Relevant:

• Character string

1. Common functions (refer to function libraries in more detail or note tips when using compilers)
2. String Formatting
3. Original string

• list

1. Common Functions
2. List Generation

• Dictionaries

1. Common Functions

• aggregate

1. Common Functions

Additional:

Many of python's compilers provide code completion and prompting when filling in parameters

Character string

 

1. Common functions:

Strings are immutable objects and methods of strings do not change the data of the original string

s=" hEllo world!\t "
print("s.capitalize():",s.capitalize())#Title formatting
print("s.center(20,'-'):",s.center(20,'-'))#Centers the string, fills the element with the specified number of characters
print("s.count('l'):",s.count('l'))#Count occurrences of a string
print("s.endswith:",s.endswith('d!'))#Determines whether the string is d!Ending
print("s.find('o'):",s.find('o'))#Find the specified element,Find the index that returns it
print("s.index('o'):",s.index('o'))#Find the specified element,Find the index that returns it
sep='ABC'
print("s.join(sep):",s.join(sep))#Put original character s Insert in the middle of every character (or element object) of the target
print("s.lower():",s.lower())#Convert all to lowercase
print("s.replace('l','j',1):",s.replace('l','j',1))#Replace the specified character, and the last is the number of replacements
print("s.split():",s.split())#Cut string, cutting character as specified character
print("s.strip():",s.strip())#Remove the left and right space elements, rstrip Only the right side is removed. lstrip Is to remove only the right side
print("s.upper():",s.upper())#Full capitalization

"""is Series:
isdigit()->Is it a number, isalnum()->Is it a letter or a number, isalpha()->Is it an English letter
islower()->Is it all lowercase?
"""

Appeal Code Result:

s.capitalize():  hello world!     
s.center(20,'-'): -- hEllo world!     ---
s.count('l'): 3
s.endswith: False
s.find('o'): 5
s.index('o'): 5
s.join(sep): A hEllo world!     B hEllo world!     C
s.lower():  hello world!     
s.replace('l','j',1):  hEjlo world!     
s.split(): ['hEllo', 'world!']
s.strip(): hEllo world!
s.upper():  HELLO WORLD!

 

2. String formatting:

python string formatting --This seems to be all given by the parameter

>>> s="%d is 250"
>>> s%250
'250 is 250'
>>> b = "%(name)+10s----%(age)-10d----"%{'name':'xx','age':20}
>>> print(b)
        xx----20        ----
>>> s="{:d} is a 250"
>>> s.format(250)
'250 is a 250'


>>> a1 = "numbers: {:b},{:o},{:d},{:x},{:X}, {:%},{:c}".format(15, 15, 15, 15, 15, 15.87623,65)
>>> print(a1)
numbers: 1111,17,15,f,F, 1587.623000%,A
>>> s="{} {} 250"
>>> s.format(250,"500-250")
'250 500-250 250'
>>> s.format(250,"500-250")
'250 500-250 250'

 

3. Original string:

Cause: To avoid excessive use of \ to escape, when the string format is "string", all the characters inside are treated as characters, such as \n no longer wrapped.

>>> print("a\tb")
a    b
>>> print(r"a\tb")
a\tb

But the string cannot be processed if it ends with a \:

>>> print(r"c:\a\b")
c:\a\b
>>> print(r"c:\a\b\")
      
SyntaxError: EOL while scanning string literal
>>> print(r"c:\a\b\\")
c:\a\b\\
>>> print(r"c:\a\b"+"\\")
c:\a\b\

This is best handled using string splicing.

list

 

1. Common functions:

print("check".center(20,'-'))
list_find=['apple','banana','pen',1,2,3]
#Find subscripts for specified elements
print(list_find.index('apple'))
#Find the number of elements that exist
print(list_find.count("apple"))

print("increase".center(20,'-'))
list_add=['apple','banana']
#Append element to end
list_add.append("Hami melon")
print(list_add)
#Insert element to specified location
list_add.insert(0,"Apple")
print(list_add)

print("Delete".center(20,'-'))
list_del=[1,2,3,4,5,6,7,8,9]
#Remove the element from the list (delete it), pop You can fill in parameters that are subscripts to deleted elements
list_del.pop()
print(list_del)
#Delete the element with the specified element name
list_del.remove(4)
print(list_del)
#Delete corresponding element space
del list_del[0]
print(list_del)

print("Other".center(20,'-'))
list_test4=['a','b','d','c']
list_test5=[1,2,3,4]
#Extended List
list_test5.extend(list_test4)
print(list_test5)
#Sort Lists
list_test4.sort()
print(list_test4)#Note: py3 Cannot sort elements of different types
#Reverse List
list_test5.reverse()
print(list_test5)

 

The results of the above code run:

---------check----------
0
1
---------increase----------
['apple', 'banana', 'Hami melon']
['Apple', 'apple', 'banana', 'Hami melon']
---------Delete----------
[1, 2, 3, 4, 5, 6, 7, 8]
[1, 2, 3, 5, 6, 7, 8]
[2, 3, 5, 6, 7, 8]
---------Other---------
[1, 2, 3, 4, 'a', 'b', 'd', 'c']
['a', 'b', 'c', 'd']
['c', 'd', 'b', 'a', 4, 3, 2, 1]

2. List Generation:

#exp = Expression
# Process: 1.iteration iterable Each element in;
# 2.Assign results to each iteration first iter_var，Then pass exp Get a new calculated value;
#3. Final Pass All exp The resulting calculated value is returned as a new list.

#[exp for iter_var in iterable]
print("1.[exp for iter_var in iterable]")
list1=[i for i in range(10)]
print(list1)
list2=[i*i for i in range(10,20)]
print(list2)
print("\n")
#[exp for iter_var in iterable if_exp]
print("2.[exp for iter_var in iterable if_exp]")
list3=[i for i in range(10) if i%2==0]
print(list3)
print("\n")
#[exp for iter_var_A in iterable_A for iter_var_B in iterable_B]
print("3.[exp for iter_var_A in iterable_A for iter_var_B in iterable_B]")
list4=[x*y for x in range(5) for y in range(5)]
print(list4)
print("\n")

The results of the above code run:

1.[exp for iter_var in iterable]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
[100, 121, 144, 169, 196, 225, 256, 289, 324, 361]


2.[exp for iter_var in iterable if_exp]
[0, 2, 4, 6, 8]


3.[exp for iter_var_A in iterable_A for iter_var_B in iterable_B]
[0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 0, 2, 4, 6, 8, 0, 3, 6, 9, 12, 0, 4, 8, 12, 16]

 

Dictionaries

 

1. Common functions:

d1={1:"Apple","Sprite":"Snow pear"}
d1.clear()#Empty Dictionary
print(d1)
d1={1:"Apple","Sprite":"Snow pear"}
print(d1.get(1))#Gets the result of the specified key in the dictionary
print(d1.get(3))#If you get a key that does not exist, return None
print(d1.items())#Get all key-value pairs of a dictionary
print(d1.keys())#Get the key to the dictionary
print(d1.values())#Get the value of the dictionary
print(d1.pop(1))#Remove the specified subscript result
print(d1.popitem())#Pop-up results without index
print(d1)
d1={1:"Apple","Sprite":"Snow pear"}
d1.update({1:'apple',3:'pen'})#Update results, update same key names, and add results for new key names
print(d1)

The results of the above code run:

{}
//Apple
None
dict_items([(1, 'Apple'), ('Sprite', 'Snow pear')])
dict_keys([1, 'Sprite'])
dict_values(['Apple', 'Snow pear'])
//Apple
('Sprite', 'Snow pear')
{}
{1: 'apple', 'Sprite': 'Snow pear', 3: 'pen'}

 

aggregate

 

1. Common functions:

s1=set(['a','b','c'])
print(s1.pop())#Randomly deletes an element from the set and returns its value
print(s1)
s3={'a','d'}
s1.update(s3)#To update
print(s1)
s1.add('f')#Add Elements
print(s1)
s1.clear()#empty
s1=set(['a','b','c','f'])
print(s1)
s1.remove('a')#Delete the target element, but error occurs if there are no elements in the collection
print(s1)
s1.discard('g')#If there are no elements in the collection, no error is reported; if there are elements, delete
print(s1)
b={'a','b','g'}
print("s1.difference(b)")
print(s1.difference(b))# Take Set s Yes, b And returns the set of elements
print("s1.interscetion(b)")
print(s1.intersection(b))#Intersection, two s and b Intersection in, return s,b A collection of elements that exist in all
print("s1.issubset(b)")
print(s1.issubset(b))#judge s Is it b A subset of
print("s1.issuperset(b)")
print(s1.issuperset(b)) #judge s Is it b The parent set of
print("s1.symmetric_difference(b)")
print(s1.symmetric_difference(b)) #Take the difference set and create a new set
print("s1.union(b)")
print(s1.union(b)) #Union
print("symmetric_difference_update")
print(s1)
s1.symmetric_difference_update(b)#No return value
print(s1)
"""
xxxx_update Meeting Coverage s1 A value of, for example:
s1.symmetric_difference_update()
//The value of s1 is overridden when the result of symmetric_difference is obtained
"""

 

The above code results:

a
{'c', 'b'}
{'c', 'b', 'd', 'a'}
{'c', 'b', 'd', 'f', 'a'}
{'a', 'c', 'b', 'f'}
{'c', 'b', 'f'}
{'c', 'b', 'f'}
s1.difference(b)
{'c', 'f'}
s1.interscetion(b)
{'b'}
s1.issubset(b)
False
s1.issuperset(b)
False
s1.symmetric_difference(b)
{'a', 'g', 'c', 'f'}
s1.union(b)
{'g', 'c', 'b', 'f', 'a'}
symmetric_difference_update
{'c', 'b', 'f'}
None
{'g', 'c', 'f', 'a'}