Solution for processing one character each time: Create a list [python] thestring = 'abcdefg' thelist = list (thestring) print thelist [python] [python] result [python] ['A ', 'B', 'C', 'D', 'E', 'F ', 'G'] use the for statement to traverse [python] thestring = 'abcdefg' for c in thestring: print c using the list derivation formula. (Note that the ord statement is used to convert the character to the character value, for example, convert a to 97) [python] thestring = 'abcdefg' results = map (ord, thestring) print results: to obtain a set of all characters in a string, call sets. set [python] import sets magic_chars = sets. set ('a Bracadabra ') poppins_chars = sets. set ('supercalifragilisticexpialidocious ') print ''. how to convert the join (magic_chars & poppins_chars) character to the character value: Use the ord and chr [python] print ord ('A') print chr (97) Functions) test whether an object is a class string (whether the object has a string behavior pattern) solution: Use isinstance and basestring to check [python] def isAString (anobj): return isinstance (anobj, basestring) anobj = 'abcde' print isAString (anobj) otherobj = list (anobj) print isAString (otherobj) string alignment (Left-aligned, center-aligned, and right-aligned) solution: Use the string object's ljust, must, and center. The parameter specifies the width [python] print '|', 'hs '. ljust (20), '|', 'hs '. must ust (20), '|', 'H '. center (20), '|' discussion: It can be printed with other characters without spaces. You only need to add the second parameter [python] print 'H '. center (20, '+') removes spaces at both ends of the string. Solution: Use the string object's lstrip, rstrip, and strip [python] x = 'hes' print '|', x. lstrip (), '|', x. rstrip (), '|', x. strip (), '|' merge strings. join [python] x = ['I', 'love', 'python'] largestring = ''using the string operator ''. join (x) Print largestring Similarly, using the most basic % can also achieve this effect [python] x = ('I', 'love', 'python ') largestring = '% s! '% X print largestring: Of course, the + operations using strings seem to be more concise, but in python, strings cannot be changed, any change will create a copy of the current string. When a large number of short strings are added, the created copy is proportional to its square, in this case, using the join method is a necessary choice. When you need to add additional content to the new string you created, it is more convenient to use %. How to reverse a string by character or word: Use the slicing method with step-1 [python] astring = 'I Love python' revchars = astring [:: -1] print revchars result nohtyP evoL I is reversed by words, you need to create a word list and reverse the list, finally, use join to merge [python] astring = 'I Love python' revwords = ''. join (astring. split () [:-1]) print revwords result Python Love I wants to reverse word by word but does not want to change the original space, use a regular expression to separate the original string [python] import re astring = 'I Love python' revwords = ''. join (re. split (R' (\ s +) ', astring) [:-1]) print revw Ords result Python Love I check whether the string contains characters in a character set. solution: the simplest method is as follows: [python] def containAny (seq, aset): for c in seq: if c in aset: return True return False seq = 'abc' aset = 'hjkyuia 'print containAny (seq, aset) can also use methods based on the itertools module of the standard library, however, in essence, the same method [python] import itertools def containAny (seq, aset): for item in itertools. ifilter (aset. _ contains __, seq): return True return False seq = 'abc' aset = 'ghjka 'print contai NAny (seq, aset) checks whether a string is a text or binary solution: no precise algorithms are available, but some heuristic methods can be used, if the string contains a null value or a high value of more than 30% is set to 1 or a strange control code, it is considered that this data segment is a binary data case-sensitive (case-sensitive conversion) solution: using the upper and lower methods is relatively simple, but generally the capitalize and title methods [python] view plaincopyprint? Print 'one tWo thrEe '. capitalize () print 'one tWo thrEe'. title () Result one two three one Two Three