Circularly shifting (or rotating) the digits the digits of a number in Python

问题

Suppose I have the following input:

How can I get the following output?

This is obtained by circularly shifting (or rotating) the digits of the input twice.

I have tried the following code:

number = 1234
bin(number >> 1)

but it is not producing the results I was expecting.

回答1:

The >> operator does a binary bitshift.

It moves the binary representation of 1234 on place to the right, discarding the rightmost (least significant) bit.

Therefore you code does not result in 3412.

You probably want string rotation instead:

>>> def rotr(string, n):
...     return string[n:] + string[:n]
... 
>>> rotr("1234", 2)
'3412'

You can also convert it back to an integer afterwards

>>> int('3412')
3412

回答2:

I would convert to string to be able to slice it.

number=1234
right_shift_no = 2
new_number = int(str(number)[right_shift_no:]+str(number)[:right_shift_no])

回答3:

Here's the lazy man's version:

>>> from collections import deque
>>> number = 1234
>>> d = deque(str(number))
>>> d.rotate(2)
>>> result = int(''.join(d))
>>> result
3412

回答4:

If you must stick with numbers (though I'd go with the string option first)

from math import log10, floor
s = 2  # digits to shift by
p = 10 ** s  # that as a power of 10
n = 1234
rhs = n // p  # right hand side of result (here 12)
rhs_n = floor(log10(rhs)) + 1  # number of digits in rhs
rhs + (n % p) * 10 ** rhs_n  # add other digits, n % p, shifted by size of rhs

and all together in a function

from math import log10, floor

def rotate(n, s):
    p = 10 ** s
    rhs = n // p
    return rhs + (n % p) * 10 ** (floor(log10(rhs)) + 1)

来源：https://stackoverflow.com/questions/52555368/circularly-shifting-or-rotating-the-digits-the-digits-of-a-number-in-python

标签

python

bit-shift

digits