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I tried assigning three expressions in Python but I'm an unexpected result.

Let's begin with simple swaps. You probably know the result of this assignment:

A = [10, 11, 12]
p = 0
A[p + 1], A[p] = A[p], A[p + 1]    # <--
print(A)

The result is (as expected):

[11, 10, 12]

Now I wanted to be a little more daring, so tried this assignment: 

A = [10, 11, 12]
p = 0
p, A[p + 1], A[p] = p + 1, A[p], A[p + 1]   # <--
print(A)

I thought that the result would be: 

[10, 12, 11]

However, the result was:

[10, 11, 10]

Which is unexpected!

I read Python documentation regarding assignments:

Although the definition of assignment implies that overlaps between the left-hand side and the right-hand side are ‘simultaneous’ (for example a, b = b, a swaps two variables), overlaps within the collection of assigned-to variables occur left-to-right, sometimes resulting in confusion. For instance, the following program prints [0, 2]:

x = [0, 1]
i = 0
i, x[i] = 1, 2         # i is updated, then x[i] is updated
print(x)

I did not get similar results for my swap. I don't understand the logic behind my swap. What's going on?

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  • 1
    The output you got was the output you were supposed to get! Commented Apr 2, 2019 at 15:02
  • 1
    Not directly related to your question, but docs.python.org/2.0/ref/assignment.html is considerably out of date. Refer to docs.python.org/3/reference/… or docs.python.org/2.7/reference/… for 3.X and 2.7 documentation respectively. Commented Apr 2, 2019 at 15:03
  • 1
    This is somewhat similar to writing C code like x[i++]=i++. Don't do it; even if the behavior is well defined, it may not be especially clear. Increment p, then perform the swap. Commented Apr 2, 2019 at 15:19

1 Answer 1

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4

You can see it as a shorthand for consecutive assignments from left to right using temporary variables:

p, A[p + 1], A[p] = p + 1, A[p], A[p + 1]

is equivalent to

temp1  = p + 1  # 1
temp2  = A[p]   # 10  (A[0])
temp3  = A[p+1] # 11  (A[1])
p      = temp1   # p    = 1
A[p+1] = temp2   # A[2] = 10
A[p]   = temp3   # A[1] = 11

so A = [10,11,10]

If you placed p at the end of the list, you probably would get closer to your expected result:

A[p + 1], A[p], p = A[p], A[p + 1], p + 1

A is now [11,10,12]
P is now 1

In other words, post-increment is possible but pre-increment will not work in this kind of scenario (where the pre-incremented index is used in the source data)

You could do it by manually computing the offset in the source data but that would be somewhat counter-intuitive:

p, A[p+1], A[p] = p+1, A[p+1], A[p+2]
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4 Comments

Relevant documentation for whom it may interest: docs.python.org/3/reference/expressions.html#evaluation-order
Thanks for editing my question and thank for your relevant document. @TrebledJ
thanks for great answer.I like you. Nice answer. Is it possible to explain those two things: 1. pre-incerment and post-incerment 2.counter-intuitive @alain-t
Pre-increment is when you increase the index (p) before using it. post-increment is when you increase the index after using it. What I felt is counter intuitive is that , in my last (pre-increment) example, the assignment's left side assumes a value of p that has been incremented but the right side of the assignment indexes the list with p before it is incremented (despite having p+1 appear before the A[..] source data).

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