String comparison technique used by Python

Question

I\'m wondering how Python does string comparison, more specifically how it determines the outcome when a less than (<) or greater than (>) op

Answer 1

From the docs:

The comparison uses lexicographical ordering: first the first two items are compared, and if they differ this determines the outcome of the comparison; if they are equal, the next two items are compared, and so on, until either sequence is exhausted.

Also:

Lexicographical ordering for strings uses the Unicode code point number to order individual characters.

or on Python 2:

Lexicographical ordering for strings uses the ASCII ordering for individual characters.

As an example:

>>> 'abc' > 'bac'
False
>>> ord('a'), ord('b')
(97, 98)

The result False is returned as soon as a is found to be less than b. The further items are not compared (as you can see for the second items: b > a is True).

Be aware of lower and uppercase:

>>> [(x, ord(x)) for x in abc]
[('a', 97), ('b', 98), ('c', 99), ('d', 100), ('e', 101), ('f', 102), ('g', 103), ('h', 104), ('i', 105), ('j', 106), ('k', 107), ('l', 108), ('m', 109), ('n', 110), ('o', 111), ('p', 112), ('q', 113), ('r', 114), ('s', 115), ('t', 116), ('u', 117), ('v', 118), ('w', 119), ('x', 120), ('y', 121), ('z', 122)]
>>> [(x, ord(x)) for x in abc.upper()]
[('A', 65), ('B', 66), ('C', 67), ('D', 68), ('E', 69), ('F', 70), ('G', 71), ('H', 72), ('I', 73), ('J', 74), ('K', 75), ('L', 76), ('M', 77), ('N', 78), ('O', 79), ('P', 80), ('Q', 81), ('R', 82), ('S', 83), ('T', 84), ('U', 85), ('V', 86), ('W', 87), ('X', 88), ('Y', 89), ('Z', 90)]