为什么在进入for循环后立即将p [1]的值设为'0'？

Question

The python function is here:

def rot(p):
    q = p

    print "P val: ", p, "len(p): ", len(p)
    for i in range(len(p)):
        if (i == (len(p)-1)):
            q[0] = p[i]
            print "P val if: ", p, "p(i) : " , p[i]

        else:
            q[i+1] = p[i]
            print "P val else: ", p, "p(i) : " , p[i]


    return q

And the output is here:

P val:  [0, 1, 0, 0, 0] len(p):  5
P val else:  [0, 0, 0, 0, 0] p(i) :  0
P val else:  [0, 0, 0, 0, 0] p(i) :  0
P val else:  [0, 0, 0, 0, 0] p(i) :  0
P val else:  [0, 0, 0, 0, 0] p(i) :  0
P val if:  [0, 0, 0, 0, 0] p(i) :  0

*The question is, why is the list p changing its p[1] value to 0 after entering the for loop?? *

Answer 1

Because q and p reference the same list. Thus any change made to the list through variable q will also be reflected in variable p because both represent the same underlying list. For example:

>>> p = [1, 2, 3]
>>> q = p
>>> q
[1, 2, 3]
>>> q[0] = 0
>>> q
[0, 2, 3]
>>> p
[0, 2, 3]

This shows that updates made through q are also visible in p . This is because p and q are bound to the same list variable:

>>> p is q
True

To fix it

Work on a copy of p using slice notation:

def rot(p):
    q = p[:]    # copy p
    ...         # rest of your function...


>>> l = [0, 1, 0, 0, 0]
>>> rot(l)
P val:  [0, 1, 0, 0, 0] len(p):  5
P val else:  [0, 1, 0, 0, 0] p(i) :  0
P val else:  [0, 1, 0, 0, 0] p(i) :  1
P val else:  [0, 1, 0, 0, 0] p(i) :  0
P val else:  [0, 1, 0, 0, 0] p(i) :  0
P val if:  [0, 1, 0, 0, 0] p(i) :  0
[0, 0, 1, 0, 0]
>>> l
[0, 1, 0, 0, 0]

l is unchanged.