Python – list transformation

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Does anyone knows what magic I have to use to change x list:

x = [1,2,3,4,5,11]

into y list?

y = [’01’,’02’,’03’,’04’,’05’,’11’]

Thank you all in advance for helping me…

You can use a list comprehension (Python 2.6+):

y = ["{0:0>2}".format(v) for v in x]

Or for Python prior to 2.6:

y = ["%02d" % v for v in x]

Or for Python 3.6+ using f-strings:

y = [f'{v:02}' for v in x] 

Edit: Missed the fact that you wanted zero-padding…

You want to use the built-in map function:

>>> x = [1,2,3,4,5]
>>> x
[1, 2, 3, 4, 5]
>>> y = map(str, x)
>>> y
['1', '2', '3', '4', '5']

EDIT You changed the requirements on me! To make it display leading zeros, you do this:

>>> x = [1,2,3,4,5,11]
>>> y = ["%02d" % v for v in x]
>>> y
['01', '02', '03', '04', '05', '11'] 

I would use a list comprehension myself, but here is another solution using map for those interested…

map(lambda v: "%02d" %v, x)

y = ['%02d'%v for v in x]

Try this:

>>> [str(v).rjust(2,'0') for v in x]
['01', '02', '03', '04', '05', '11']

rjust as a method of string class, takes an integer argument(result length) and an optional padding character

to get the 0’s:

y = ['%02d' % i for i in x]

An alternative to format strings would be to use the string’s zfill() method:

y = [str(i).zfill(2) for i in x]

Another thing: you might be after padding based on the largest item in the list, so instead of just using 2, you could do:

pad_length = len(str(max(x)))
y = [str(i).zfill(pad_length) for i in x]


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