Why does this iterative list-growing code give IndexError: list assignment index out of range?
PythonListExceptionPython Problem Overview
Please consider the following code:
i = [1, 2, 3, 5, 8, 13]
j = []
k = 0
for l in i:
j[k] = l
k += 1
print j
The output (Python 2.6.6 on Win 7 32-bit) is:
> Traceback (most recent call last):
> j[k] = l IndexError: list assignment index out of range
I guess it's something simple I don't understand. Can someone clear it up?
Python Solutions
Solution 1 - Python
j is an empty list, but you're attempting to write to element [0] in the first iteration, which doesn't exist yet.
Try the following instead, to add a new element to the end of the list:
for l in i:
j.append(l)
Of course, you'd never do this in practice if all you wanted to do was to copy an existing list. You'd just do:
j = list(i)
Alternatively, if you wanted to use the Python list like an array in other languages, then you could pre-create a list with its elements set to a null value (None in the example below), and later, overwrite the values in specific positions:
i = [1, 2, 3, 5, 8, 13]
j = [None] * len(i)
#j == [None, None, None, None, None, None]
k = 0
for l in i:
j[k] = l
k += 1
The thing to realise is that a list object will not allow you to assign a value to an index that doesn't exist.
Solution 2 - Python
Your other option is to initialize j:
j = [None] * len(i)
Solution 3 - Python
Do j.append(l) instead of j[k] = l and avoid k at all.
Solution 4 - Python
You could also use a list comprehension:
j = [l for l in i]
or make a copy of it using the statement:
j = i[:]
Solution 5 - Python
j.append(l)
Also avoid using lower-case "L's" because it is easy for them to be confused with 1's
Solution 6 - Python
I think the Python method insert is what you're looking for:
Inserts element x at position i. list.insert(i,x)
array = [1,2,3,4,5]
# array.insert(index, element)
array.insert(1,20)
print(array)
# prints [1,20,2,3,4,5]
Solution 7 - Python
You could use a dictionary (similar to an associative array) for j
i = [1, 2, 3, 5, 8, 13]
j = {} #initiate as dictionary
k = 0
for l in i:
j[k] = l
k += 1
print(j)
will print :
{0: 1, 1: 2, 2: 3, 3: 5, 4: 8, 5: 13}
Solution 8 - Python
One more way:
j=i[0]
for k in range(1,len(i)):
j = numpy.vstack([j,i[k]])
In this case j will be a numpy array
Solution 9 - Python
Maybe you need extend()
i=[1,3,5,7]
j=[]
j.extend(i)