Il 20/02/2021 15:40, C W ha scritto:
Hello everyone,
I'm curious if there is a way take number and back each digit by 3 ?
2342 becomes 9019
8475 becomes 5142
5873 becomes 2540
The tricky part is that 2 becomes 9, not -1.
Here's my toy example and what I attempted,
test_series = pd.Series(list(['2342', '8475', '5873']))
test_series
0 2342
1 8475
2 5873
dtype: object
test_series.str.split('')
[, 2, 3, 4, 2, ]
[, 8, 4, 7, 5, ]
[, 5, 8, 7, 3, ]
dtype: object
What a good approach to this? Is there a method or function that should be
handling this?
Thanks so much!
Mike
>>> num='0123456789'
>>> n=8475
>>> sn = ''
>>> for x in str(n):
sn += num[(int(x) - 3) % 10]
>>> int(sn)
5142
>>>
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