CODEWITHSUNDEEP
pythonpython-3.x
temporary_user_name
temporary_user_name

Reputation: 37018

What's going on in this code?

x,y,z = [1,2,3], [4,5,6], [7,8,9]
for a,b,c in x,y,z:
    print(a,b,c)

The output is :

1 2 3
4 5 6
7 8 9

I can't mentally navigate whatever logic is going on here to produce this output. I am aware of the zip function to make this code behave in the way I clearly intend it to; but I'm just trying to understand why it works this way when you don't use the zip function.

Is this a deliberate functionality, a feature, that you can successively iterate through multiple lists this way? Sort of?

Upvotes: 7

Views: 250

Answers (7)

neves
neves

Reputation: 39153

It is an unusual python thing. The implicit creation of tuple.

Here, you create an anonymous tuple at the right

x,y,z = [1,2,3], [4,5,6], [7,8,9]

This is a similar code:

a, b = 1, 2

that is the same:

a, b =  (1, 2)

or 

   a = 1
   b = 2

It allows a common python trick (idiom). You can swap values without an temporary variable:

a, b = b, a

The same happens interacting the key and values of a dictionary:

for i, j in  my_dict.items():
   print i, j

In your code, the another temporary tuple is being created in the for loop:

for a,b,c in (x,y,z):
    print(a,b,c)

That means:

for a, b,c in ([1,2,3], [4,5,6], [7,8,9]):
    print(a,b,c)

In other words: rewrite this code for something more legible. Python isn't following its own mantra: Explicit is better than implicit..

BTW, see the fun Python Zen typing import this in a Python shell.

Upvotes: 2

Bi Rico
Bi Rico

Reputation: 25813

Oh man this is a mess. This is simply too much use of python's iterable unpacking. The statement a, b, c = iterable simply assigns the elements of iterable to the variables a, b, and c. In this case iterable must have 3 elements.

First you have:

x,y,z = [1,2,3], [4,5,6], [7,8,9]
# Which is:
x = [1, 2, 3]
y = [4, 5, 6]
z = [7, 8, 9]

then you have:

for a, b, c in x, y, z:
    print(a, b, c)
# Which is:
temp = (x, y, z)
for item in temp:
    a = item[0]
    b = item[1]
    c = item[2]
    print(a, b, c)

One more thing to note is that the statement mytuple = 1, 2, 3 is the same as mytuple = (1, 2, 3).

Upvotes: 6

Tim Peters
Tim Peters

Reputation: 70582

You have good answers already, but I think considering this equivalent variation will help to make it clearer:

x,y,z = [1,2,3], [4,5,6], [7,8,9]
for t in x,y,z:
    a, b, c = t
    print(a,b,c)

You're not surprised that t is successively bound to x, y and z, right? Exactly the same thing is happening in your original code, except that the:

a, b, c = t

part isn't as obvious.

Upvotes: 8

inspectorG4dget
inspectorG4dget

Reputation: 113905

x,y,z = [1,2,3], [4,5,6], [7,8,9]

is the same as saying

x = [1,2,3]
y = [4,5,6]
z = [7,8,9]

Next

for a,b,c in x,y,z:

is equivalent to

for a,b,c in [x,y,z]:

which just says "take x, y and z in turn. Assign their contents, respectively to a, b and c; i.e. a=x[0], b=x[1], c=x[2].
Ultimately, this turns into

a,b,c = x
print(a,b,c)
a,b,c = y
print(a,b,c)
a,b,c = z
print(a,b,c)

Upvotes: 2

Free Monica Cellio
Free Monica Cellio

Reputation: 2290

Everything is explained here: http://docs.python.org/2/tutorial/datastructures.html#tuples-and-sequences

The first part is realizing that commas implicitly create tuples. So the first line is equivalent to:

x,y,z = ([1,2,3], [4,5,6], [7,8,9])

This also means that your for loop is equivalent to:

for a,b,c in ([1,2,3], [4,5,6], [7,8,9]):

The second part to understand is sequence unpacking. This means that if you assign a sequence of length n to n variables, Python assigns the items in the sequence appropriately. So the first part is effectively:

x = [1,2,3]
y = [4,5,6]
z = [7,8,9]

And the for loop is the same as:

for t in (x,y,z):
   a = t[0]
   b = t[1]
   c = t[2]
   print(a,b,c)

Upvotes: 1

Dennis Flagg
Dennis Flagg

Reputation: 664

It looks like you have 3 arrays

x = [1,2,3]
y = [4,5,6]
z = [7,8,9]

and a,b,c represent the elements in each array. So it looks like the for loop is iterating over 3 arrays and mapping the elements to a,b, and c. Then printing them out. Then again, I don't know python

Upvotes: 0

user764357
user764357

Reputation:

Its pretty straight forward code really.

  1. This assigns the three lists to x, y, and z.

    x,y,z = [1,2,3], [4,5,6], [7,8,9]

  2. This creates a tuple of (x,y,z) and will iterate over each element.

    for a,b,c in x,y,z:

    However, the a,b,c means that the iterables are expected to have 3 objects

  3. Then, this prints a,b and c.

        print(a,b,c)

If you want to see what is happening, I'd suggest altering one of the elements in y:

x,y,z = [1,2,3], [3,4,5,6], [7,8,9]
for a,b,c in x,y,z:
    print(a,b,c)

1 2 3
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: too many values to unpack

Or, by removing one from x:

x,y,z = [1,2], [4,5,6], [7,8,9]
for a,b,c in x,y,z:
    print(a,b,c)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
ValueError: need more than 2 values to unpack

Upvotes: 2

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