How to reverse "groups" (not list) of strings? For example I would like to convert "123456789" into "321654987" given that the "group" size equals to 3.
Followings are my code but it turns out to print empty strings:
string = "12345678901234567890"
new_string = ""
for i in range(0, len(string), 5):
new_string = new_string + string[i:i+5:-1]
print (new_string)
Thanks for any input and advice.
When using a negative step, swap the start and end values too:
new_string += string[i + 4:i - 1 if i else None:-1]
Note that the end value is excluded, and None should be used to include the first character (-1 would slice from the end again and thus string[4:-1:-1] would be empty).
Demo:
>>> string = "12345678901234567890"
>>> new_string = ""
>>> for i in range(0, len(string), 5):
... new_string = new_string + string[i + 4:i - 1 if i else None:-1]
...
>>> new_string
'54321098765432109876'
However, it may be easier to first slice, then reverse, and save yourself the headaches of such slicings:
new_string += string[i:i + 4][::-1]
A simple one liner for your problem would go something like.
>>> from itertools import chain
>>> testString = "123456789"
>>> group = 3
>>> "".join(chain.from_iterable([reversed(elem) for elem in zip(*[iter(testString)]*group)]))
'321654987'
Another Example ->
>>> testString = "12345678901234567890"
>>> group = 5
>>> "".join(chain.from_iterable([reversed(elem) for elem in zip(*[iter(testString)]*group)]))
'54321098765432109876'
This can be done in one line:
newline = ''.join([string[i:i+n:1][::-1] for i in range(0, len(string), n)])
Where n is the size of the "group". In your case, it would be 3. An Explanation is below:
i in range(0, len(string), n) is responsible for counting by three
string[i+i+n:1] is responsible for getting each chunk
[::-1] will reverse each chunk
''.join() joins each now-reversed chunk back into a single string.
Another option is:
''.join([string[i+n-1:i-1 if i else None:-1] for i in range(0, len(string), n)])
Where the selecton and reversing is done in one step.
