String double quotation marks "" or apostrophe '' can be.
(A) string concatenation
= S1 " numbers: " A = 33 is B = STR (A) # use str () converts the value into a string c = 55 Print (S1 + B) Print (A + C) Print (S1 + STR (A)) Print (S1 + A) # string values directly spliced, program error
result:
Digital: 33 88 Number: 33 Traceback (most recent call last): File "D:/untitled/demo.py", line 296, in <module> print(s1+a) TypeError: can only concatenate str (not "int") to str
(B) the string sections
Code:
= s ' 0123456789987654321 ' Print (s [0: 5 ]) # obtain the substring s index from 0 to 5 at index (no) of Print (s [0: -5 ]) # get a substring s from index 0 to the penultimate 5 (not included) of the string Print (s [-8: -4 ]) # get a substring s 8 from the penultimate to the penultimate string 4 (not included) of the string Print (s [-18: 6 ]) # get a substring s 18 from the penultimate to the string index 6 (not included) of the string Print (S [:: 2 ]) # every character takes a Print (S [::. 3 ]) # every two, takes a character Print (s [. 3 :]) # get a substring from index s to the end of the string of three Print (s [-7 :]) # get a substring s from the reciprocal of the seventh character to the end Print (s [: 7 ]) # get a substring s 7 (not included) from the index 0 to the index Print (s [: - 7 ]) # get a substring s index from 0 to 7, the reciprocal (not included) of
result:
01234 01234567899876 8765 12345 0246897531 0369741 3456789987654321 7654321 0123456 012345678998
(C) len () function
len () function to get the string length or number of bytes.
Code:
= b ' 123 I was little cute? ' Print (len (B)) # calculates the length of the character Print (len (b.encode ())) # string after encoding acquired its bytes # characters plus Chinese punctuation of 7, 7 * 3 accounting for 21 bytes, and the letters and numbers, and punctuation, 3 bytes, occupies a total of 24 bytes. = A ' No you're not! ' Print (len (A)) Print (len (a.encode ()))
result:
10 24 6 18
(D) count () function
Statistical functions frequency occurrence of the string count ().
Code:
= STR ' 01234567890123456789 ' # search string at the specified index Print (str.count ( ' 12345678 ' , 0, -2 )) # from index 0 to the inverse of a second (not included) queries the query string designated number (the number to find 012345678901234567 12345678) Print (str.count ( ' 1234567 ' , 0, -2 )) # from the index 0 to the inverse of a second (not including) the query string specified number (012345678901234567 find the number 1234567) Print (str.count ( ' 3456 ' , 3,7 )) # (3456 in the number of queries 3456) from the index to the index 7 3 (not included) query string specified number Print (str.count ( ' 3456 ' , 3,6 )) # (3456 Discover number 345) from the index of the index 3 to 6 (not included) query string specified number Print (str.count ( ' 3456 ' , 3,17 )) # (3456 in the number of queries 34567890123456) from the index to the index 17 3 (not included) query string specified number Print (str.count ( ' 3456 ' , 3,16 )) # (3456 in the number 3456789012345 query) from the index to the index 16 3 (not included) query string specified number Print (str.count ( ' 23456 ' , -18, -3 )) # from the first to the third countdown at 18 reciprocal (not included) query string specified number (the number of queries 234567890123456 23456) Print (str.count ( ' 23456 ' , -18, -4 )) # from the first to the third countdown at 18 reciprocal (not included) query string specified number (the number of queries 23456789012345 23456)
result:
1
2
1
0
2
1
2
1
Five) split () function
split () function is used to split a string.
Code:
= str " Hello! Welcome to embark on this road of no return yards agriculture. bailing? >>> see trick! hahaha " List1 = str.split () # default separator is divided (by default be in the form of a space) Print (list1) List2 = str.split ( ' >>> ' ) # using a plurality of characters divided Print (List2) list3 = str.split ( ' . ' ) # adopted. No. split Print (list3) list4 = str.split ( ' ' , 4) # use space is divided, and only up to a predetermined sequence is divided into four sub- Print (list4) list5 = str.split ( ' > ' ) # employed> characters divided Print (list5)
result:
[ ' Hello! Welcome to embark on this road of no return yards agriculture. ' , ' Bailing? ' , ' >>> ' , ' see trick! hahaha ' ] [ ' Hello! Welcome to embark on this road of no return yards agriculture. Bailing? ' , ' See trick! hahaha ' ] [ ' Hello! Welcome to embark on this road of no return yards agriculture ' , ' bailing? >>> see trick! hahaha ' ] [ ' Hello! Welcome to embark on this road of no return yards agriculture. ' , ' Bailing? ' , ' >>> ' , ' see trick! hahaha ' ] [ ' Hello! Welcome to embark on this road of no return yards agriculture. Bailing? ' , '' , '' , ' See trick! hahaha ' ]
(Vi) format () function
format () method of the string format.
Code:
# Displayed in monetary Print ( " monetary: {D} :, " .format (1000000 )) # scientific notation Print ( " scientific notation: {:} E " .format (1200.12 )) # hexadecimal hexadecimal representation Print ( " hex 100: {: # X} " .format (100 )) # output percentage Print ( " percentage represents 0.01: 0 {%} :. " .format (0.01 )) Print ( " monetary Output: {D} :, " .format (875 638 768 )) Print ( " in hexadecimal output: {: # X} " .format (100 )) Print ( " form output scientific notation: {:} E " .format (1567.786 )) Print ( " percentage represents 0.19: 0 {%} :. " .format (0.19))
result:
Monetary: 000,000 scientific notation: 1.200120E + 03 100 hex: 0x64 percentage representation 0.01: 1% monetary output: 875,638,768 hexadecimal output: 0x64 forms output scientific notation: 1.567786E + 03 0.19 percentage represents: 19%