Convert range(r) to list of strings of length 2 in python
Question:
I just want to change a list (that I make using range(r)) to a list of strings, but if the length of the string is 1, tack a 0 on the front.
I know how to turn the list into strings using
ranger= map(str,range(r))
but I want to be able to also change the length of those strings.
Input:
r = 12
ranger = range(r)
ranger = magic_function(ranger)
Output:
print ranger
>>> ['00','01','02','03','04','05','06','07','08','09','10','11']
And if possible, my final goal is this:
I have a matrix of the form
numpy.array([[1,2,3],[4,5,6],[7,8,9]])
and I want to make a set of strings such that the first 2 characters are the row, the second two are the column and the third two are ’01’, and have matrix[row,col] of each one of these.
so the above values would look like such:
000001 since matrix[0,0] = 1
000101 since matrix[0,1] = 2
000101 since matrix[0,1] = 2
000201
000201
000201
etc
Answers:
Use string formatting and list comprehension:
>>> lst = range(11)
>>> ["{:02d}".format(x) for x in lst]
['00', '01', '02', '03', '04', '05', '06', '07', '08', '09', '10']
or format:
>>> [format(x, '02d') for x in lst]
['00', '01', '02', '03', '04', '05', '06', '07', '08', '09', '10']
Here’s my take on it:
>>> map('{:02}'.format, xrange(12))
['00', '01', '02', '03', '04', '05', '06', '07', '08', '09', '10', '11']
For your own enlightenment, try reading about the format string syntax.
Use string formatting:
>>> sr = []
>>> for r in range(11):
... sr.append('%02i' % r)
...
>>> sr
['00', '01', '02', '03', '04', '05', '06', '07', '08', '09', '10']
zfill does exactly what you want and doesn’t require you to understand an arcane mini-language as with the various types of string formatting. There’s a place for that, but this is a simple job with a ready-made built-in tool.
ranger = [str(x).zfill(2) for x in range(r)]
One liner in Python 3.x
>>> ranger = [f'{i:>02}' for i in range(1, 12)]
>>> print (ranger)
['01', '02', '03', '04', '05', '06', '07', '08', '09', '10', '11']
I just want to change a list (that I make using range(r)) to a list of strings, but if the length of the string is 1, tack a 0 on the front.
I know how to turn the list into strings using
ranger= map(str,range(r))
but I want to be able to also change the length of those strings.
Input:
r = 12
ranger = range(r)
ranger = magic_function(ranger)
Output:
print ranger
>>> ['00','01','02','03','04','05','06','07','08','09','10','11']
And if possible, my final goal is this:
I have a matrix of the form
numpy.array([[1,2,3],[4,5,6],[7,8,9]])
and I want to make a set of strings such that the first 2 characters are the row, the second two are the column and the third two are ’01’, and have matrix[row,col] of each one of these.
so the above values would look like such:
000001 since matrix[0,0] = 1
000101 since matrix[0,1] = 2
000101 since matrix[0,1] = 2
000201
000201
000201
etc
Use string formatting and list comprehension:
>>> lst = range(11)
>>> ["{:02d}".format(x) for x in lst]
['00', '01', '02', '03', '04', '05', '06', '07', '08', '09', '10']
or format:
>>> [format(x, '02d') for x in lst]
['00', '01', '02', '03', '04', '05', '06', '07', '08', '09', '10']
Here’s my take on it:
>>> map('{:02}'.format, xrange(12))
['00', '01', '02', '03', '04', '05', '06', '07', '08', '09', '10', '11']
For your own enlightenment, try reading about the format string syntax.
Use string formatting:
>>> sr = []
>>> for r in range(11):
... sr.append('%02i' % r)
...
>>> sr
['00', '01', '02', '03', '04', '05', '06', '07', '08', '09', '10']
zfill does exactly what you want and doesn’t require you to understand an arcane mini-language as with the various types of string formatting. There’s a place for that, but this is a simple job with a ready-made built-in tool.
ranger = [str(x).zfill(2) for x in range(r)]
One liner in Python 3.x
>>> ranger = [f'{i:>02}' for i in range(1, 12)]
>>> print (ranger)
['01', '02', '03', '04', '05', '06', '07', '08', '09', '10', '11']