# What does x[x &lt; 2] = 0 mean in Python?

I came across some code with a line similar to

```    x[x<2]=0
```

Playing around with variations, I am still stuck on what this syntax does.

Examples:

```    >>> x = [1,2,3,4,5]
>>> x[x<2]
1
>>> x[x<3]
1
>>> x[x>2]
2
>>> x[x<2]=0
>>> x
[0, 2, 3, 4, 5]
```

This only makes sense with NumPy arrays. The behavior with lists is useless, and specific to Python 2 (not Python 3). You may want to double-check if the original object was indeed a NumPy array (see further below) and not a list.

But in your code here, x is a simple list.

Since

```    x < 2
```

is False i.e 0, therefore

`x[x<2]` is `x`

`x` gets changed.

Conversely, `x[x>2]` is `x[True]` or `x`

So, `x` gets changed.

Why does this happen?

The rules for comparison are:

1. When you order two strings or two numeric types the ordering is done in the expected way (lexicographic ordering for string, numeric ordering for integers).

2. When you order a numeric and a non-numeric type, the numeric type comes first.

3. When you order two incompatible types where neither is numeric, they are ordered by the alphabetical order of their typenames:

So, we have the following order

numeric < list < string < tuple

See the accepted answer for How does Python compare string and int?.

If x is a NumPy array , then the syntax makes more sense because of boolean array indexing. In that case, `x < 2` isn't a boolean at all; it's an array of booleans representing whether each element of `x` was less than 2. `x[x < 2] = 0` then selects the elements of `x` that were less than 2 and sets those cells to 0. See Indexing.

```    >>> x = np.array([1., -1., -2., 3])
>>> x < 0
array([False,  True,  True, False], dtype=bool)
>>> x[x < 0] += 20   # All elements < 0 get increased by 20
>>> x
array([  1.,  19.,  18.,   3.]) # Only elements < 0 are affected
```

From: stackoverflow.com/q/36603042