CODEWITHSUNDEEP
pythonlist
Patriots299
Patriots299

Reputation: 365

Index and Lists - Index out of range

Shouldn't the following code print? 100 100

price = 100  # assigns 'price' reference to 100
price = [price] # creates a 'price' list with 1 element: [100]

for i in range(1, 3):
    print(price[0]) # prints 100
    price[i] = price[i - 1]
    price.append(price[i])
    print(price[i])

Getting a IndexError: list assignment index out of range error at line price[i] = price[i - 1], but the line right before prints 100 successfully. Shouldnt price[i] simply be getting assigned price[0] value?

Upvotes: 0

Views: 65

Answers (5)

smci
smci

Reputation: 33940

You're trying to append items to a list, or more precisely to initialize a list with repeated copies of something. Here are Pythonic ways to do that:

# Use a list comprehension
>>> price = [100 for _ in range(3)]
[100, 100, 100]

# Use itertools.repeat
>>> import itertools
>>> list(itertools.repeat(100, 3))
[100, 100, 100]

These are both faster than (repeatedly) doing append(), which is O(N), so repeatedly doing append() is O(N^2) on a long list, which gets very slow.

(Btw if you know a priori the list will have at least N elements, and N is large, you could initialize it to price = [None] * N and you get [None, None, None...]. Now you can directly assign to them. But, explicitly doing append is better practice for beginners.)

Upvotes: 2

GKE
GKE

Reputation: 1000

That will not print out:

100

100

You initialized a list with 1 element, the size of that list is 1. However your range starts at 1 for the for loop , what's really happening is:

price = 100  # assigns 'price' to 100
price = [price] # creates a 'price' list with 1 element: [100]

for i in range(1, 3): # The list is 0-indexed, meaning price[0] contains 100
    print(price[0]) # prints 100 as it should
    price[i] = price[i - 1] # i is 1, price[i] is not an assigned value, i.e: you never assigned price[1]
    price.append(price[i]) # This doesn't execute because an exception was thrown
    print(price[i]) # Neither does this

To get the result you're looking for, this would work:

price = [100] # creates a 'price' list with 1 element: [100]

for i in range(0, 2): # Start at index 0
    print(price[i]) # Print current index
    price.append(price[i]) # Append current value of price[i] to the price list

To ensure everything appended as you expected you can test it with len:

print(len(price))

Output:3

However, it is a preferred way of appending as @smci has shown in his/her answer.

Upvotes: 0

Isac Moura
Isac Moura

Reputation: 6888

You can't assign a value to a list directly why this list has note a previous size defined. You have to use append to add elements to the position you want.

Check:

# Check the value on our initial position
print(price[0])
for i in range(1, 3):
    price.append(price[i-1])
    print(price[i])

Upvotes: 0

Ryan Rapp
Ryan Rapp

Reputation: 1713

The problem is you are assigning at an index too great for the length of the array.

for i in range(1, 3):

This initializes i to 1. Since arrays are zero-indexed, and the length of your array is 1 on the first pass, you will hit an assignment out of range error at the line you mentioned (when i=1).

Here is a minimal example showing the issue:

my_array = ["foo"]
my_array[1] = "bar" # throws assignment out of range error

Upvotes: 0

C.Nivs
C.Nivs

Reputation: 13106

If you are just trying to append to a list, trying to do that with an index won't exactly work because that index isn't present in the list:

somelist = []
somelist[0] = 1
IndexError

So just use append

for i in range(1,3):
    price.append(price[i-1])

Upvotes: 0

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