IndexError: Index Out of Bounds - How to Fix It
Answer
This error means you're trying to access an array element at a position that doesn't exist. Fix it by checking the array's shape with .shape before indexing. Remember that NumPy uses zero-based indexing, so an array of length 5 has valid indices 0 through 4.
Why This Happens
NumPy arrays have fixed dimensions. If your array has 10 elements and you try to access index 10, it fails because valid indices are 0-9. This commonly happens with off-by-one errors, hardcoded indices that don't match the actual data size, or after operations that change array dimensions.
Solution
The rule: always check .shape or len() before indexing. Use negative indices for accessing elements from the end, and use slicing when you want to avoid errors on out-of-bounds access.
import numpy as np
arr = np.array([10, 20, 30, 40, 50])
# โ Problematic: index 5 doesn't exist in a 5-element array
arr[5]
# IndexError: index 5 is out of bounds for axis 0 with size 5
# โ
Fixed: check shape first
print(arr.shape) # (5,)
print(len(arr)) # 5
arr[4] # last valid index
# โ
Fixed: use negative indexing for last elements
arr[-1] # last element (50)
arr[-2] # second to last (40)
# โ
Fixed: use bounds checking before access
idx = 5
if idx < len(arr):
value = arr[idx]
else:
value = None # or handle the error
# โ
For 2D arrays, check both dimensions
arr_2d = np.array([[1, 2, 3], [4, 5, 6]])
print(arr_2d.shape) # (2, 3)
arr_2d[1, 2] # valid: row 1, column 2
# arr_2d[2, 0] # invalid: only 2 rows (0 and 1)
# โ
Safe slicing (never raises IndexError)
arr[10:20] # returns empty array if out of boundsBetter Workflow
In Zerve, array shapes are always visible in block outputs. No need for print statements to debug. When an IndexError appears, check the upstream block to see the actual shape, create a new block to test safe indexing approaches, and verify the fix instantly. Zerve's variable inspector shows shape and type metadata automatically, so you always know what you're working with before you index into it.
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