# Floating Point, Overflow and NaN — NumPy / Pandas / scikit-learn

Source: https://www.skillbyai.com/en/numpy-pandas-sklearn/c-float

> Numbers that surprise.

## Know the limits of the types

Floating-point numbers are binary approximations: `0.1 + 0.2 != 0.3`, and adding very large and very small numbers loses precision. Compare floats with `np.isclose`. Fixed-size integers **wrap around** on overflow (uint8 200 + 100 gives 44) without an error. Division by zero gives `inf` with a warning, and **NaN** (not a number) is not equal to itself and spreads through calculations unless you use nan-aware functions such as `np.nanmean`.

## Precision, overflow and NaN behaviour, run

I ran this with Python 3.12.3 and NumPy 2.5.3. Each line shows a classic surprise. Warnings were silenced for the run; NumPy normally prints a RuntimeWarning for division by zero and the mean containing NaN.

```python
import numpy as np

print(0.1 + 0.2 == 0.3)
print(np.isclose(0.1 + 0.2, 0.3))
a = np.array([1e16, 1.0, -1e16])
print(a.sum(), "vs exact 1.0")
print(np.array([200], dtype=np.uint8) + np.uint8(100))
print(np.array([1, 2, 3]) / 0)
print(np.nan == np.nan, np.isnan(np.nan))
print(np.mean([1.0, np.nan, 3.0]), np.nanmean([1.0, np.nan, 3.0]))
```

Output:

```
False
True
0.0 vs exact 1.0
[44]
[inf inf inf]
False True
nan 2.0
```

## Use int64 and float64 by default

Choose smaller dtypes only for memory reasons and after checking that values fit.

**Quiz:** What is np.mean([1.0, np.nan, 3.0])?

- [ ] 2.0
- [x] nan
- [ ] 4.0
- [ ] An error

*Answer:* nan. NaN spreads; use np.nanmean to ignore it.