3. Supplement of strings, lists, tuples, dictionaries, and collections, and string dictionaries

3. Supplement of strings, lists, tuples, dictionaries, and collections, and string dictionaries
Related:

• String

• List

• Dictionary

• Set

Additional:

Many python compilers provide code complementing functions and prompt functions when entering parameters.

String

 

1. Common functions:

The string is an immutable object, and the string method does not change the data of the original string.

S = "hEllo world! \ T "print (" s. capitalize (): ", s. capitalize () # format the title print ("s. center (20, '-'): ", s. center (20, '-') # Fill the string in the center, fill in the element to specify characters, the total number of characters is specified print ("s. count ('l'): ", s. count ('l') # count the number of occurrences of a string print ("s. endswith: ", s. endswith ('d! ') # Judge whether the string is d! Print ("s. find ('O'): ", s. find ('O') # find the specified Element and return its index print ("s. index ('O'): ", s. index ('O') # Find the specified Element and return its index sep = 'abc' print ("s. join (sep): ", s. join (sep) # Insert the original character s to print ("s. lower (): ", s. lower () # convert all to lowercase print ("s. replace ('l', 'J', 1): ", s. replace ('l', 'J', 1) # replace the specified character. The last one is to replace the Count print ("s. split (): ", s. split () # cut the string. The cut character is the specified print ("s. strip (): ", s. strip () # Remove left and right space elements. rstrip removes only the right side, and lstrip removes only the right side of print ("s. upper (): ", s. upper () # convert all to uppercase "is series: isdigit ()-" whether it is a number, isalnum ()-"whether it is a letter or a number, isalpha () -whether it is an English letter islower ()-whether it is all lowercase ,"""

Appeal code result:

s.capitalize():  hello world!     s.center(20,'-'): -- hEllo world!     ---s.count('l'): 3s.endswith: Falses.find('o'): 5s.index('o'): 5s.join(sep): A hEllo world!     B hEllo world!     Cs.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 parameter seems to give a good result

>>> 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 \ escape, when the string format is r "string", all characters in it are considered as characters, for example, \ n is no longer the original line feed.

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

However, the string cannot be processed. If the end is \:

>>> 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\

In this case, it is best to use String concatenation for processing.

List

 

1. Common functions:

Print ("query ". center (20, '-') list_find = ['apple', 'Banana ', 'pen, 3] # search for the subscript print (list_find.index ('apple') of a specified Element) # search for the number of existing print (list_find.count ("apple") print ("add ". center (20, '-') list_add = ['apple', 'bana'] # append an element to the end of list_add.append ("Hami melon") print (list_add) # insert an element to the specified position list_add.insert (0, "apple") print (list_add) print ("delete ". center (20, '-') list_del = [1, 2, 3, 4, 5, 6, 7, 8, 9] # retrieve elements from the List (delete). pop can be set as a parameter, the parameter is the subscript list_del.pop () print (list_del) of the deletion element # Delete the element list_del.remove (4) print (list_del) of the specified element name) # Delete the 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 the list list_test4.sort () print (list_test4) # Note: py3 cannot sort different element types # reverse list list_test5.reverse () print (list_test5)

 

The code execution result is as follows:

--------- Query ---------- 01 --------- add ---------- ['apple', 'Banana ', 'cantalou'] ['apple', 'apple', 'Banana ', 'cantalanga'] --------- Delete ---------- [1, 2, 3, 4, 5, 6, 7, 8] [1, 2, 3, 5, 6, 7, 8] [2, 3, 5, 6, 7, 8] --------- others --------- [1, 2, 3, 4, 'A', 'B', 'D ', 'C'] ['A', 'B', 'C', 'D'] ['C', 'D', 'B', 'A', 'A', 4, 3, 2, 1]

2. List generation:

# Exp = expression # process: 1. iterate every element in iterable; #2. each iteration first assigns the result to iter_var, and then obtains a new calculated value through exp; #3. finally, all the calculated values obtained through exp are returned in the form of 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 code execution result is as follows:

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]

 

Dictionary

 

1. Common functions:

D1 = {1: "apple", "sprite": "Sydney"} d1.clear () # Clear the dictionary print (d1) d1 = {1: "apple", "sprite ": "Sydney"} print (d1.get (1) # obtain the print (d1.get (3) Result of the specified dictionary key # If the key does not exist, return Noneprint (d1.items ()) # obtain all key-value pairs of the dictionary print (d1.keys () # obtain the dictionary key print (d1.values () # obtain the dictionary value print (d1.pop (1 )) # print (d1.popitem () # print (d1) d1 = {1: "apple", "sprite ": "Sydney"} d1.update ({1: 'apple', 3: 'pen '}) # update result, same key name updated, new key name added result print (d1)

The code execution result is as follows:

{} Apple Nonedict_items ([(1, 'apple'), ('sprid', 'sydney ')]) dict_keys ([1, 'sprid']) dict_values (['apple ', 'sydney']) apple ('sprite ', 'sydney') {}{ 1: 'apple', 'sprite ': 'sydney ', 3: 'pen '}

 

Set

 

1. Common functions:

S1 = set (['A', 'B', 'C']) print (s1.pop () # randomly delete an element in the set, after an element is obtained, the returned value is print (s1) s3 = {'A', 'D'} s1.update (s3) # update print (s1) s1.add ('F ') # add element print (s1) s1.clear () # Clear s1 = set (['A', 'B', 'C', 'F']) print (s1) s1.remove ('A') # Delete the target element. If no element exists in the Set, the following error occurs: print (s1) s1.discard ('G') # If no element exists in the Set, no error is returned; if an element exists, print (s1) B = {'A', 'B', 'G'} print ("s1.difference (B)") print (s1.difference (B) is deleted )) # retrieve the elements in s and B in the set and return the print ("s1.interscetion (B)") print (s1.intersection (B) # intersection of the elements, returns the print ("s1.issubset (B)") (s1.issubset (B) set composed of elements in s and B )) # determine if s is a subset of B print ("s1.issuperset (B)") print (s1.issuperset (B) # determine if s is a parent set of B print ("s1.20.ric _ difference (B) ") print (s1.equalric _ difference (B) # Get the difference set and create a new set print (" s1.union (B) ") print (s1.union (B )) # Union set print ("symmetric_difference_update") print (s1) s1.equalric _ difference_update (B) # print (s1) without return value "xxxx_update will overwrite the s1 value, for example: s1.symmetric _ difference_update (): The result of resolving ric_difference will overwrite the value of s1 """

 

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)Falses1.issuperset(b)Falses1.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'}