问题
I have code like this:
squares = []
for value in range(1, 5):
squares.insert(value+1,value**2)
print(squares)
print(squares[0])
print(len(squares))
And the output is :
[1, 4, 9, 16]
1
4
So even if I ask python to insert '1' at index '2', it inserts at the first available index. So how does 'insert' makes the decision?
回答1:
From the Python3 doc:
list.insert(i, x)
Insert an item at a given position. The first argument is the index of the element before which to insert, so a.insert(0, x) inserts at the front of the list, and a.insert(len(a), x) is equivalent to a.append(x).
What is not mentionned is that you can give an index that is out of range and Python will then append to the list.
If you dig into the Python implementation you find the following in the ins1
function that does the insertion:
if (where > n)
where = n;
So basically Python will max out your index to the length of the list.
回答2:
Basically, it's similar to append, except that it allows you to insert a new item at any position in the list, as opposed to just at the end.
来源:https://stackoverflow.com/questions/47614538/what-does-list-insert-in-actually-do-in-python