Unexpected Behavior of Extend with a list in Python
Question:
I am trying to understand how extend works in Python and it is not quite doing what I would expect. For instance:
>>> a = [1, 2, 3]
>>> b = [4, 5, 6].extend(a)
>>> b
>>>
But I would have expected:
[4, 5, 6, 1, 2, 3]
Why is that returning a None instead of extending the list?
Answers:
list methods operate in-place for the most part, and return None.
>>> a = [1, 2, 3]
>>> b = [4, 5, 6]
>>> b.extend(a)
>>> b
[4, 5, 6, 1, 2, 3]
The extend() method appends to the existing array and returns None. In your case, you are creating an array — [4, 5, 6] — on the fly, extending it and then discarding it. The variable b ends up with the return value of None.
extend extends its operand, but doesn’t return a value. If you had done:
b = [4, 5, 6]
b.extend(a)
Then you would get the expected result.
Others have pointed out many list methods, particularly those that mutate the list, return None rather than a reference to the list. The reason they do this is so that you don’t get confused about whether a copy of the list is made. If you could write a = b.extend([4, 5, 6]) then is a a reference to the same list as b? Was b modified by the statement? By returning None instead of the mutated list, such a statement is made useless, you figure out quickly that a doesn’t have in it what you thought it did, and you learn to just write b.extend(...) instead. Thus the lack of clarity is removed.
I had this problem and while the other answers provide correct explanations, the solution/workaround I liked isn’t here. Using the addition operator will concatenate lists together and return the result. In my case I was bookkeeping color as a 3-digit list and opacity as a float, but the library needed color as a 4 digit list with opacity as the 4th digit. I didn’t want to name a throwaway variable, so this syntax suited my needs:
color = [1, 1, 0]
opacity = 0.75
plot.setColor(color + [opacity])
This creates a new list for opacity on the fly and a new list after the concatenation, but that’s fine for my purposes. I just wanted compact syntax for extending a list with a float and returning the resulting list without affecting the original list or float.
I am trying to understand how extend works in Python and it is not quite doing what I would expect. For instance:
>>> a = [1, 2, 3]
>>> b = [4, 5, 6].extend(a)
>>> b
>>>
But I would have expected:
[4, 5, 6, 1, 2, 3]
Why is that returning a None instead of extending the list?
list methods operate in-place for the most part, and return None.
>>> a = [1, 2, 3]
>>> b = [4, 5, 6]
>>> b.extend(a)
>>> b
[4, 5, 6, 1, 2, 3]
The extend() method appends to the existing array and returns None. In your case, you are creating an array — [4, 5, 6] — on the fly, extending it and then discarding it. The variable b ends up with the return value of None.
extend extends its operand, but doesn’t return a value. If you had done:
b = [4, 5, 6]
b.extend(a)
Then you would get the expected result.
Others have pointed out many list methods, particularly those that mutate the list, return None rather than a reference to the list. The reason they do this is so that you don’t get confused about whether a copy of the list is made. If you could write a = b.extend([4, 5, 6]) then is a a reference to the same list as b? Was b modified by the statement? By returning None instead of the mutated list, such a statement is made useless, you figure out quickly that a doesn’t have in it what you thought it did, and you learn to just write b.extend(...) instead. Thus the lack of clarity is removed.
I had this problem and while the other answers provide correct explanations, the solution/workaround I liked isn’t here. Using the addition operator will concatenate lists together and return the result. In my case I was bookkeeping color as a 3-digit list and opacity as a float, but the library needed color as a 4 digit list with opacity as the 4th digit. I didn’t want to name a throwaway variable, so this syntax suited my needs:
color = [1, 1, 0]
opacity = 0.75
plot.setColor(color + [opacity])
This creates a new list for opacity on the fly and a new list after the concatenation, but that’s fine for my purposes. I just wanted compact syntax for extending a list with a float and returning the resulting list without affecting the original list or float.