list.insert()¶

We can use the function list.insert() in order to insert elements into a list. Consider the following list:

In [1]:
L = [42, 43, 45, 46, 47]

Suppose we want to insert 99 between 43 and 45. In order to do that, we use list.insert(). L.insert(ind, e) means "insert e before index ind in the list L"

In [2]:
L = [42, 43, 45, 46, 47]
L.insert(2, 99)
In [3]:
print(L)

[42, 43, 99, 45, 46, 47]

We inserted 99 before 45, which used to be at index 4.

How can we insert an element at the beginning? Insert before index 0:

In [4]:
L = [42, 43, 45, 46, 47]
L.insert(0, 9)
print(L)

[9, 42, 43, 45, 46, 47]

How about inserting an element at the very end? Technically, any large-enough index will work:

In [5]:
L = [42, 43, 45, 46, 47]
L.insert(10000000, 99)
print(L)

[42, 43, 45, 46, 47, 99]

"Insert before index 10000000" just means "insert at the end" here. It's much nicer, however, to insert before index len(L). The index len(L)-1 is the index of the last element in the list (the indices are 0, 1, 2, 3..., len(L), for a total of len(L) indices). That means that inserting before index len(L) is the same as inserting the element at the very end:

In [6]:
L = [42, 43, 45, 46, 47]
L.insert(len(L), 99)
print(L)

[42, 43, 45, 46, 47, 99]

list.append()¶

"Inserting at the end of the list" has a name -- appending. We can (and should) use list.append to append to the end of a list.

In [7]:
L = [42, 43, 45, 46, 47]
L.append(99)
print(L)

[42, 43, 45, 46, 47, 99]

list.index()¶

Suppose we want to find the index of a particular element of L -- to find the location of the element equal to e in the list L. This can be done using list.index().

In [8]:
L = [4, 42, 35, 20]
L.index(20)
Out[8]:
3

20 is found at index 3 of L. We can check this:

In [9]:
L[L.index(20)]
Out[9]:
20

Of course, this would work for any element found in the list L, since L.index(e) is the index of an element equal to e!

In [10]:
L[L.index(4)]
Out[10]:
4

What if the list L has more than one element that is equal to e? In that case, L.index(e) will return the smallest index of an element in L equal to e:

In [11]:
L = [4, 5, 7, 5, 20]
L.index(5)
Out[11]:
1

What if the element is not found in L? In that case, we get an error, and the program crashes (unless we use exception handling, which we haven't covered...)

In [12]:
L.index(42)

---------------------------------------------------------------------------
ValueError                                Traceback (most recent call last)
<ipython-input-12-71974f648f1f> in <module>()
----> 1 L.index(42)

ValueError: 42 is not in list

in, not in¶

We can use in and not in in order to check whether an element equal to e is in L or not.

e in L is True iff one of the elements of L is equal to e e not in L is True iff none of the elements of L are equal to e

Of course, e not in L and not (e in L) are just the same.

Here are a few examples:

In [13]:
4 in [5, 6, 10]
Out[13]:
False
In [14]:
3 in [10, 3, 1]
Out[14]:
True
In [15]:
3 not in [10, 3, 1]
Out[15]:
False

Finding the index of an element which may be present in L¶

We are now in a position to try to find the index of an element which may or may not be present in L:

In [16]:
L = [10, 5, 4, 20, 4]
e = 3
if e in L:
    print("The index of the first occurence of", e, "in L is", L.index(e))
else:
    print(e, "is not in L")

3 is not in L
In [17]:
L = [10, 5, 4, 20, 4]
e = 4
if e in L:
    print("The index of the first occurence of", e, "in L is", L.index(e))
else:
    print(e, "is not in L")

The index of the first occurence of 4 in L is 2