Learn to Program 17

Posted by Derek Banas on Aug 13, 2016 in Web Design | 0 comments

In this part of my tutorial I continue covering Regular Expressions. We’ll look at Back References, Back Reference Substitutions, Look Ahead, Look Behind, and Negative Look Ahead & Behind. These are thought to be confusing and I hope to clear them up.

If you missed previous parts of my Regular Expression tutorial they are here Part 1 and Part 2. The code and transcript are found below.

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Code & Transcript

import re

# Did you find a match
# if re.search("REGEX", yourString)

# Get list of matches
# print("Matches :", len(re.findall("REGEX", yourString)))

# Get a pattern object
# regex = re.compile("REGEX")

# Substitute the match
# yourString = regex.sub("substitution", yourString)

# [ ]   : Match what is in the brackets
# [^ ]  : Match anything not in the brackets
# ( )   : Return surrounded submatch
# .     : Match any 1 character or space
# +     : Match 1 or more of what proceeds
# ?     : Match 0 or 1
# *     : Match 0 or More
# *?    : Lazy match the smallest match
# \b    : Word boundary
# ^     : Beginning of String
# $     : End of String
# \n    : Newline
# \d    : Any 1 number
# \D    : Anything but a number
# \w    : Same as [a-zA-Z0-9_]
# \W    : Same as [^a-zA-Z0-9_]
# \s    : Same as [\f\n\r\t\v]
# \S    : Same as [^\f\n\r\t\v]
# {5}   : Match 5 of what proceeds the curly brackets
# {5,7} : Match values that are between 5 and 7 in length
# ($m)  : Allow ^ on multiline string

# ---------- Back References ----------
# A back reference allows you to to reuse the expression
# that proceeds it

# Grab a double word
randStr = "The cat cat fell out the window"

# Match a word boundary, 1 or more characters followed
# by a space if it is then followed by the same
# match that is surrounded by the parentheses
regex = re.compile(r"(\b\w+)\s+\1")

matches = re.findall(regex, randStr)

print("Matches :", len(matches))

for i in matches:
    print(i)

# ---------- Back Reference Substitutions ----------

# Replace the bold tags in the link with no tags
randStr = "<a href='#'><b>The Link</b></a>"

# Regex matches bold tags and grabs the text between
# them to be used by the back reference
regex = re.compile(r"<b>(.*?)</b>")

# Replace the tags with just the text between them
randStr = re.sub(regex, r"\1", randStr)

print(randStr)

# ---------- Another Back Reference Substitution ----------

# Receive this string
randStr = "412-555-1212"

# Match the phone number using multiple subexpressions
regex = re.compile(r"([\d]{3})-([\d]{3}-[\d]{4})")

# Output (412)555-1212
randStr = re.sub(regex, r"(\1)\2", randStr)

print(randStr)

# ---------- PROBLEM ----------
# Receive a string like this

randStr = "https://www.youtube.com http://www.google.com"

# Grab the URL and then provide the following output
# using a back reference substitution
# <a href='https://www.youtube.com'>www.youtube.com</a>
# <a href='https://www.google.com'>www.google.com</a>

regex = re.compile(r"(https?://([\w.]+))")

randStr = re.sub(regex, r"<a href='\1'>\2</a>\n", randStr)

print(randStr)

# ---------- Look Ahead ----------
# A look ahead defines a pattern to match but not return
# You define the expression to look for but not return
# like this (?=expression)

randStr = "One two three four"

# Grab all letters and numbers of 1 or more separated
# by a word boundary but don't include it
regex = re.compile(r"\w+(?=\b)")

matches = re.findall(regex, randStr)

for i in matches:
    print(i)

# ---------- Look Behind ----------
# The look behind looks for what is before the text
# to return, but doesn't return it
# It is defined like (?<=expression)

randStr = "1. Bread 2. Apples 3. Lettuce"

# Find the number, period and space, but only return
# the 1 or more letters or numbers that follow
regex = re.compile(r"(?<=\d.\s)\w+")

matches = re.findall(regex, randStr)

for i in matches:
    print(i)

# ---------- Look Ahead & Behind ----------

randStr = "<h1>I'm Important</h1> <h1>So am I</h1>"

# Use the look behind, get 1 or more of anything,
# and use the look ahead
regex = re.compile(r"(?<=<h1>).+?(?=</h1>)")

matches = re.findall(regex, randStr)

for i in matches:
    print(i)

import re

# ---------- Negative Look Ahead & Behind ----------
# These are used to look for text that doesn't match
# the pattern

# (?!expression) : Negative Look Ahead
# (?<!expression) : Negative Look Behind

randStr = "8 Apples $3, 1 Bread $1, 1 Cereal $4"

# Grab the total number of grocery items by ignoring the $
regex = re.compile(r"(?<!\$)\d+")

matches = re.findall(regex, randStr)

print(len(matches))

# Convert from a string list to an int list
matches = [int(i) for i in matches]

from functools import reduce

# Sum the items in the list with reduce
print("Total Items {}".format(reduce((lambda x, y: x + y), matches)))

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import re

# Did you find a match

# if re.search("REGEX", yourString)

# Get list of matches

# print("Matches :", len(re.findall("REGEX", yourString)))

# Get a pattern object

# regex = re.compile("REGEX")

# Substitute the match

# yourString = regex.sub("substitution", yourString)

# [ ] : Match what is in the brackets

# [^ ] : Match anything not in the brackets

# ( ) : Return surrounded submatch

# . : Match any 1 character or space

# + : Match 1 or more of what proceeds

# ? : Match 0 or 1

# * : Match 0 or More

# *? : Lazy match the smallest match

# \b : Word boundary

# ^ : Beginning of String

# $ : End of String

# \n : Newline

# \d : Any 1 number

# \D : Anything but a number

# \w : Same as [a-zA-Z0-9_]

# \W : Same as [^a-zA-Z0-9_]

# \s : Same as [\f\n\r\t\v]

# \S : Same as [^\f\n\r\t\v]

# {5} : Match 5 of what proceeds the curly brackets

# {5,7} : Match values that are between 5 and 7 in length

# ($m) : Allow ^ on multiline string

# ---------- Back References ----------

# A back reference allows you to to reuse the expression

# that proceeds it

# Grab a double word

randStr = "The cat cat fell out the window"

# Match a word boundary, 1 or more characters followed

# by a space if it is then followed by the same

# match that is surrounded by the parentheses

regex = re.compile(r"(\b\w+)\s+\1")

matches = re.findall(regex, randStr)

print("Matches :", len(matches))

for i in matches:

print(i)

# ---------- Back Reference Substitutions ----------

# Replace the bold tags in the link with no tags

randStr = "<a href='#'><b>The Link</b></a>"

# Regex matches bold tags and grabs the text between

# them to be used by the back reference

regex = re.compile(r"<b>(.*?)</b>")

# Replace the tags with just the text between them

randStr = re.sub(regex, r"\1", randStr)

print(randStr)

# ---------- Another Back Reference Substitution ----------

# Receive this string

randStr = "412-555-1212"

# Match the phone number using multiple subexpressions

regex = re.compile(r"([\d]{3})-([\d]{3}-[\d]{4})")

# Output (412)555-1212

randStr = re.sub(regex, r"(\1)\2", randStr)

print(randStr)

# ---------- PROBLEM ----------

# Receive a string like this

randStr = "https://www.youtube.com http://www.google.com"

# Grab the URL and then provide the following output

# using a back reference substitution

# <a href='https://www.youtube.com'>www.youtube.com</a>

# <a href='https://www.google.com'>www.google.com</a>

regex = re.compile(r"(https?://([\w.]+))")

randStr = re.sub(regex, r"<a href='\1'>\2</a>\n", randStr)

print(randStr)

# ---------- Look Ahead ----------

# A look ahead defines a pattern to match but not return

# You define the expression to look for but not return

# like this (?=expression)

randStr = "One two three four"

# Grab all letters and numbers of 1 or more separated

# by a word boundary but don't include it

regex = re.compile(r"\w+(?=\b)")

matches = re.findall(regex, randStr)

for i in matches:

print(i)

# ---------- Look Behind ----------

# The look behind looks for what is before the text

# to return, but doesn't return it

# It is defined like (?<=expression)

randStr = "1. Bread 2. Apples 3. Lettuce"

# Find the number, period and space, but only return

# the 1 or more letters or numbers that follow

regex = re.compile(r"(?<=\d.\s)\w+")

matches = re.findall(regex, randStr)

for i in matches:

print(i)

# ---------- Look Ahead & Behind ----------

randStr = "<h1>I'm Important</h1> <h1>So am I</h1>"

# Use the look behind, get 1 or more of anything,

# and use the look ahead

regex = re.compile(r"(?<=<h1>).+?(?=</h1>)")

matches = re.findall(regex, randStr)

for i in matches:

print(i)

import re

# ---------- Negative Look Ahead & Behind ----------

# These are used to look for text that doesn't match

# the pattern

# (?!expression) : Negative Look Ahead

# (?<!expression) : Negative Look Behind

randStr = "8 Apples $3, 1 Bread $1, 1 Cereal $4"

# Grab the total number of grocery items by ignoring the $

regex = re.compile(r"(?<!\$)\d+")

matches = re.findall(regex, randStr)

print(len(matches))

# Convert from a string list to an int list

matches = [int(i) for i in matches]

from functools import reduce

# Sum the items in the list with reduce

print("Total Items {}".format(reduce((lambda x, y: x + y), matches)))

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Learn to Program 17

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