Question or problem about Python programming:
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 = 
for n in l: i += n print(i)
Output: [‘t’, ‘a’, ‘b’, ‘l’, ‘e’]
for n in l: i = i + n print(i)
What is causing this strange error?
How to solve the problem:
They don’t have to be the same.
+ 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
__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
Demo 1: using
>>> 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
>>> l =  >>> l = l + 'table' [...] TypeError: can only concatenate list (not "str") to list
+= 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
>>> 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]
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
Ned Batchelder’s challenge to find this in the docs
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.
If in the second case, you wrap a list around
n to avoid errors:
for n in l: i = i + [n] print(i)
So they are different operations.