What does x[x < 2] = 0 mean in Python?

PythonPython 2.7Numpy

Python Problem Overview


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]

Python Solutions


Solution 1 - Python

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[0]

x[0] gets changed.

Conversely, x[x>2] is x[True] or x[1]

So, x[1] 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 https://stackoverflow.com/questions/3270680/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

Solution 2 - Python

>>> x = [1,2,3,4,5]
>>> x<2
False
>>> x[False]
1
>>> x[True]
2

The bool is simply converted to an integer. The index is either 0 or 1.

Solution 3 - Python

The original code in your question works only in Python 2. If x is a list in Python 2, the comparison x < y is False if y is an integer. This is because it does not make sense to compare a list with an integer. However in Python 2, if the operands are not comparable, the comparison is based in CPython on the alphabetical ordering of the names of the types; additionally all numbers come first in mixed-type comparisons. This is not even spelled out in the documentation of CPython 2, and different Python 2 implementations could give different results. That is [1, 2, 3, 4, 5] < 2 evaluates to False because 2 is a number and thus "smaller" than a list in CPython. This mixed comparison was eventually deemed to be too obscure a feature, and was removed in Python 3.0.


Now, the result of < is a bool; and bool is a subclass of int:

>>> isinstance(False, int)
True
>>> isinstance(True, int)
True
>>> False == 0
True
>>> True == 1
True
>>> False + 5
5
>>> True + 5
6

So basically you're taking the element 0 or 1 depending on whether the comparison is true or false.


If you try the code above in Python 3, you will get TypeError: unorderable types: list() < int() due to a change in Python 3.0:

> Ordering Comparisons > > Python 3.0 has simplified the rules for ordering comparisons: > > The ordering comparison operators (<, <=, >=, >) raise a TypeError exception when the operands don’t have a meaningful natural ordering. Thus, expressions like 1 < '', 0 > None or len <= len are no longer valid, and e.g. None < None raises TypeError instead of returning False. A corollary is that sorting a heterogeneous list no longer makes sense – all the elements must be comparable to each other. Note that this does not apply to the == and != operators: objects of different incomparable types always compare unequal to each other.


There are many datatypes that overload the comparison operators to do something different (dataframes from pandas, numpy's arrays). If the code that you were using did something else, it was because x was not a list, but an instance of some other class with operator < overridden to return a value that is not a bool; and this value was then handled specially by x[] (aka __getitem__/__setitem__)

Solution 4 - Python

This has one more use: code golf. Code golf is the art of writing programs that solve some problem in as few source code bytes as possible.

return(a,b)[c<d]

is roughly equivalent to

if c < d:
    return b
else:
    return a

except that both a and b are evaluated in the first version, but not in the second version.

c<d evaluates to True or False.
(a, b) is a tuple.
Indexing on a tuple works like indexing on a list: (3,5)[1] == 5.
True is equal to 1 and False is equal to 0.

  1. (a,b)[c<d]
  2. (a,b)[True]
  3. (a,b)[1]
  4. b

or for False:

  1. (a,b)[c<d]
  2. (a,b)[False]
  3. (a,b)[0]
  4. a

There's a good list on the stack exchange network of many nasty things you can do to python in order to save a few bytes. https://codegolf.stackexchange.com/questions/54/tips-for-golfing-in-python

Although in normal code this should never be used, and in your case it would mean that x acts both as something that can be compared to an integer and as a container that supports slicing, which is a very unusual combination. It's probably Numpy code, as others have pointed out.

Solution 5 - Python

In general it could mean anything. It was already explained what it means if x is a list or numpy.ndarray but in general it only depends on how the comparison operators (<, >, ...) and also how the get/set-item ([...]-syntax) are implemented.

x.__getitem__(x.__lt__(2))      # this is what x[x < 2] means!
x.__setitem__(x.__lt__(2), 0)   # this is what x[x < 2] = 0 means!

Because:

  • x < value is equivalent to x.__lt__(value)
  • x[value] is (roughly) equivalent to x.__getitem__(value)
  • x[value] = othervalue is (also roughly) equivalent to x.__setitem__(value, othervalue).


This can be customized to do anything you want. Just as an example (mimics a bit numpys-boolean indexing):

class Test:
    def __init__(self, value):
        self.value = value
        
    def __lt__(self, other):
        # You could do anything in here. For example create a new list indicating if that 
        # element is less than the other value
        res = [item < other for item in self.value]
        return self.__class__(res)
    
    def __repr__(self):
        return '{0} ({1})'.format(self.__class__.__name__, self.value)
    
    def __getitem__(self, item):
        # If you index with an instance of this class use "boolean-indexing"
        if isinstance(item, Test):
            res = self.__class__([i for i, index in zip(self.value, item) if index])
            return res
        # Something else was given just try to use it on the value
        return self.value[item]

    def __setitem__(self, item, value):
        if isinstance(item, Test):
            self.value = [i if not index else value for i, index in zip(self.value, item)]
        else:
            self.value[item] = value

So now let's see what happens if you use it:

>>> a = Test([1,2,3])
>>> a
Test ([1, 2, 3])
>>> a < 2  # calls __lt__
Test ([True, False, False])
>>> a[Test([True, False, False])] # calls __getitem__
Test ([1])
>>> a[a < 2] # or short form
Test ([1])

>>> a[a < 2] = 0  # calls __setitem__
>>> a
Test ([0, 2, 3])

Notice this is just one possibility. You are free to implement almost everything you want.

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Solution 1 - Pythontrans1st0rView Answer on Stackoverflow
Solution 2 - PythonKaroly HorvathView Answer on Stackoverflow
Solution 3 - PythonAntti Haapala -- Слава УкраїніView Answer on Stackoverflow
Solution 4 - PythonFilip HaglundView Answer on Stackoverflow
Solution 5 - PythonMSeifertView Answer on Stackoverflow