Python Cookbook

1. Filtering a String for a Set of Characters
   
2. Credit: Jürgen Hermann, Nick Perkins, Peter Cogolo
   
3. 
   
4. Problem
   
5. Given a set of characters to keep, you need to build a filtering function that, applied to any string s, returns a copy of s that contains only characters in the set.
   
6. 
   
7. Solution
   
8. The translate method of string objects is fast and handy for all tasks of this ilk. However, to call translate effectively to solve this recipe’s task, we must do some advance preparation. The first argument to translate is a translation table: in this recipe, we do not want to do any translation, so we must prepare a first argument that specifies “no translation”. The second argument to translate specifies which characters we want to delete: since the task here says that we’re given, instead, a set of characters to keep (i.e., to not delete), we must prepare a second argument that gives the set complement—deleting all characters we must not keep. A closure is the best way to do this advance preparation just once, obtaining a fast filtering function tailored to our exact needs:
   
9. 
   
10. import string
    
11. # Make a reusable string of all characters, which does double duty
    
12. # as a translation table specifying "no translation whatsoever"allchars = string.maketrans('', '')
    
13. def makefilter(keep):
    
14.     """ Return a function that takes a string and returns a partial copy
    
15.         of that string consisting of only the characters in 'keep'.
    
16.         Note that `keep' must be a plain string.
    
17.     """
    
18.     # Make a string of all characters that are not in 'keep': the "set
    
19.     # complement" of keep, meaning the string of characters we must delete
    
20.     delchars = allchars.translate(allchars, keep)
    
21.     # Make and return the desired filtering function (as a closure)
    
22.     def thefilter(s):
    
23.         return s.translate(allchars, delchars)
    
24.     return thefilter
    
25. if _ _name_ _ == '_ _main_ _':
    
26.     just_vowels = makefilter('aeiouy')
    
27.     print just_vowels('four score and seven years ago')
    
28. # emits: ouoeaeeyeaao
    
29.     print just_vowels('tiger, tiger burning bright')
    
30. # emits: ieieuii

复制代码

1. Checking Whether a String Is Text or Binary
   
2. Credit: Andrew Dalke
   
3. 
   
4. Problem
   
5. Python can use a plain string to hold either text or arbitrary bytes, and you need to determine (heuristically, of course: there can be no precise algorithm for this) which of the two cases holds for a certain string.
   
6. 
   
7. Solution
   
8. We can use the same heuristic criteria as Perl does, deeming a string binary if it contains any nulls or if more than 30% of its characters have the high bit set (i.e., codes greater than 126) or are strange control codes. We have to code this ourselves, but this also means we easily get to tweak the heuristics for special application needs:
   
9. 
   
10. from _ _future_ _ import division           # ensure / does NOT truncate
    
11. import string
    
12. text_characters = "".join(map(chr, range(32, 127))) + "\n\r\t\b"
    
13. _null_trans = string.maketrans("", "")
    
14. def istext(s, text_characters=text_characters, threshold=0.30):
    
15.     # if s contains any null, it's not text:
    
16.     if "\0" in s:
    
17.         return False
    
18.     # an "empty" string is "text" (arbitrary but reasonable choice):
    
19.     if not s:
    
20.         return True
    
21.     # Get the substring of s made up of non-text characters
    
22.     t = s.translate(_null_trans, text_characters)
    
23.     # s is 'text' if less than 30% of its characters are non-text ones:
    
24.     return len(t)/len(s) <= threshold

复制代码

1. Controlling Case
   
2. Credit: Luther Blissett
   
3. 
   
4. Problem
   
5. You need to convert a string from uppercase to lowercase, or vice versa.
   
6. 
   
7. Solution
   
8. That’s what the upper and lower methods of string objects are for. Each takes no arguments and returns a copy of the string in which each letter has been changed to upper- or lowercase, respectively.
   
9. 
   
10. big = little.upper( )
    
11. little = big.lower( )

复制代码

1. Accessing Substrings
   
2. Credit: Alex Martelli
   
3. 
   
4. Problem
   
5. You want to access portions of a string. For example, you’ve read a fixed-width record and want to extract the record’s fields.
   
6. 
   
7. Solution
   
8. Slicing is great, but it only does one field at a time:
   
9. 
   
10. afield = theline[3:8]
    
11. If you need to think in terms of field lengths, struct.unpack may be appropriate. For example:
    
12. 
    
13. import struct
    
14. # Get a 5-byte string, skip 3, get two 8-byte strings, then all the rest:
    
15. baseformat = "5s 3x 8s 8s"
    
16. # by how many bytes does theline exceed the length implied by this
    
17. # base-format (24 bytes in this case, but struct.calcsize is general)
    
18. numremain = len(theline) - struct.calcsize(baseformat)
    
19. # complete the format with the appropriate 's' field, then unpack
    
20. format = "%s %ds" % (baseformat, numremain)
    
21. l, s1, s2, t = struct.unpack(format, theline)

复制代码

1. Changing the Indentation of a Multiline String
   
2. Credit: Tom Good
   
3. 
   
4. Problem
   
5. You have a string made up of multiple lines, and you need to build another string from it, adding or removing leading spaces on each line so that the indentation of each line is some absolute number of spaces.
   
6. 
   
7. Solution
   
8. The methods of string objects are quite handy, and let us write a simple function to perform this task:
   
9. 
   
10. def reindent(s, numSpaces):
    
11.     leading_space = numSpaces * ' '
    
12.     lines = [ leading_space + line.strip( )
    
13.               for line in s.splitlines( ) ]
    
14.     return '\n'.join(lines)

复制代码

1. Expanding and Compressing Tabs
   
2. Credit: Alex Martelli, David Ascher
   
3. 
   
4. Problem
   
5. You want to convert tabs in a string to the appropriate number of spaces, or vice versa.
   
6. 
   
7. Solution
   
8. Changing tabs to the appropriate number of spaces is a reasonably frequent task, easily accomplished with Python strings’ expandtabs method. Because strings are immutable, the method returns a new string object, a modified copy of the original one. However, it’s easy to rebind a string variable name from the original to the modified-copy value:
   
9. 
   
10. mystring = mystring.expandtabs( )
    
11. This doesn’t change the string object to which mystring originally referred, but it does rebind the name mystring to a newly created string object, a modified copy of mystring in which tabs are expanded into runs of spaces. expandtabs, by default, uses a tab length of 8; you can pass expandtabs an integer argument to use as the tab length.
    
12. 
    
13. Changing spaces into tabs is a rare and peculiar need. Compression, if that’s what you’re after, is far better performed in other ways, so Python doesn’t offer a built-in way to “unexpand” spaces into tabs. We can, of course, write our own function for the purpose. String processing tends to be fastest in a split/process/rejoin approach, rather than with repeated overall string transformations:
    
14. 
    
15. def unexpand(astring, tablen=8):
    
16.     import re
    
17.     # split into alternating space and non-space sequences
    
18.     pieces = re.split(r'( +)', astring.expandtabs(tablen))
    
19.     # keep track of the total length of the string so far
    
20.     lensofar = 0
    
21.     for i, piece in enumerate(pieces):
    
22.         thislen = len(piece)
    
23.         lensofar += thislen
    
24.         if piece.isspace( ):
    
25.             # change each space sequences into tabs+spaces
    
26.             numblanks = lensofar % tablen
    
27.             numtabs = (thislen-numblanks+tablen-1)/tablen
    
28.             pieces[i] = '\t'*numtabs + ' '*numblanks
    
29.     return ''.join(pieces)

复制代码

非常好的书，谢谢分享！