Python: Short way of creating a sequential list with a prefix

Question

how can I generate a string (or Pandas series) like below:

a1,a2,a3,a4,...,a19

the following works, but I would like to know a shorter way

my_str = ""
for i in range(1, 20):
   comma = ',' if i!= 19 else ''
   my_str += "d" + str(i) + comma

Asked By: Emmet B

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Source

Answer 1

You can just use a list comprehension to generate the individual elements (using string concatenation or str.format), and then join the resulting list on the separator string (the comma):

>>> ['a{}'.format(i) for i in range(1, 20)]
['a1', 'a2', 'a3', 'a4', 'a5', 'a6', 'a7', 'a8', 'a9', 'a10', 'a11', 'a12', 'a13', 'a14', 'a15', 'a16', 'a17', 'a18', 'a19']
>>> ','.join(['a{}'.format(i) for i in range(1, 20)])
'a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15,a16,a17,a18,a19'

Answered By: poke

Answer 2

What about:

s = ','.join(["a%d" % i for i in range(1,20)])
print(s)

Output:

a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12,a13,a14,a15,a16,a17,a18,a19

~~(This uses a generator expression, which is more efficient than joining over a list comprehension — only slightly, with such a short list, however)~~ This now uses a list comprehension! See comments for an interesting read about why list comprehensions are more appropriate than generator expressions for join.

Answered By: jedwards

Answer 3

Adding onto what poke said (which I found to be the most helpful), but bringing it into 2022:

[f'a{i}' for i in range(1, 21)]

Answered By: hNomad

Python: Short way of creating a sequential list with a prefix

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