You wrote arr[5, 10] and NumPy threw IndexError: index 10 is out of bounds for axis 1 with size 8. The error names exactly which axis and what size, so debugging is mechanical: print(arr.shape) and verify your indexes fit.
📌 Quick answer: print(arr.shape) first. The error message tells you which axis (0=rows, 1=columns for 2D, higher for tensors) and what valid range is. arr[5, 10] on shape (6, 8) means axis 1 (size 8) does not have index 10. Either bound your indexes or check why the array is smaller than expected.
Cause 1: Hardcoded index larger than array
The most common: your code assumed a fixed size but data is variable.
import numpy as np
arr = np.zeros((6, 8))
arr[5, 10] # ❌ index 10 out of bounds for axis 1 with size 8
# Safe pattern
row, col = 5, 10
if row < arr.shape[0] and col < arr.shape[1]:
val = arr[row, col]
else:
val = NoneCause 2: Array smaller than expected after filter
You filtered an array with a boolean mask and tried to access by old index.
arr = np.array([10, 20, 30, 40, 50])
filtered = arr[arr > 25] # shape (3,): [30, 40, 50]
filtered[4] # ❌ index 4 out of bounds for size 3Cause 3: Axis confusion in higher-dim arrays
For a 3D tensor of shape (batch, height, width), arr[0, 5, 10] means batch 0, row 5, col 10. Mixing up the order is the bug.
import numpy as np
batch = np.zeros((4, 28, 28)) # 4 images, 28x28 each
batch[5, 0, 0] # ❌ axis 0 (batch) has size 4
batch[0, 5, 10] # ✓ batch 0, row 5, col 10Cause 4: Reshaping changed dimensions
arr.reshape(-1) flattens to 1D. Now arr[i, j] raises because there is no axis 1.
arr = np.zeros((4, 8)) # 2D
flat = arr.reshape(-1) # 1D, shape (32,)
flat[0, 5] # ❌ too many indices for array: array is 1-dimensional
flat[5] # ✓Cause 5: Negative index too negative
Negative indexing wraps: arr[-1] is the last element. But arr[-100] on a 5-element array fails.
arr = np.array([1, 2, 3, 4, 5])
arr[-1] # ✓ 5 (last)
arr[-5] # ✓ 1 (first via wrap)
arr[-6] # ❌ index -6 out of bounds for size 5Prevention
- Always
print(arr.shape)before indexing in new code - Use
np.clip(idx, 0, arr.shape[0]-1)to bound indexes to valid range - Use boolean masks instead of integer indexing when filtering
- Use
arr[..., -1]for “last along last axis” — works for any rank
Related Guides
- List index out of range (full guide)
- String index out of range
- All IndexError fixes
- Python Tutorial hub
Frequently Asked Questions
What does ‘index N is out of bounds for axis N with size N’ mean in NumPy?
You tried to access an array element at a position that doesn’t exist. The error names exactly which axis (0=rows, 1=cols for 2D) and what the valid range is. arr[5, 10] on shape (6, 8) fails because axis 1 has size 8 (valid indexes 0-7).
How do I find out an array’s dimensions before indexing?
Use arr.shape (returns a tuple), arr.ndim (number of axes), arr.size (total elements). print(arr.shape) is the first debugging step for any IndexError on a NumPy array.
Why does flat[0, 5] fail when arr[0, 5] worked?
Because flat = arr.reshape(-1) is 1-dimensional. A 1D array only supports single-axis indexing. Use flat[5] or reshape back to 2D: flat.reshape(4, 8)[0, 5].
How do I safely index when the array shape varies?
Use np.clip(idx, 0, arr.shape[0]-1) to bound any index. Or use boolean masks (arr[arr > threshold]) which never raise IndexError. Or use try/except IndexError with a sensible fallback.
What’s the difference between IndexError and ValueError in NumPy?
IndexError: index out of bounds for an existing axis. ValueError: shape mismatch in operations (broadcast, dot product). They’re distinct, but both common in tensor code. Both go away once you print shapes first and verify operations match.
Adrian Mercurio
Full-Stack Developer at PIES IT Solution
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