57

One block of code works but the other does not. Which would make sense except the second block is the same as the first only with an operation written in shorthand. They are practically the same operation.

l = ['table']
i = []

Version 1

for n in l:
    i += n
print(i)

Output: ['t', 'a', 'b', 'l', 'e']

Version 2

for n in l:
    i = i + n
print(i)

Output:

TypeError: can only concatenate list (not "str") to list

What is causing this strange error?

Jonathon Reinhart
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Luke Bakare
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    No, not the same for lists. `+=` extends a list. `+` concatenates two lists into a new list. – khelwood Oct 10 '18 at 19:58
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    Ideally, if you're using this idea in code, it's probably safer to use the `append()` and `extend()` methods for adding elements and concatenating lists, respectively, to avoid ambiguity like this. – Green Cloak Guy Oct 10 '18 at 20:01

3 Answers3

81

They don't have to be the same.

Using the + operator calls the method __add__ while using the += operator calls __iadd__. It is completely up to the object in question what happens when one of these methods is called.

If you use x += y but x does not provide an __iadd__ method (or the method returns NotImplemented), __add__ is used as a fallback, meaning that x = x + y happens.

In the case of lists, using l += iterable actually extends the list l with the elements of iterable. In your case, every character from the string (which is an iterable) is appended during the extend operation.

Demo 1: using __iadd__

>>> l = []
>>> l += 'table'
>>> l
['t', 'a', 'b', 'l', 'e']

Demo 2: using extend does the same

>>> l = []
>>> l.extend('table')
>>> l
['t', 'a', 'b', 'l', 'e']

Demo 3: adding a list and a string raises a TypeError.

>>> l = []
>>> l = l + 'table'
[...]
TypeError: can only concatenate list (not "str") to list

Not using += gives you the TypeError here because only __iadd__ implements the extending behavior.

Demo 4: common pitfall: += does not build a new list. We can confirm this by checking for equal object identities with the is operator.

>>> l = []
>>> l_ref = l # another name for l, no data is copied here
>>> l += [1, 2, 3] # uses __iadd__, mutates l in-place
>>> l is l_ref # confirm that l and l_ref are names for the same object
True
>>> l
[1, 2, 3]
>>> l_ref # mutations are seen across all names
[1, 2, 3]

However, the l = l + iterable syntax does build a new list.

>>> l = []
>>> l_ref = l # another name for l, no data is copied here
>>> l = l + [1, 2, 3] # uses __add__, builds new list and reassigns name l
>>> l is l_ref # confirm that l and l_ref are names for different objects
False
>>> l
[1, 2, 3]
>>> l_ref
[]

In some cases, this can produce subtle bugs, because += mutates the original list, while
l = l + iterable builds a new list and reassigns the name l.

BONUS

Ned Batchelder's challenge to find this in the docs

timgeb
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3

If in the second case, you wrap a list around n to avoid errors:

for n in l:
    i = i + [n]
print(i)

you get

['table']

So they are different operations.

Jake
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3

No.

7.2.1. Augmented assignment statements:

An augmented assignment expression like x += 1 can be rewritten as x = x + 1 to achieve a similar, but not exactly equal effect. In the augmented version, x is only evaluated once. Also, when possible, the actual operation is performed in-place, meaning that rather than creating a new object and assigning that to the target, the old object is modified instead.

msc
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