No Fluff Guide to Python - P9 - String & Regex

 

String compare

print('abc' == 'abc')
# True

print('abc' == 'xyz')
# False

This operation is case-sensitive

print('abc' == 'ABC')
# False

print('abc' != 'xyz')
# True

print('abc' != 'abc')
# False

Partial match: in, not in

in operator, which determines if one string contains another string

print('bbb' in 'aaa-bbb-ccc')
# True

print('xxx' in 'aaa-bbb-ccc')
# False

print('abc' in 'aaa-bbb-ccc')
# False

print('xxx' not in 'aaa-bbb-ccc')
# True

print('bbb' not in 'aaa-bbb-ccc')
# False

the in and not in operators test membership in lists, tuples, dictionaries, and so on

print(1 in [0, 1, 2])
# True

print(100 in [0, 1, 2])
# False

in tuple, set, and range

print(1 in (0, 1, 2))
# True

print(1 in {0, 1, 2})
# True

print(1 in range(3))
# True

in operator for dictionaries (dict) checks for the presence of a key.

d = {'key1': 'value1', 'key2': 'value2', 'key3': 'value3'}

print('key1' in d)
# True

print('value1' in d)
# False

Use the values() and items() methods to test for the presence of values or key-value pairs.

print('value1' in d.values())
# True

print(('key1', 'value1') in d.items())
# True

print(('key1', 'value2') in d.items())
# False

test whether list B is a subset of list A

l1 = [0, 1, 2, 3, 4]
l2 = [0, 1, 2]
l3 = [0, 1, 5]
l4 = [5, 6, 7]

print(set(l2) <= set(l1))
# True

print(set(l3) <= set(l1))
# False

String

s = 'aaa-bbb-ccc'

print(s.startswith('aaa'))
# True

print(s.startswith('bbb'))
# False

print(s.endswith('ccc'))
# True

print(s.endswith('bbb'))
# False

print(s.endswith(('aaa', 'bbb', 'ccc')))
# True

Regex: re.search(), re.fullmatch()

Use re.search() for partial, forward, and backward matching

import re

s = 'aaa-AAA-123'

print(re.search('aaa', s))
# <re.Match object; span=(0, 3), match='aaa'>

print(re.search('xxx', s))
# None

metacharacter ^ matches the start of the string, and $ matches the end of the string.

print(re.search('^aaa', s))
# <re.Match object; span=(0, 3), match='aaa'>

print(re.search('^123', s))
# None

print(re.search('aaa$', s))
# None

print(re.search('123$', s))
# <re.Match object; span=(8, 11), match='123'>

[A-Z] represents any single uppercase alphabet letter, and + means that the previous pattern is repeated one or more times. Thus, [A-Z]+ matches any substring that consists of one or more consecutive uppercase alphabetic characters.

print(re.search('[A-Z]+', s))
# <re.Match object; span=(4, 7), match='AAA'>

Use re.fullmatch() to check whether the entire string matches a regular expression pattern

s = '012-3456-7890'

print(re.fullmatch(r'\d{3}-\d{4}-\d{4}', s))
# <re.Match object; span=(0, 13), match='012-3456-7890'>

s = 'tel: 012-3456-7890'

print(re.fullmatch(r'\d{3}-\d{4}-\d{4}', s))
# None

\d represents a number and {n} represents n repetitions. Since backslash \ is used in special sequences of regular expressions, such as \d, it is useful to use raw strings (r'' or r"") that treat backslashes as literal characters

re.fullmatch() was added in Python 3.4. In earlier versions, you can use re.search() with ^ and $ to achieve the same result. You can also use re.match() and $, although it is not shown here.

s = '012-3456-7890'

print(re.search(r'^\d{3}-\d{4}-\d{4}$', s))
# <re.Match object; span=(0, 13), match='012-3456-7890'>

s = 'tel: 012-3456-7890'

print(re.search('^\d{3}-\d{4}-\d{4}$', s))
# None

Fetch last char of string

s = '90G'
print(s[len(s) - 1])


v = '90G+'
print(v[len(v) - 1])

G

+

Remove Last Character from String

string_val = "Python Guide"

new_string = string_val[:-1]

print(new_string)

helper function convert G,T to GB and TB

s = '90G'
lchar = s[len(s) - 1]
slchar = ''
multiplier = 1
size_val = 0
if lchar == '+':
    slchar = s[len(s) - 2]


if slchar == '':
    if lchar == 'G':
        multiplier = 1
    if lchar == 'T':
        multiplier = 1000
    size_val = int(s[:-1])
else:
    if slchar == 'G':
        multiplier = 1
    if slchar == 'T':
        multiplier = 1000
    size_val = int(s[:-2])