One 、 The derived type

res = []
for i in range(1, 11):
    res.append(i ** 2)
print(res)  # [1, 4, 9, 16, 25, 36, 49, 64, 81, 100]

res = [x ** 2 for x in range(1, 11)]  #  The outer square brackets indicate the list derivation 
print(res)

res = []
for i in range(1, 11):
    if i ** 2 % 2 == 0:
        res.append(i ** 2)
print(res)  # [4, 16, 36, 64, 100]

res = [x ** 2 for x in range(1, 11) if x ** 2 % 2 == 0]
print(res)

res = [x for x in "python"]  #  Split string 
print(res)  # ['p', 'y', 't', 'h', 'o', 'n']

res = [x + y for x in "python" for y in "123" if x + y != "y1"]  #  Split string 
print(res)
# ['p1', 'p2', 'p3', 'y2', 'y3', 't1', 't2', 't3', 'h1', 'h2', 'h3', 'o1', 'o2', 'o3', 'n1', 'n2', 'n3']

res = (x ** 2 for x in range(1, 11))  # () Is defined as a generator 
print(res)  # <generator object <genexpr> at 0x7fdb743e9570>

res = tuple(x ** 2 for x in range(1, 11))  #  keyword tuple Deducing tuples 
print(res)

dic = {
    x: x ** 2 for x in range(1, 11)}  # {} Brace Dictionary   With colon :
print(dic)

ss = {
    x for x in range(1, 11)}  #  Unordered and unrepeatable sets   No colon 
print(ss)

ss = {
    x for x in "aaa123"}  #  Unordered and unrepeatable sets   No colon 
print(ss)

Two 、 Regular form

import re
#  Regular expressions   character string   Focus on handling rules 
result = re.findall("py", "python")  # "python"  Whether contains py
print(result)
result = re.findall("pn", "python")  # "python"  Whether contains pn
print(result)
result = re.findall("python", "I like python")  #  Whether the sentence contains words 
print(result)
result = re.findall("o", "I love python")  #  Whether the sentence contains letters 
print(result)
result = re.findall("2", "1234567890abcdef")  #  Whether the sentence contains numbers 
print(result)

#  Predefined characters 
# \d \s \w \D \S \W  Case complements each other 
# \d  All figures  0123456789
# \D  The digital 
# \w  Underline included  a-z A-Z 0-9
# \W  Special characters   Underline without numbers, uppercase and lowercase letters 
# \s  blank   Tabulation   Line break 
# \S  The blank   Tabulation   Line break 
result = re.findall("\d", "12a789bc345f67890de")  #  All the numbers in the sentence 
print(result)
result = re.findall(r"\d", "12a789bc345f67890de")  #  All the numbers in the sentence   Don't want escape processing to make r Original output 
print(result)
result = re.findall(r"\D", "1de2abc345f678__90de")  #  Not all numbers in the sentence 
print(result)
result = re.findall(r"\w", "1de2abc345f6__7890A$%&Ade")  #  All sentences include underscores  a-z A-Z 0-9
print(result)
result = re.findall(r"\W", "1de2abc345f6__7890A$%&Ade")  #  All special characters in the sentence   Underline without numbers, uppercase and lowercase letters 
print(result)
result = re.findall(r"\s", "1de2abc 345f\n6__7890A\t$%&Ade")  #  All blanks in the sentence   Tabulation   Line break 
print(result)
result = re.findall(r"\S", "1de2abc 345f\n6__7890A\t$%&Ade")  #  All non blanks in the sentence   Tabulation   Line break 
print(result)

#  Metacharacters 
# []  Match a character   Or relationship 
result = re.findall(r"[123]", "1e3deabc45f\n6__7890A\t$%&Ade")  # 1 or 2 or 3
print(result)
result = re.findall(r"[\d\s]", "1e3deabc45f\n6__7890A\t$%&Ade")  # \d or \s
print(result)
result = re.findall(r"[^\d\s]", "1e3deabc45f\n6__7890A\t$%&Ade")  # \d or \s Take the opposite  ^ Only take the opposite in square brackets 
print(result)
# -  Section 
result = re.findall(r"[1-7]", "1e3deabc45f\n6__7890A\t$%&Ade")  # 1 To 7
print(result)
result = re.findall(r"[b-f]", "1e3deaCCCbc45f\n6__7890A\t$%&Ade")  # b To f
print(result)
result = re.findall(r"[1-7b-f]", "1e3deabc45f\n6__7890A\t$%&Ade")  # 1-7 or b-f All together 
print(result)

#  All the above concepts are one character 
# () On behalf of the group 
result = re.findall(r"a([abc])", "aaabacad")
print(result)
result = re.findall(r"a[abc]", "aaabacad")  # a It can be followed by any or one of the relationships   lookup aa ab ac
print(result)

#  Repeat match 
result = re.findall(r"\d\d\d", "123456789abcdefg")  #  Three consecutive numbers in a sentence 
print(result)
# {n} The first character is repeated n Time 
result = re.findall(r"\d{3}", "123456789abcdefg")  #  Three consecutive numbers in a sentence 
print(result)
# {n, m} The first character repeats at least n Time   at most m Time 
result = re.findall(r"\d{2,5}", "12e3456789abc6defg")  #  Continuous in a sentence 2-5 A digital 
print(result)  #  He will match as many as possible   Greedy matching 
result = re.findall(r"\d{2,5}?", "12e3456789abc6defg")  #  Continuous in a sentence 2-5 A digital 
print(result)  #  Type an English half angle ?  Non greedy matching   and {n} No difference between 
result = re.findall(r"\d{2,}", "12e3456789abc6defg")  #  Numbers appear at least twice in a row in a sentence 
print(result)  #  Greedy matching 

# ?  Leading characters appear 0 or 1 Time  {0,1}
result = re.findall(r"\d?", "12e3456789abc6defg")
print(result)  #  Non numeric places are filled with empty strings " "  There will be another end at the end " "

# +  The preceding characters appear at least 1 Time  {1,}
result = re.findall(r"\d+", "12e3456789abc6defg")
print(result)  #  greedy   Block mixing 
result = re.findall(r"\d+?", "12e3456789abc6defg")
print(result)  #  Not greed   The numbers are extracted one by one 

# *  The preceding character does not appear   appear 0 Time   Appear any number of times 
result = re.findall(r"\d*", "12e3456789abc6defg")  #  Numbers appear at least twice in a row in a sentence 
print(result)  #  greedy   Block mixing   And each non numeric character has " " placeholder 

#  Escape symbol *+?
result = re.findall(r"\*", "12e*345a*b67*89*abc6d*efg")  #  look for * Must escape 
print(result)
result = re.findall(r"a\*b", "12e*345a*b67*89*abc6d*efg")  #  look for * Must escape 
print(result)
result = re.findall(r"a*b", "12e*345a*b67*89*abc6d*efg")  #  look for b front a It doesn't matter   Print if you have 
print(result)
result = re.findall(r"\\d", "12e*345\da*b67*\d89*abc6d*efg")  #  look for "\d"
print(result)  #  The print should be \\d instead of \d  Although I'm looking for \d

#  Greed and non greed 
result = re.findall(r"d\w+d", "dxxxxxxxxdxxxxxd")  #  Greedy with d Begin with d ending 
print(result)  # dxxxxxxxxdxxxxxd
result = re.findall(r"d\w+?d", "dxxxxxxxxdxxxxxd")  #  Not greedy with d Begin with d ending 
print(result)  # dxxxxxxxxd

htmlstr = "<td>python</td><td>$123</td><td>[email protected]</td>"
result = re.findall(r"<td>.+</td>", htmlstr)  # . Everything except line breaks 
print(result)  # ['<td>python</td><td>$123</td><td>[email protected]</td>']

htmlstr = "<td>python</td><td>$123</td><td>[email protected]</td>"
result = re.findall(r"<td>.+?</td>", htmlstr)  #  Not greed 
print(result)  # ['<td>python</td>', '<td>$123</td>', '<td>[email protected]</td>']

htmlstr = "<td>python</td><td>$123</td><td>[email protected]</td>"
result = re.findall(r"<td>(.+?)</td>", htmlstr)  #  Just within parentheses 
print(result)  # ['python', '$123', '[email protected]']

#  backreferences 
wordstr = """'hello' "python" 'love" "haha'"""
#  Extract words with the same symbols on both sides 
result = re.findall(r"['\"]\w+['\"]", wordstr)  #  Just within parentheses   If not + It defaults to the middle letter   Can't print out 
print(result)  # ["'hello'", '"python"', '\'love"', '"haha\'']

result = re.findall(r"(['\"])(\w+)(\1)", wordstr)  #  Front side () Put a group  (\1) Is the content of the first group 
print(result)  # [("'", 'hello', "'"), ('"', 'python', '"')]  The matched three parts in parentheses form a tuple 

print([x[1] for x in result])  # ['hello', 'python']

#  The verification password cannot appear continuously 3 Subnumeral 
result = re.findall(r"(\d)(\1{2,})", "12222223")  #  Match a number   Repeat more than twice 
print(result)  # [('2', '22222')]
print([x[0]+x[1] for x in result])  # ['222222']


#  Position matching  ^( Not in brackets ) start  $ ending 
result = re.findall(r"\d{11}", "13578617265123123")
print(result)
result = re.findall(r"^\d{11}", "13578617265123123")  #  Match from front 
print(result)
result = re.findall(r"\d{11}$", "13578617265123123")  #  Match from behind 
print(result)
result = re.findall(r"^\d{11}$", "13578617265123123")  #  Must happen to 11 Numbers 
print(result)