2 variant

The relevant doc link for python2 is: http://docs.python.org/2/library/string.html#format-specification-mini-language

Answer 2

a = 5
print '%02d' % a
# output: 05

The '%' operator is called string formatting operator when used with a string on the left side. '%d' is the formatting code to print out an integer number (you will get a type error if the value isn't numeric). With '%2d you can specify the length, and '%02d' can be used to set the padding character to a 0 instead of the default space.

Answer 3

>>> print '{0}'.format('5'.zfill(2))
05

Read more here.

Answer 4

>>> a=["%02d" % x for x in range(24)]
>>> a
['00', '01', '02', '03', '04', '05', '06', '07', '08', '09', '10', '11', '12', '13', '14', '15', '16', '17', '18', '19', '20', '21', '22', '23']
>>> 

It is that simple

Answer 5

In Python 3.6 you can use so called f-strings. In my opinion this method is much clearer to read.

>>> f'{a:02d}'
'05'

Answer 6

In Python3, you can:

print("%02d" % a)

Answer 7

Based on what @user225312 said, you can use .zfill() to add paddng to numbers converted to strings.

My approach is to leave number as a number until the moment you want to convert it into string:

>>> num = 11
>>> padding = 3
>>> print(str(num).zfill(padding))
011

Answer 8

Branching off of Mohommad's answer:

str_years = [x for x in range(24)]
#[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23]

#Or, if you're starting with ints:
int_years = [int(x) for x in str_years]

#Formatted here
form_years = ["%02d" % x for x in int_years]

print(form_years)
#['00', '01', '02', '03', '04', '05', '06', '07', '08', '09', '10', '11', '12', '13', '14', '15', '16', '17', '18', '19', '20', '21', '22', '23']

Answer 9

df["col_name"].str.rjust(4,'0')#(length of string,'value') --> ValueXXX --> 0XXX  
df["col_name"].str.ljust(4,'0')#(length of string,'value') --> XXXValue --> XXX0

Answer 10

If you are an analyst and not a full stack guy, this might be more intuitive:

[(str('00000') + str(i))[-5:] for i in arange(100)]

breaking that down, you:

• start by creating a list that repeats 0's or X's, in this case, 100 long, i.e., arange(100)

• add the numbers you want to the string, in this case, numbers 0-99, i.e., 'i'

• keep only the right hand 5 digits, i.e., '[-5:]' for subsetting

• output is numbered list, all with 5 digits

Answer 11

This is a dumb solution but I was getting type errors with the other solutions above. So if all else fails, yolo:

images3digit = []

for i in images:
        if len(i)==1:
            i = '00'+i
            images3digit.append(i)
        elif len(i)==2:
            i = '0'+i
            images3digit.append(i)
        elif len(i)==3:
            images3digit.append(i)